Pages: 1, 2, 3, 4, 5, 6, 7

jeremyebert
QUOTE (Raphie Frank+Nov 8 2011, 12:39 PM)
"Not surprisingly, there is a lot of crossing of lines going on at divisor locations. Perhaps there is more to be learned from studying this overlay."
http://divisorplot.com/3.html

"Upon first discovering these parabolas, one gets the intuitive sense that something interesting might be going at in these locations. It is almost as if an infinite series of fingers along the square root boundary are saying, 'look here, look here!'"
http://divisorplot.com/4.html

Score one for Jeffrey, Art Education Major with a minor in Art History and a double MFA in Computer Graphics and Visible Language.

(As for me, a Psychology and Dramatic Writing Double major.)

- RF

BTW, Jeffrey started a new blog.
http://eyemath.wordpress.com/
jeremyebert
QUOTE (jeremyebert+Nov 20 2011, 04:31 PM)
jeremyebert
Some very trivial but useful information

Extending

n + ((n-1)/2)^2 = ((n+1)/2)^2

to

n + ((n-k^2)/(2k))^2 = ((n+k^2)/(2k))^2

SUM[(n+k^2)/(2k)] where 1<=k<=n = 1/4 (2n H(n)+n^2+n)

1/4 (n+n^2+2 n HarmonicNumber(n))

http://www.wolframalpha.com/input/?i=1%2F4...ashPrefs_*Math-

SUM[(n-k^2)/(2k)] where 1<=k<=n = 1/4 (2n H(n)-n^2-n)

1/4 (-n-n^2+2 n HarmonicNumber(n))

http://www.wolframalpha.com/input/?i=1%2F4...ashPrefs_*Math-

1/4 (2n H(n)+n^2+n) + 1/4 (2n H(n)-n^2-n) = n H(n)

http://www.wolframalpha.com/input/?i=1%2F4...HeavisideTheta-

1/4 (2n H(n)+n^2+n) - 1/4 (2n H(n)-n^2-n) = T(n) Triagular numbers

http://www.wolframalpha.com/input/?i=1%2F4...n%29-n%5E2-n%29
Raphie Frank
QUOTE (jeremyebert+Nov 21 2011, 12:41 PM)
BTW, Jeffrey started a new blog.
http://eyemath.wordpress.com/

Interesting stuff, and I very much agree...

"Because of my terrible grades in math, never getting beyond Algebra II, and then failing the lowest possible math class in college, I concluded that I was a complete math failure.

But I now know that mathematics encompasses a much broader set of activities than what most of us have been lead to believe. The curriculum is what failed. Thanks to my skills in visual thinking, drawing, and spatial logic (talents often associated with dyslexia) I was able to leap over that crufty heap of equations and explore mathematical concepts with my visual brain. And I didn’t even know I was doing math…until I got accepted into MIT, all expenses paid, to earn a master’s degree. MIT doesn’t pay full tuition and stipend to math flunkies. So, how do you explain this?"

- RF
Raphie Frank
QUOTE (jeremyebert+Nov 22 2011, 11:04 PM)
Some very trivial but useful information

It's only "trivial" if you know it already. Just because you could have known it if you had only had the imagination to ask the question in the first place doesn't count.

REFERENCE FOR THE SKEPTICAL: "Metamath" by Gregory Chaitin.

(any number of passages from the gut of the guy who re-solved Godel's Incompleteness Theorem)

And by the way, Jeremy, "metamathematics" is not just for those who happen to know the already institutionally sanctioned "language" for it. If it were, IMHO, the very conception of it never would have arisen. But here's an informative post from RPenner, just to familiarize you with how one might begin to talk about such things with a pro.

A Simple Problem In My Textbook
http://www.physforum.com/index.php?showtopic=29766&hl=

- RF
Raphie Frank
In the spirit of "trivial but useful" information, 2* the Harmonic Average of any two triangular numbers = n^2 -1.

To know what n is, just add together the indices of the two Triangular Numbers, add 1, and divide by two.

e.g. 2*H-bar(T_9, T_10) = 10^2 - 1 = 99

- RF
Raphie Frank
(x + u)/sqrt (x - u)
~
(2*Me*Mg)/(h-bar*Mg)^2

x = 67092478
u = 4, -12, 32, -88

Relative Weighting:
2/9 --> 4
2/9 --> -12
2/9 --> 32
3/9 --> -88
(Why? Currently unexplained. In other words... I don't know)

Me ~ 9.10938291*10^-31 Kilograms (Electron Mass)
Mg ~ 2*10^34 Kilograms (Giant Molecular Cloud Mass)
h-bar ~ 6.62606959*10^-34 Joule-Seconds/(2*pi) (Planck's Constant/(2*pi))

Independently derives all 5 determinations of Planck's Constant within known ranges of uncertainty.

4, 12, 32, 88 --> n-step (self-avoiding) walks on a Kagome Lattice.
4, 12, 32, 88 --> Convergents to 1 + sqrt (3)
67092478 == sqrt ((67092480 - 2)*(67092482 - 4))

for p = "prime number"

67092480 == 1 * 2 * 4 * 8 * 15 * 26 * 42 * 64 (Product (0, 1, 2, 3, 4, 5, 6, 7) Divisions of 3-Space)
2 = -2 ("-2" Divisions of 3-Space)
67092482 == 8191^2 + 8191(mod 2); 8191 == p_(2^(T_4) + 4) == 2^13 - 1
4 == 2^2 - 2(mod 2); 2 = p_(2^T_0 + 0) == 2^0 + 1
u == p_2 + 1, p_6 - 1, p_11 +1, p_24 - 1
delta u == 0^2 + 4, 1^2 + 4, 3^2 + 4

All values derived from an Interpretation of the Solar System based upon the Geometric Division of Space, in tandem with the hypothesized assumption that a fractal architecture underlies both prime number and physical constant distributions.

2010 Co-Data Value for Planck's Constant = 6.62606957(29)*10^-34 Joule-Seconds.
Geometrically derived Value weighted as per above = 6.62606959*10^-34 Joule-Seconds

(One of many) Prediction(s) this model makes:
The Magnetic Resonance Determination of Planck's Constant is a tad low [and may well be just a different way of measuring the same phenomena (or nearly so) as the x-Ray Crystal Density Method].

But, insofar, Jeremy, as I was able to tell you to look at the Dirichlet Divisor Sum in relation to the Visual Prime Pattern you identified, and it turned out to be exactly (at least part of) what you had discovered, the thinking behind this model has already been predictive.

Best,
Raphie

P.S.

Delta (c*s/m, G_(24-6(n(mod 2))*L_(24+6(n(mod 2))/((p_24 - 1)+(p_6 - 1))) = floor [sqrt (x + 4)] = = sqrt (67092478/2 + 67092480)/2) = 8191

c*s/m = 299792458
G_24 = 289154
G_18 = 16114
L_24 = 103682
L_30 = 1860498

G --> Fibonacci (2, 5) aka "The Golden Scale"
L --> Fibonacci (2, 1) aka "Lucas Numbers"

Delta (G_24*L_24, G_18*L_30)/4 = |sigma u| = 64
Raphie Frank
=====================================================
A002605
a(n)=2*(a(n-1)+a(n-2)), a(0)=0, a(1)=1

0, 1, 2, 6, 16, 44, 120, 328, 896, 2448, 6688, 18272, 49920, 136384, 372608, 1017984, 2781184, 7598336, 20759040, 56714752, 154947584, 423324672, 1156544512, 3159738368, 8632565760, 23584608256, 64434348032, 176037912576, 480944521216, 1313964867584

Individually, both this sequence and A028859 are convergents to 1+sqrt(3). Mutually, both sequences are convergents to 2+sqrt(3) and 1+sqrt(3)/2.- Klaus E. Kastberg, Nov 04 2001
http://oeis.org/A002605
=====================================================

-2 + 6 = 4
6 - 16 = -12
-12 + 44 = 32
32 - 120 = -88

Tick-tock, tick tock... just like a pendulum set to (idealized Mersenne...) little g=pi^2 m/s^2. Or (idealized Sophie Germain) Big G set to...

10^-11*(((4*pi^2)/sqrt (35)) + ((4*pi^2/(sqrt (35)10^4) + (4*pi^2/(sqrt (35)*10^5) + (4*pi^2/(sqrt (35)*10^13)))
~ 6.673804*10^11

10^-11*(((4*pi^2)/sqrt (35)) - ((4*pi^2/(sqrt (35)10^4) - (4*pi^2/(sqrt (35)*10^5) - (4*pi^2/(sqrt (35)*10^13)))
~ 6.672470*10^-11

Gravitational Constant
6.67384(80) * 10^-11 m^3*kg^-1*s^-2
http://en.wikipedia.org/wiki/Gravitational_constant

Data re: Variations in Gravitational Constant Measurements over time
http://blazelabs.com/f-u-massvariation.asp
(I don't agree with the researcher's extrapolations, but the data is solid)

delta (2, 6) = 4 = 0^2 + 4
delta (6, 11) = 5 = 1^2 + 4
delta (11, 24) = 13 = 3^2 + 4
(see previous post...)
pi (35) = 11

The full sets of numbers I am currently working with...

q = 2, 2, 6, 11, 24, 128
u = -2, 4, -12, 32, -88, 720

u = (p_q + (-1)^(n-1))(-1)^(n-1)

delta q = 0, 4, 5, 13, 104, which all follow the form n^2 + 4*sgn(n)

- RF
Raphie Frank
QUOTE (Raphie Frank+Nov 24 2011, 09:23 PM)
The full sets of numbers I am currently working with...

q = 2, 2, 6, 11, 24, 128
u = -2, 4, -12, 32, -88, 720

u = (p_q + (-1)^(n-1))(-1)^(n-1)

delta q = 0, 4, 5, 13, 104, which all follow the form n^2 + 4*sgn(n)

q' = floor[log 2(q)] = 1, 1, 2, 3, 4, 7

e.g.
PRODUCT (CUT(n,3)) for range 1 --> 7 = 67092480
PRODUCT (CUT(n,3)) for range 1 --> 1 = 2
Delta (67092480, 2) = x = 67092478

2^1*2^7 = Delta (G_24*L_24, G_18*L_30)

q-range = 2^1 --> 2^7
u-range = 2! --> 6! or, alternatively, K_2/(3*T_1) --> K_24/(3*T_13) = 6/2 --> 196560/273

(2^13 - 1)^2 + (2^13 - 1)mod (2) == 67092482
(2^1 - 0)^2 + (2^1 - 0)mod (2) == 4
Delta (67092482, 4) = x = 67092478
1 == F_1, 13 == F_7

720/2, of course, gives us 360, the first integer with 24 divisors and, in degrees, the fundamental domain addressed by the Crystallographic Restriction Theorem.

More than a little relevant, since the derivation of u corresponds 1:1 with the set of n in N such that 2*cos(2*pi/(n^sgn(n)) is in N, aka the set of integers such that -1 < totient(n) < 3

{0, 1, 2, 3, 4, 6}

(1/2((0 - 2)^2 + (0 - 2)))^2 = 001 + 4 = 05
(1/2((1 - 2)^2 + (1 - 2)))^2 = 000 + 0 = 00
(1/2((2 - 2)^2 + (2 - 2)))^2 = 000 + 4 = 04
(1/2((3 - 2)^2 + (3 - 2)))^2 = 001 + 4 = 05
(1/2((4 - 2)^2 + (4 - 2)))^2 = 009 + 4 = 13
(1/2((6 - 2)^2 + (6 - 2)))^2 = 100 + 4 = 104

SUM (5, 5, 9, 14, 27, 131) - 3
= 2, 2, 6, 11, 24, 128

Give a guy a little c = {0, 1, 2, 3, 4, 6}, Jeremy, and there's no telling, what beautiful things may arise.

In the past 3 posts I just gave you Geometrically based equivalents for:

G, c, Me, Mg and all 5 determinations of Planck's Constant, h-bar that A ) are 100% in-line with uncertainty ranges for these physical constants, and B ) derived in 100% Symmetrical Fashion from the Planetary orbits for the range [Mercury,Pluto]

To place a little different spin on the term "quack," if it looks like a duck, talks like a duck and walks like a duck, then, as Richard Darman was fond of saying, who's to say it's not a duck?

But, for good measure, let's just throw in B and Mp as well.

(totient^11)*SUM (CUT(n, 3))) for range 0 --> (6 + 24) = 364560 = (B*Mp/(meters*kilograms))

for B = The Balmer Constant and Mp ~ Mass of a Primordial Black Hole.

With the nice little bonus that p_11/totient(11)^pi(11)*ceiling[e^11/11] gives a super nice little estimate of the minimum achievable Planetary Positioning ratio at n = totient (11), as does the sum of the main diagonals of the square number spiral, also divided by 10^5 = sqrt (10^10) at n = p_11 = 31

- RF
Raphie Frank
Via Wikipedia

KAGOME LATTICE
"Some minerals, namely jarosites and herbertsmithite, contain layers with kagome lattice arrangement of atoms in their crystal structure. These minerals display novel physical properties connected with geometrically frustrated magnetism."
http://en.wikipedia.org/wiki/Kagome_lattice

GEOMETRICAL FRUSTRATION
"[I]n simple terms, the substance can never be completely frozen, because the structure it forms does not have a single minimal-energy state, so motion on a molecular scale continues even at absolute zero and even without input of energy."

1 Magnetic ordering
2 Mathematical definition
3 Water ice
4 Spin ice
5 Extension of Pauling’s model: General frustration
6 Artificial geometrically frustrated ferromagnets
7 Geometric Frustration without Lattice
7.1 Simple two-dimensional examples
7.2 Dense structures and tetrahedral packings
7.3 Regular packing of tetrahedra: the polytope {3,3,5}
8 Literature
9 References
http://en.wikipedia.org/wiki/Geometrically_frustrated_magnet

Number of n-step walks on Kagome lattice.
1, 4, 12, 32, 88, 240, 652, 1744, 4616, 12208, 32328, 85408, 224608, 588832
http://oeis.org/A001665

Here's a question for you. What ratio does this series converge to?
(4/1) = 4, (88/32) = 2.75, (588832/224608) = 2.62159851830

RELATED:
Self-Avoiding Walk Connective Constant
http://mathworld.wolfram.com/Self-Avoiding...veConstant.html

- RF
Raphie Frank
QUOTE (Raphie Frank+Nov 24 2011, 10:31 PM)
720/2, of course, gives us 360, the first integer with 24 divisors and, in degrees, the fundamental domain addressed by the Crystallographic Restriction Theorem.

More than a little relevant, since the derivation of u corresponds 1:1 with the set of n in N such that 2*cos(2*pi/(n^sgn(n)) is in N, aka the set of integers such that -1 < totient(n) < 3

It just gets weirder and weirder...

-2 = = p_2 - 1 = u_-1
-12 =p_6 - 1 = u_1
32 = p_11 + 1 = u_2
720 = p_128 + 1 = u_4

(p_ceiling[720/|-2|] - p_ceiling [32/|-12|]))/10^3 = 2.423 - 0.005 = 2.418

Or roughly the refraction index of a diamond.

Source:
Bailey, Lowell. Diamonds are Forever. The Georgia Mineral Society, 2003.
via Hypertextbook.com
http://hypertextbook.com/facts/2005/JaminBennett.shtml
"Light travels through space, a vacuum, at 186,282 miles per second, that is the base refractive index of 1.00, once it hits the atmosphere of earth it is slowed to 186,232 miles per second, that gives air, an index of 1.02 and in water it is slowed down to 140,061 miles per second, water has the refractive index of 1.33, if it goes through glass, such as a window pane it is slowed to 122.554 miles per second, glass has a refractive index of 1.52, when that light hits Diamond the speed is cut drastically to 77,056 miles per second, the diamond has a refractive index of 2.4175."

Source:
All About Jewels Glossary R. Enchantedlearning. May 2005.
Diamonds have a hardness of 10, a specific gravity of 3.5 and a refractive index of 2.417-2.419"

3.5 = 35/10, both integers I posted about earlier in relation to Big G

The standardized results vary from 2.417 to 2.419. 2.418 is right there in the middle.

2417 is the 359th prime
2423 is the 360th prime

360 = (19^2 - 1), as you may or may not be aware, is the second simple group. 168 = (13^2 - 1) is the first.

2 and 720 are both Highly Composite numbers (the 1st and 13th).

- RF

Note: And for whatever it's worth...
360 = 19^2 - 1 = (p_ceiling [88/-12])^2 - 1
jeremyebert
QUOTE (Raphie Frank+Nov 24 2011, 06:40 PM)
Via Wikipedia

KAGOME LATTICE
"Some minerals, namely jarosites and herbertsmithite, contain layers with kagome lattice arrangement of atoms in their crystal structure. These minerals display novel physical properties connected with geometrically frustrated magnetism."
http://en.wikipedia.org/wiki/Kagome_lattice

GEOMETRICAL FRUSTRATION
"[I]n simple terms, the substance can never be completely frozen, because the structure it forms does not have a single minimal-energy state, so motion on a molecular scale continues even at absolute zero and even without input of energy."

1 Magnetic ordering
2 Mathematical definition
3 Water ice
4 Spin ice
5 Extension of Pauling’s model: General frustration
6 Artificial geometrically frustrated ferromagnets
7 Geometric Frustration without Lattice
7.1 Simple two-dimensional examples
7.2 Dense structures and tetrahedral packings
7.3 Regular packing of tetrahedra: the polytope {3,3,5}
8 Literature
9 References
http://en.wikipedia.org/wiki/Geometrically_frustrated_magnet

Number of n-step walks on Kagome lattice.
1, 4, 12, 32, 88, 240, 652, 1744, 4616, 12208, 32328, 85408, 224608, 588832
http://oeis.org/A001665

Here's a question for you. What ratio does this series converge to?
(4/1) = 4, (88/32) = 2.75, (588832/224608) = 2.62159851830

RELATED:
Self-Avoiding Walk Connective Constant
http://mathworld.wolfram.com/Self-Avoiding...veConstant.html

- RF

Is gemotrical frustration a source of energy then? Thats just amazing to think about. Very cool stuff Raphie.
Raphie Frank
QUOTE (jeremyebert+Nov 25 2011, 02:36 AM)
Is gemotrical frustration a source of energy then? Thats just amazing to think about. Very cool stuff Raphie.

Personally, I think more in terms of consciousness. I really don't know the relationship to energy, but you can't help but wonder how that works into fantastical notions of perpetual motion machines, which all reasonable evidence has shown to be impossible. Which I don't argue with, by the way, at non-atomic scales.

I was always under the apparently wrong impression that at absolute zero, all movement stops, so it's certainly interesting to me as well
Raphie Frank
Jeremy, I just threw a lot of information your way, so let me help you sort it. Below is the key post. In a way, all the rest is "backup" even though many of those observations preceded this one.

Fundamentally, what the quoted post "predicts" is this: A (geometrically-based) relationship between "quanta of action" at large and small scales alike. And if there is a relationship between the large and small, it logically follows that there is a relationship at all levels in between (which includes you and me).

Heretical thought, "scientifically" speaking I know. Because we all "know" that "God" made the "small things" with one set of rules and the "large things" with another. And "He" especially made humankind by another set of rules all together (Pre-Copernican thought yet still rules...)...

[Counter-point Reference: "The Elementary Forms of Religious Life" by Emile Durkheim (1912)]

Emile Durkheim was one of the three "Founding Fathers" of Modern Day Sociology. Marx and Weber were the other two. It was Durkheim, not Jung, who coined the phrase "Collective Unconscious" But on a "root" level, Modern Day Sociology actually finds it origins in the work and ideas of A ) Condorcet (according to E.O. Wilson, evolutionary biologist at Harvard University) who conceived the idea of there being a "Physics of Society," and B ) Quetelet, who conceived the notion of the "Average Man."

And thus the Bell Curve was born.

As for Condorcet, he was a "Failure" as a scientist, but a brilliant success as a man and it was only in January of 2009 that Physicists and Sociologists finally got together again in "De-Humpty-Dumptification" vein, at the Santa Fe Institute in New Mexico, to ask the (ultra-obvious since already proven by Barabasi and Bianconi in 1999) question "Is there a Physics of Society?"

The (seemingly) utterly absurd conclusion of the professionals? "Yes, no and Maybe."

- RF

QUOTE (Raphie Frank+Nov 24 2011, 08:30 PM)
(x + u)/sqrt (x - u)
~
(2*Me*Mg)/(h-bar*Mg)^2

x = 67092478
u = 4, -12, 32, -88

Relative Weighting:
2/9 --> 4
2/9 --> -12
2/9 --> 32
3/9 --> -88
(Why? Currently unexplained. In other words... I don't know)

Me ~ 9.10938291*10^-31 Kilograms (Electron Mass)
Mg ~ 2*10^34 Kilograms (Giant Molecular Cloud Mass)
h-bar ~ 6.62606959*10^-34 Joule-Seconds/(2*pi) (Planck's Constant/(2*pi))

Independently derives all 5 determinations of Planck's Constant within known ranges of uncertainty.

4, 12, 32, 88 --> n-step (self-avoiding) walks on a Kagome Lattice.
4, 12, 32, 88 --> Convergents to 1 + sqrt (3)
67092478 == sqrt ((67092480 - 2)*(67092482 - 4))

for p = "prime number"

67092480 == 1 * 2 * 4 * 8 * 15 * 26 * 42 * 64 (Product (0, 1, 2, 3, 4, 5, 6, 7) Divisions of 3-Space)
2 = -2 ("-2" Divisions of 3-Space)
67092482 == 8191^2 + 8191(mod 2); 8191 == p_(2^(T_4) + 4) == 2^13 - 1
4 == 2^2 - 2(mod 2); 2 = p_(2^T_0 + 0) == 2^0 + 1
u == p_2 + 1, p_6 - 1, p_11 +1, p_24 - 1
delta u == 0^2 + 4, 1^2 + 4, 3^2 + 4

All values derived from an Interpretation of the Solar System based upon the Geometric Division of Space, in tandem with the hypothesized assumption that a fractal architecture underlies both prime number and physical constant distributions.

2010 Co-Data Value for Planck's Constant = 6.62606957(29)*10^-34 Joule-Seconds.
Geometrically derived Value weighted as per above = 6.62606959*10^-34 Joule-Seconds

(One of many) Prediction(s) this model makes:
The Magnetic Resonance Determination of Planck's Constant is a tad low [and may well be just a different way of measuring the same phenomena (or nearly so) as the x-Ray Crystal Density Method].

But, insofar, Jeremy, as I was able to tell you to look at the Dirichlet Divisor Sum in relation to the Visual Prime Pattern you identified, and it turned out to be exactly (at least part of) what you had discovered, the thinking behind this model has already been predictive.

Best,
Raphie

P.S.

Delta (c*s/m, G_(24-6(n(mod 2))*L_(24+6(n(mod 2))/((p_24 - 1)+(p_6 - 1))) = floor [sqrt (x + 4)] = = sqrt (67092478/2 + 67092480)/2) = 8191

c*s/m = 299792458
G_24 = 289154
G_18 = 16114
L_24 = 103682
L_30 = 1860498

G --> Fibonacci (2, 5) aka "The Golden Scale"
L --> Fibonacci (2, 1) aka "Lucas Numbers"

Delta (G_24*L_24, G_18*L_30)/4 = |sigma u| = 64
jeremyebert
QUOTE (Raphie Frank+Nov 25 2011, 04:44 AM)
Jeremy, I just threw a lot of information your way, so let me help you sort it. Below is the key post. In a way, all the rest is "backup" even though many of those observations preceded this one.

Fundamentally, what the quoted post "predicts" is this: A (geometrically-based) relationship between "quanta of action" at large and small scales alike. And if there is a relationship between the large and small, it logically follows that there is a relationship at all levels in between (which includes you and me).

Heretical thought, "scientifically" speaking I know. Because we all "know" that "God" made the "small things" with one set of rules and the "large things" with another. And "He" especially made humankind by another set of rules all together (Pre-Copernican thought yet still rules...)...

[Counter-point Reference: "The Elementary Forms of Religious Life" by Emile Durkheim (1912)]

Emile Durkheim was one of the three "Founding Fathers" of Modern Day Sociology. Marx and Weber were the other two. It was Durkheim, not Jung, who coined the phrase "Collective Unconscious" But on a "root" level, Modern Day Sociology actually finds it origins in the work and ideas of A ) Condorcet (according to E.O. Wilson, evolutionary biologist at Harvard University) who conceived the idea of there being a "Physics of Society," and B ) Quetelet, who conceived the notion of the "Average Man."

And thus the Bell Curve was born.

As for Condorcet, he was a "Failure" as a scientist, but a brilliant success as a man and it was only in January of 2009 that Physicists and Sociologists finally got together again in "De-Humpty-Dumptification" vein, at the Santa Fe Institute in New Mexico, to ask the (ultra-obvious since already proven by Barabasi and Bianconi in 1999) question "Is there a Physics of Society?"

The (seemingly) utterly absurd conclusion of the professionals? "Yes, no and Maybe."

- RF

Raphie all of this is extremely thought provoking. I cant remember how I came across this article but I just re-read it today. I think its applicable to our current topic. It presents an interesting view of the Riemann Hypothesis.

http://www.integralworld.net/collins18.html

One thing that I didn't catch before that I thought interesting:

"Gauss showed the chance that a number t is prime can be given by 1/log t. A simple formula exists for counting the frequency of non-trivial zeros along the critical line in the vicinity of the value t, which is 2pi/log(t/2pi) This would suggest for example an average gap of 2.27 (approx) between zeros in the region of 100! Now 2pi represents the circumference of the circle of unit radius. So if we replace the circle (i.e. the circumference 2pi) with its linear radius (1) in the formula we get 1/log t, i.e. the same as the chance that the number t is prime. So seen in this way there are close connections between both results (that are linear and circular with respect to each other). "
Raphie Frank
The Rydberg Constant
R = 1.0973731568539(55)*10^7 m^-1
R = (Me*c*alpha^2)/(2*h)

SUM 3(c + 1)/10^(|8-n| - n) for range n = 4 --> -1
c = 0, 1, 2, 3, 4, 6, indexed from -1

= 21.1512090603

Let U_z = (x - 21.1512090603)/(x + 21.1512090603)
x = 67092478
Let U_z' = (2*Me*Mg)/(h-bar*Mg)^2

----------------------------------
Me' ~ 9.10938291*10^-31 kg
Mg' ~ 2*10^34 kg
h-bar' ~ 6.6260957698*10^-34 J-s/(2*pi)
compared to Co-Data Value of...
h-bar ~ 6.6260957(29)*10^-34 J-s
----------------------------------

U_z = 8190.9959435346
U_z' = 8190.9959435343

R = (Me*c*alpha^2)/(2*h)

If one inserts Co-Data values into that formula for the Rydberg Constant (e.g. alpha^-1 = 137.035999074), one doesn't actually get the Rydberg Constant. Actually, what you get is 1.0973731576765 * 10^7 m^-1, which is close, but no cigar since the Rydberg Constant has been measured with extreme accuracy.

If, on the other hand, you extract h-bar from U_z' and use our exploratory value for alpha^-1 = 137.035999053, then, leaving units out for the sake of simplicity...

R ~ (9.10938291*10^-31*299792458*137.035999053^(-2))/(2*6.62606957698*10^-34) m^-1
R ~ 1.0973731568569 * 10^7 * m^-1
versus...
R = 1.0973731568539(55)*10^7 m^-1 (2010 Co-Data)

Accurate to all significant digits...

RYDBERG CONSTANT
"The Rydberg constant, symbol R∞, named after the Swedish physicist Johannes Rydberg, is a physical constant relating to atomic spectra in the science of spectroscopy. Rydberg initially determined its value empirically from spectroscopy, but Niels Bohr later showed that its value could be calculated from more fundamental constants by using quantum mechanics. As of 2010, it is the most accurately measured fundamental physical constant."

"The Rydberg constant represents the limiting value of the highest wavenumber (the inverse wavelength) of any photon that can be emitted from the hydrogen atom, or, alternatively, the wavenumber of the lowest-energy photon capable of ionizing the hydrogen atom from its ground state. The spectrum of hydrogen can be expressed simply in terms of the Rydberg constant, using the Rydberg formula"
http://en.wikipedia.org/wiki/Rydberg_constant

Expansion of chi(q)^3 / chi(q^3) in powers of q where chi() is a Ramanujan theta function
1, 3, 3, 3, 6, 9, 12, 15, 21, 30, 36, 45, 60, 78, 96, 117, 150, 189, 228, 276, 342, 420, 504
https://oeis.org/A132972

Ramanujan Theta Functions
"(Sloane's A010815; Berndt 1985, pp. 36-37; Berndt et al. 2000), where is a q-Pochhammer symbol. The identities above are equivalent to the pentagonal number theorem."
http://mathworld.wolfram.com/RamanujanThetaFunctions.html

- RF

P.S. The Rydberg Constant is roughly 4/B, where B is the Balmer Constant. But that's not exact, actually. The Balmer Series is just one of six named series pertaining to the hydrogen emission spectrum. If there's anything to the above, then there's a good chance that there would be a 1:1 correspodence between the elements of c = 0, 1, 2, 3, 4, 6 and those series...

But, of course, that would be rather "out there", don't you think, to assert that hydrogen might have anything to do with crystals?

"As shown by X-ray crystallography, the hexagonal symmetry of snowflakes results from the tetrahedral arrangement of hydrogen bonds about each water molecule. The water molecules are arranged similarly to the silicon atoms in the tridymite polymorph of SiO2. The resulting crystal structure has hexagonal symmetry when viewed along a principal axis."
http://en.wikipedia.org/wiki/X-ray_crystallography
See" Snow flakes by Wilson Bentley
http://en.wikipedia.org/wiki/File:Snowflake8.png
jeremyebert
QUOTE (Raphie Frank+Nov 25 2011, 03:15 PM)

Expansion of chi(q)^3 / chi(q^3) in powers of q where chi() is a Ramanujan theta function
1, 3, 3, 3, 6, 9, 12, 15, 21, 30, 36, 45, 60, 78, 96, 117, 150, 189, 228, 276, 342, 420, 504
https://oeis.org/A132972

Ramanujan Theta Functions
"(Sloane's A010815; Berndt 1985, pp. 36-37; Berndt et al. 2000), where  is a q-Pochhammer symbol. The identities above are equivalent to the pentagonal number theorem."
http://mathworld.wolfram.com/RamanujanThetaFunctions.html

I wonder if we can draw a relation to from the Ramanujan Theta function

f( a , b ) = sum( a^(n(n+1)/2) * b^(n(n-1)/2) )

extending

n + ((n-1)/2)^2 = ((n+1)/2)^2

to

n + ((n-k^2)/(2k))^2 = ((n+k^2)/(2k))^2

also a high resolution recursion scatter plot for

(n+1)/t * e^(i cos^(-1)(1-k(2/(n+1))))

1<=k<=n
1<=n<=t
1<=t<=100

http://dl.dropbox.com/u/13155084/Recusion100.png
jeremyebert
QUOTE (jeremyebert+Nov 25 2011, 10:40 PM)
I wonder if we can draw a relation to from the Ramanujan Theta function

f( a , b ) = sum( a^(n(n+1)/2) * b^(n(n-1)/2) )

extending

n + ((n-1)/2)^2 = ((n+1)/2)^2

to

n + ((n-k^2)/(2k))^2 = ((n+k^2)/(2k))^2

also a high resolution recursion scatter plot for

(n+1)/t * e^(i cos^(-1)(1-k(2/(n+1))))

1<=k<=n
1<=n<=t
1<=t<=100

http://dl.dropbox.com/u/13155084/Recusion100.png

If you have the free CDF player.
http://dl.dropbox.com/u/13155084/Recusion.cdf
Raphie Frank
QUOTE (jeremyebert+Nov 26 2011, 03:40 AM)
I wonder if we can draw a relation to from the Ramanujan Theta function

f( a , b ) = sum(  a^(n(n+1)/2) * b^(n(n-1)/2) )

extending

n + ((n-1)/2)^2 = ((n+1)/2)^2

to

n + ((n-k^2)/(2k))^2 = ((n+k^2)/(2k))^2

also a high resolution recursion scatter plot for

(n+1)/t * e^(i cos^(-1)(1-k(2/(n+1))))

1<=k<=n
1<=n<=t
1<=t<=100

http://dl.dropbox.com/u/13155084/Recusion100.png

I don't know if "we" can draw that relationship, Jeremy; I posted it in the hopes that maybe you could.

I have yet to come across you making a (self-) educated guess that further research did not bear out.

- RF

P.S. Give me 3 pages, now, and, with white space to spare, I can give you a complete description of some of the most important fundamental constants of nature. Not saying it's "right", per se, only that it is 100% accurate within known ranges of uncertainty. Woe is unto me, that those same 3 pages also describe, with a correlation coefficient of .99999, the planetary orbits from Mercury through Pluto. And I use the term "woe is unto me," because the relationships I have been posting are, contradictorily, both too simple to be "true" and too simple not to be.
Raphie Frank
QUOTE (Raphie Frank+Nov 25 2011, 08:15 PM)
R ~ (9.10938291*10^-31*299792458*137.035999053^(-2))/(2*6.62606957698*10^-34) m^-1
R ~ 1.0973731568569 * 10^7 * m^-1
versus...
R = 1.0973731568539(55)*10^7 m^-1 (2010 Co-Data)

Accurate to all significant digits...

- RF

What this amounts to, essentially, is a 4-body problem, with one of those bodies defined precisely, namely the speed of light. From here it's a simple matter to program in to a spreadsheet or whatever how changing two of the three undefined variables would affect the third with the constraint that if you "miss" on the Rydberg Constant it's more or less as "worthless," except as a starting point, as having a formula for Planetary Positioning that flat out misses on Neptune.

And FWIW, the arithmetic average of the last two measurements of Inverse Alpha are 137.035999055. The funny part being that I actually far "prefer" the following value for inverse alpha 137.0359990065 which is based on pure mathematics:

((e^(i*(3010349/1860498))^2)/10^1 + (1 + sqrt 3)/10^6 + (7.0359990065)/10^9)^-1
= 137.0359990065...

3010349/1860498 ~ The Golden Ratio = L_31/L_30 where 31 is the maximal number of areas of a fully connected hexagon with Rotational Symmetry not preserved, and 30 equals same, but with Rotational Symmetry preserved.

((e^(i*(Golden Ratio))^2)/10^1 + (1 + sqrt 3)/10^6 + (7.0359990068)/10^9)^-1
= 137.0359990068...

But neither version quite works with empirical reality.

Close, but no cigar.

Still, 1 + sqrt 3 seems to be turning up in some interesting places...

- RF

=======================
Example using 137.035999055 (arithmetic average of last two determinations by physicists) as our value for Alpha^-1

(9.10938291*10^-31*299792458*137.035999055^(-2))/(2*6.62606957681*10^-34) = 1.0973731568529 * 10^7

Then Planck's Constant is roughly... 6.62606957681*10^-34 Joule-Seconds as opposed to 6.62606957698*10^-34, the value given above.

Since this is not me making geometric guesses, but scientists making guesses based on pretty sophisticated techniques, it becomes a fair statement to suggest that 6.626069577*10^-34 Joule-Seconds, is a better approximation of Planck's Constant than the current approximation which is 6.62606957*10^-34 Joule Seconds.

A value which just happens to agree with (Planetary/Lattice) Geometry as well...
(9.10938291*10^-31*299792458*137.035999053^(-2))/(2*6.626069577*10^-34)

~ 1.0973731568535 * 10^7
vs.1.0973731568539(55)*10^7 m^-1 (2010 Co-Data)
=======================
jeremyebert
QUOTE (Raphie Frank+Nov 26 2011, 12:16 PM)
What this amounts to, essentially, is a 4-body problem, with one of those bodies defined precisely, namely the speed of light. From here it's a simple matter to program in to a spreadsheet or whatever how changing two of the three undefined variables would affect the third with the constraint that if you "miss" on the Rydberg Constant it's more or less as "worthless," except as a starting point, as having a formula for Planetary Positioning that flat out misses on Neptune.

And FWIW, the arithmetic average of the last two measurements of Inverse Alpha are 137.035999055. The funny part being that I actually far "prefer" the following value for inverse alpha 137.0359990065 which is based on pure mathematics:

((e^(i*(3010349/1860498))^2)/10^1 + (1 + sqrt 3)/10^6 + (7.0359990065)/10^9)^-1
= 137.0359990065...

3010349/1860498 ~ The Golden Ratio = L_31/L_30 where 31 is the maximal number of areas of a fully connected hexagon with Rotational Symmetry not preserved, and 30 equals same, but with Rotational Symmetry preserved.

((e^(i*(Golden Ratio))^2)/10^1 + (1 + sqrt 3)/10^6 + (7.0359990068)/10^9)^-1
= 137.0359990068...

But neither version quite works with empirical reality.

Close, but no cigar.

Still, 1 + sqrt 3 seems to be turning up in some interesting places...

- RF

=======================
Example using 137.035999055 (arithmetic average of last two determinations by physicists) as our value for Alpha^-1

(9.10938291*10^-31*299792458*137.035999055^(-2))/(2*6.62606957681*10^-34) = 1.0973731568529 * 10^7

Then Planck's Constant is roughly... 6.62606957681*10^-34 Joule-Seconds as opposed to 6.62606957698*10^-34, the value given above.

Since this is not me making geometric guesses, but scientists making guesses based on pretty sophisticated techniques, it becomes a fair statement to suggest that 6.626069577*10^-34 Joule-Seconds, is a better approximation of Planck's Constant than the current approximation which is 6.62606957*10^-34 Joule Seconds.

A value which just happens to agree with (Planetary/Lattice) Geometry as well...
(9.10938291*10^-31*299792458*137.035999053^(-2))/(2*6.626069577*10^-34)

~ 1.0973731568535 * 10^7
vs.1.0973731568539(55)*10^7 m^-1 (2010 Co-Data)
=======================

nice link to the hexagon kagome lattice Raphie
Raphie Frank
Jeremy, there are a zillion ways to interpret both the Fine Structure Constant and the Rydberg Constant, so I'm just going to give you my take on where I see both Constants being most relevant to your (and my) explorations...

RYDBERG CONSTANT

The Coefficient for the energy of the atomic orbitals of a hydrogen atom is...

E_n = - hcR*1/n^2
http://en.wikipedia.org/wiki/Rydberg_constant

Therefore, it follows that:
A ) 1/n^2 = (i^2*E_n)/(hcR)
B ) R = ((ni)^2* E_n)/(hc)

Change in Energy = R*(1/n^2(initial) - 1/n^2(final))

The Balmer Series = R*(1/2^2(initial) - 1/n^2(final)) for n > 2, meaning it's a special case of the Rydberg Formula.

FINE STRUCTURE CONSTANT

The ratio between the velocity of an electron in the Bohr Model of the atom and the speed of light.
http://en.wikipedia.org/wiki/Fine-structure_constant

Both are intimately associated with the hydrogen emission spectrum. which of course, is related to waves, perhaps of prime-influenced nature, perhaps not.

- RF
Raphie Frank
Related to the above

Hartree Energy
http://en.wikipedia.org/wiki/Hartree

The value given on the page does not accord with current Co-Data values:

= 4.35974394(22)* 10-18 J

Co-Data (2010)
(299792458)^2*(9.10938291*10^-31)*(137.035999074)^-2
= 4.35974434 * 10^-18 J

Most Recent Determinations of Fine Structure Constant (arithmetic average)
(299792458)^2*(9.10938291*10^-31)*(137.035999055)^-2
= 4.35974435 * 10^-18 J

- RF
Raphie Frank
Oh, and Jeremy, remember I showed you that relationship regarding Fibonacci Primes and Mersenne primes, part of which is:

p_17711 - p_31 = 196560 = K_24
p_00003 - p_03 = 000000 = K_00
p_00000 - p_00 = 000000 = K_00

17711 = F_(22)
00003 = F_(04)
00000 = F_(00)
31 = p_11
03 = p_02
00 = p'_-1

?

Here was the original source of that observation, meaning the observation that led to it, since what I am showing you below only makes sense to anyone but a mystic if it can somehow be related to the lattices (and/or prime architecture) upon which theoretical physics is based...

529-00011-1
000-17711-0

PANCAKE NUMBERS
00529 --> 32 cuts of a 2D space, 32 = -1^0 + p_(22/02) = 31 + 1 == 4*2^3
00011 --> 04 cuts of a 2D space, 04 = -1^0 + p_(04/02) = 03 + 1 == 4*1^3
00001 --> 00 cuts of a 2D space, 00 = -1^1 + p'_(00/02) = -1 + 1== 4*0^3

00000 --> 00th Fibonacci Number --> (-1 + sum of diagonals of Division of Space Square at row 0)
17711 --> 22nd Fibonacci Number --> (1 + sum of diagonals of Division of Space Square at row 19)
00000 --> 00th Fibonacci Number --> (-1 + sum of diagonals of Division of Space Square at row 0)

(F_00 + (T_32+1) * 10^5))/10^(0+19+0)
+
(F_22 + (T_04+1) * 10^1)//10^(0+19+0)
+
(F_00 + (T_00+0) * 10^0)/10^(0+19+0)

=
529000111/10^19
+
000177110/10^19
=
5.29177211*10^-11

Now compare...

((6.626069577*10^-34)/(2*pi))/(9.10938291*10^-31*299792458*137.035999055^(-1))
= 5.29177211011 * 10^-11

((6.62606957698*10^-34)/(2*pi))/(9.10938291*10^-31*299792458*137.035999053^(-1))
= 5.291772110022 * 10^-11

vs. 5.2917721092(17)×10^−11
2010 Co-Data Value for the Bohr Radius.

Seems nutty, I know, but the numbers keep adding up in very strange ways...

- RF

P.S. In point of fact, the Fibonacci/Mersenne Prime relation works for the first 6 iterated p'rimes = 0, 1, 2, 3, 5, 11 * 2 --> 0, 2, 4, 6, 10, 22

0, 1, 2, 3, 5, 11 + 1 --> 1, 2, 3, 4, 6, 12, the divisors of 12.

Now you know where that observation came from. Also, as you know...

pi_0(12) -1 = 12 - 1 = 11; 2^(11-1) + 4 = 1028; p_1028 = 8191; 24*(8191 - 1) = 196560 = K_24
pi_1(12) -1 = 05 - 1 = 04; 2^(04 - 1) + 3 = 0011; p_0011 = 0031; 08*(0031 - 1) = 240 = K_8
pi_2(12) -1 = 03 - 1 = 02; 2^(02 - 1) + 2 = 0004; p_0004 = 0007; 04*(0007 - 1) = 24 = K_4
pi_3(12) -1 = 02 - 1 = 01; 2^(01 - 1) + 1 = 0002; p_0002 = 0003; 01*(0003 - 1) = 2 = K_1
pi_4(12) -1 = 01 - 1 = 00; 2^(00 - 0) + 0 = 0001; p_0001 = 0002; 00*(0002 - 1) = 0 = K_0
pi_5(12) -1 = 00 - 1 = -1; 2^(-1 + 1) - 1 = 0000; p'_0000 = 0001; 00*(0001 - 1) = 0 = K_0

-1, 0, 1, 2, 4, 11 * 2 = -2, 0, 2, 4, 8, 22

That also followed from the above, which led to this observation...

(1^1 * 2^2 * 3^1 * 7^2 * 31^1) = B/2 *m^-1 = 36456/2*m^-1

To make it dimensionless, just multiply B/2 * R, the Rydberg Constant.
Raphie Frank
Jeremy, there's a mistake of one digit in the above regarding the Bohr Radius, which is irrelevant more or less. As usual I'm typing from memory and there are about as many ways you can combine triangles and Fibonacci Numbers to get the right results as there are ways to express the fine structure constant. What's far more important is what that observation about a year ago led to. It really got me thinking about primes and lattices.

And then in January Ken Ono came out with his work regarding the partition numbers, super interesting since the number of equivalence classes of Symmetric Group S_n is equal to a partition number.

- RF

P.S. Remember that the main diagonals of the CUT(n,k) Triangle sum to a Fibonacci number - 1, which is where the mistake was made.

e.g. To correct the mistake, but not read too much into it...

529-00011-1 = Concatenate T_32+1, T_04+1, T_00+1
000-17710-0 = Concatenate F_02 - 1, F_22 - 1, F_02 - 1

Combine/Overlay
529177211 --> first 9 digits of the Bohr Radius

Same principle at work here...

30.10349000 = 10^3*L_(31 + 0)/10^8
00.00017710 = 10^0*F_(31 - 9)/10^8
30.10366710 Lucas/Fibonacci Overlay
30.10366151 --> Neptune @ Semimajor

0.7000 = 10^-0*L_(04 + 0)/10^1
0.0233 = 10^-3*F_(04 + 9)/10^1
0.7233 Lucas/Fibonacci Overlay
0.7233 --> Venus@Semimajor Axis

Hopefully, it's not lost on you that x-bar (CUT(31,4) + CUT(-31, 4)) * 10^-11 * m = 3.6456 * 10^-7 m

...Or that 31 + 4, the "Sophie Germain Constant" just happens to equal 35 and 4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2 = 6.6730705216 * 10^-11 m^3*kg^-1*s^-2 just happens to give a very nice little first order approximation of Big G that's NOT correct, but is about as correct (a bit more so actually) as what the Google search engine gives you when you do a search for "Gravitational Constant."
Raphie Frank
QUOTE (Raphie Frank+Nov 27 2011, 02:03 AM)
...Or that 31 + 4, the "Sophie Germain Constant" just happens to equal 35 and 4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2 = 6.6730705216 * 10^-11 m^3*kg^-1*s^-2 just happens to give a very nice little first order approximation of Big G that's NOT correct, but is about as correct (a bit more so actually) as what the Google search engine gives you when you do a search for "Gravitational Constant."

And good enough of an approximation (NOT) to fool the Wolfram Alpha Computational Engine, anyway...

4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2

Wolfram Interpretation:
Newtonian gravitational coupling

But fooling Wolfram isn't hard to do... (6.673 + 100) *10^-11 m^3*kg^-1*s^-2 gives the same result as does 0* m^3*kg^-1*s^-2. All Wolfram is looking at is the units.

Conversely, plug in 6.67384 * 10^-11, which is the 2010 Co-Data Value for the Gravitational Constant without units, and Wolfram will basically give you a blank stare...

It does a little better with the following...

(4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2)*(m^-2/s^-2)*(kg)*(m), which is meters^2

and the following...

(4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2)*(m^-2/s^-2)*(kg), which is meters.

NOTE THE FOLLOWING:
If you plug those values into Wolfram, do you see how vastly different the results are just by multiplying, for instance, meters by meters?

Example:

Plug in the following...
(4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2)*(((103682*289154)/100 + (16114*1860498)/100 - 2*8191)*m^-2/s^-2)*(kg)

And then plug in the following...
(4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2)*(((103682*289154)/100 + (16114*1860498)/100 - 2*8191)*m^-2/s^-2)*(kg)/(m)

- RF
cbennett
Raphie Frank
QUOTE (cbennett+Nov 27 2011, 04:47 PM)

About 9/10 of which are me and Jeremy in all likelihood...
jeremyebert
QUOTE (jeremyebert+Nov 22 2011, 06:04 PM)
Some very trivial but useful information

Extending

n + ((n-1)/2)^2 = ((n+1)/2)^2

to

n + ((n-k^2)/(2k))^2 = ((n+k^2)/(2k))^2

SUM[(n+k^2)/(2k)] where 1<=k<=n = 1/4 (2n H(n)+n^2+n)

1/4 (n+n^2+2 n HarmonicNumber(n))

http://www.wolframalpha.com/input/?i=1%2F4...ashPrefs_*Math-

SUM[(n-k^2)/(2k)] where 1<=k<=n = 1/4 (2n H(n)-n^2-n)

1/4 (-n-n^2+2 n HarmonicNumber(n))

http://www.wolframalpha.com/input/?i=1%2F4...ashPrefs_*Math-

1/4 (2n H(n)+n^2+n) + 1/4 (2n H(n)-n^2-n) = n H(n)

http://www.wolframalpha.com/input/?i=1%2F4...HeavisideTheta-

1/4 (2n H(n)+n^2+n) - 1/4 (2n H(n)-n^2-n) = T(n) Triagular numbers

http://www.wolframalpha.com/input/?i=1%2F4...n%29-n%5E2-n%29

interesting relation to my model and sqrt(3).

sqrt(n) = half the radius of the circle

((n+1)/2)/2 - sqrt(n) = 0

n = 7-4 sqrt(3)
n = 7+4 sqrt(3)
jeremyebert
QUOTE (Raphie Frank+Nov 27 2011, 12:09 PM)
About 9/10 of which are me and Jeremy in all likelihood...

ha!
jeremyebert
QUOTE (jeremyebert+Nov 27 2011, 02:20 PM)
interesting relation to my model and sqrt(3).

sqrt(n) = half the radius of the circle

((n+1)/2)/2 - sqrt(n) = 0

n = 7-4 sqrt(3)
n = 7+4 sqrt(3)

sqrt(n) = half the radius of the cirlce visual:

http://dl.dropbox.com/u/13155084/12_recusion.png

((n+1)/2)/2 - sqrt(n) = 0

n = 7-4 sqrt(3)= 0.0717968 = 1/(7+4 sqrt(3))
n = 7+4 sqrt(3)= 13.9282 = 1/(7-4 sqrt(3))

extending

1/(y-x sqrt(3)) = y+x sqrt(3)

-sqrt(1+3 x^2)
+sqrt(1+3 x^2)

roots = x = (+-)i/sqrt(3)

http://www.wolframalpha.com/input/?i=-sqrt...ashPrefs_*Math-
jeremyebert
QUOTE (jeremyebert+Nov 27 2011, 06:36 PM)
sqrt(n) = half the radius of the cirlce visual:

http://dl.dropbox.com/u/13155084/12_recusion.png

((n+1)/2)/2 - sqrt(n)  = 0

n = 7-4 sqrt(3)= 0.0717968  = 1/(7+4 sqrt(3))
n = 7+4 sqrt(3)= 13.9282 = 1/(7-4 sqrt(3))

extending

1/(y-x sqrt(3)) = y+x sqrt(3)

-sqrt(1+3 x^2)
+sqrt(1+3 x^2)

roots = x = (+-)i/sqrt(3)

http://www.wolframalpha.com/input/?i=-sqrt...ashPrefs_*Math-

Interesting

1+3 x^2 =

http://www.wolframalpha.com/input/?i=table...%2C0%2C25%7D%5D

Third spoke of a hexagonal spiral.

http://oeis.org/A056107

[1-3 x^2] =

http://www.wolframalpha.com/input/?i=table...%2C0%2C25%7D%5D

http://oeis.org/A080663

"This sequence equals for n=>2 the third right hand column of triangle A165674. Its recurrence relation leads to Pascal's triangle A007318. Crowley's formula for A080663[n-1] leads to Wiggen's triangle A028421 and the o.g.f. of this sequence, without the first term, leads to Wood's polynomials A126671. See also A165676, A165677, A165678 and A165679."
Raphie Frank
QUOTE (jeremyebert+Nov 28 2011, 02:06 AM)
Interesting

Also interesting is that in Rydberg Atomic Units...

e = sqrt 2 --> Elementary Charge
K_J = (sqrt 2/pi) --> The Josephson Constant

Since h-bar = 1, then h-bar/e --> 1/sqrt 2

And since Me = 1/2 in Rydberg Atomic Units, then (h-bar/e)^2 = Me = 1/2

h-bar --> Planck's Reduced Constant
Me --> Electron Mass

Basic Geometry. And Basic Geometry that just happens to agree with that special place on the number line where one might expect to find the critical zeroes of Riemann Hypothesis fame: 1/2

NATURAL UNITS
http://en.wikipedia.org/wiki/Natural_units

lim n --> infinity (PELL_n + PELL_n+1)/(PELL_n) = sqrt 2

Recursion Rule 1A + 2B = C

0, 1, 2, 5, 12, 29, etc... (Pell Numbers)
1, 1, 3, 7, 17, 41, etc...

- RF
Raphie Frank
A Great example of how mathematicians make really simple things super confusing...

For n (even), a(n) = [ [(3 + Sqrt(8))^((n/2)+1) - (3 - Sqrt(8))^((n/2)+1)] - 2*[(3 + Sqrt(8))^((n/2)-1) - (3 - Sqrt(8))^((n/2)-1)] ] / (6*Sqrt(8)) For n (odd), a(n) = [ [(3 + Sqrt(8))^((n+1)/2) - (3 - Sqrt(8))^((n+1)/2)] - 2*[(3 + Sqrt(8))^((n-1)/2) - (3 - Sqrt(8))^((n-1)/2)] ] / (2*Sqrt(8))

1, 1, 2, 4, 11, 23, 64, 134
--> n such that n^2-1 is a triangular number - Benoit Cloitre
http://oeis.org/A006452

Why not, say, put that in terms a human being can understand:

(PELL(n+1) + PELL_(n+0) - (-1)^n*PELL_(n-1) - (-1)^n*PELL-(n-2) - 2)/4
1, 1, 2, 4, 11, 23, 64, 134...

as opposed to...

(PELL(n+1) + PELL_(n+0) - (-1)^n*PELL_(n-1) + (-1)^n*PELL-(n-2) - 0)/4
-1, 0, 1, 4, 9, 26, 55, 154...

... which gives the indices of the Sophie Germain Triangular Numbers.

--> n such that 2*T_n +1 is a triangular number

I'd provide a link, but that sequence isn't on OEIS. But you can derive it quite simply in a number of ways, since...

A ) Add the two series together and you get the summation of the Pell Numbers, (the ratio between the successive first differences, of course, when properly combined, approaches sqrt 2). So, it's easily constructible as a first difference between two well-known number progressions. (1, 1, 2, 4, 11, 23, 64, 134... is associated with how 4 dogs meet in a field)

B ) Just apply the formula (sqrt (8*Sophie Germain Triangle + 1) - 1)/2

Sophie Germain Triangles are constructible by the recursion rule:

35*a(n-2) - 1*a(n-4) + 1*a(n-6) = a(n)

For a(n) equals a Sophie Germain Triangular Number

The coefficients sum to 35, which is why I term 35 the "Sophie Germain Constant".

35 = (a(n) + 1*a(n-4) - a(n-6))/a(n-2)

- RF

=====================================
Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number.
0, 1, 10, 45, 351, 1540, 11935, 52326, 405450, 1777555, 13773376, 60384555, 467889345, 2051297326, 15894464365, 69683724540, 539943899076, 2367195337045, 18342198104230, 80414957735001, 623094791644755
https://oeis.org/A124174

1, 1, 4, 16, 121, 529, 4096, 17956... etc. is also not on OEIS. But this sequence is, which is essentially the same...

Solution to a Diophantine equation: each term is a triangular number and each term + 1 is a square.
0, 3, 15, 120, 528, 4095, 17955, 139128, 609960, 4726275, 20720703, 160554240, 703893960, 5454117903, 23911673955, 185279454480, 812293020528, 6294047334435, 27594051024015, 213812329916328, 937385441796000

a(n) = A006451(n)*(A006451(n)+1)/2
http://oeis.org/A006454

Numbers n such that n*(n+1)/2+1 is a square
0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074...
a(n)=6a(n-2)-a(n-4)+2 with a(0)=0, a(1)=2, a(2)=5, a(3)=15
http://oeis.org/A006451
=====================================
Raphie Frank
In other words Jeremy, the integer 35 is intimately related to sqrt 2 and sqrt 2 is intimately related to Rydberg Atomic units (and is also the covering radius of the Leech Lattice).

(4 * (pi^2)) / sqrt(35) = 6.67307052

- RF

QUOTE (Raphie Frank+Nov 28 2011, 06:39 PM)
Also interesting is that in Rydberg Atomic Units...

e = sqrt 2 --> Elementary Charge
K_J = (sqrt 2/pi)  --> The Josephson Constant

Since h-bar = 1, then h-bar/e --> 1/sqrt 2

And since Me = 1/2 in Rydberg Atomic Units, then (h-bar/e)^2 = Me = 1/2

h-bar --> Planck's Reduced Constant
Me --> Electron Mass
- RF
Raphie Frank
QUOTE (Raphie Frank+Nov 28 2011, 07:08 PM)
(PELL(n+1) + PELL_(n+0) - (-1)^n*PELL_(n-1) - (-1)^n*PELL-(n-2) - 2)/4
1, 1, 2, 4, 11, 23, 64, 134...

as opposed to...

(PELL(n+1) + PELL_(n+0) - (-1)^n*PELL_(n-1) + (-1)^n*PELL-(n-2) - 0)/4
-1, 0, 1, 4, 9, 26, 55, 154...

-1, 0, 1, 4, 9, 26, 55, 154 = R_0:n (Indices of "Sophie Germain Figurate Numbers of the 1st Kind")
[no sequence on OEIS]

1, 1, 2, 4, 11, 23, 64, 134 = R_1:n (Indices of "Sophie Germain Figurate Numbers of the 2nd Kind")
http://oeis.org/A006452

The relation between the two... (3*R_0:n - R_0:n-1)/2 - R_0:n+1

(3*R_0:n - R_0:n-1)/2 by itself gives this sequence:

Nonnegative integers n such that 2n^2+2n-3 is square
1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422, 15541, 43261, 90582, 252146, 527953, 1469617, 3077138, 8565558, 17934877, 49923733, 104532126, 290976842, 609257881, 1695937321
http://oeis.org/A124124

(R_0:n - R_0:n+2)/2 gives this one:

Numbers n such that n*(n+1)/2+1 is a square.
0, 2, 5, 15, 32, 90, 189, 527, 1104, 3074, 6437, 17919, 37520, 104442, 218685, 608735, 1274592, 3547970, 7428869, 20679087, 43298624, 120526554, 252362877, 702480239, 1470878640, 4094354882, 8572908965, 23863649055, 49966575152

Take the first difference between the two as an absolute value = 1, 0, 1, 2, 5, 12, 29, 70.... gives this well known sequence:

A000129
Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2)
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149
http://oeis.org/A000129

|PELL(n-1) = (3*R_0:n - R_0:n-1)/2 - (R_0:n - R_0:n+2)/2|

And since (PELL(n-1)+PELL(n))/PELL(n-1) is convergent to sqrt 2, then we've just shown how to construct the hypotenuse of a unit triangle entirely from Sophie Germain Triangles.

But, of course, it's quite plain to see you can get there from many, many more progressions than just this one.

- RF

QUOTE (Raphie Frank+Nov 28 2011, 06:39 PM)
Also interesting is that in Rydberg Atomic Units...

e = sqrt 2 --> Elementary Charge
K_J = (sqrt 2/pi)  --> The Josephson Constant

Since h-bar = 1, then h-bar/e --> 1/sqrt 2

And since Me = 1/2 in Rydberg Atomic Units, then (h-bar/e)^2 = Me = 1/2

h-bar --> Planck's Reduced Constant
Me --> Electron Mass

Basic Geometry. And Basic Geometry that just happens to agree with that special place on the number line where one might expect to find the critical zeroes of Riemann Hypothesis fame: 1/2

NATURAL UNITS
http://en.wikipedia.org/wiki/Natural_units

lim n --> infinity (PELL_n + PELL_n+1)/(PELL_n) = sqrt 2

Recursion Rule 1A + 2B = C

0, 1, 2, 5, 12, 29, etc... (Pell Numbers)
1, 1, 3, 7, 17, 41, etc...

- RF
Raphie Frank
QUOTE (Raphie Frank+Nov 28 2011, 08:37 PM)
-1, 0, 1, 4, 9, 26, 55, 154 = R_0:n (Indices of "Sophie Germain Figurate Numbers of the 1st Kind")
[no sequence on OEIS]

1, 1, 2, 4, 11, 23, 64, 134 = R_1:n (Indices of "Sophie Germain Figurate Numbers of the 2nd Kind")
http://oeis.org/A006452

The relation between the two... (3*R_0:n - R_0:n-1)/2 - R_0:n+1

-1, 0, 1, 4, 9, 26, 55, 154
1, 1, 2, 4, 11, 23, 64, 134

Can be combined into an alternating series...

R_n = -1, 1, 0, 1, 1, 2, 4, 4, 11, 9, 23, 26, 64, 55, 134, 154

p'_(R_n)| = 0, 2, 1, 2, 2, 3, 7, 7, 31... for as long as the series does not decrease...

Take the product x^sgn(x)...

0^0* 2^1* 1^1 * 2^1 * 2^1 * 3^1 * 7^1 * 7^1 * 31^1 = 36456

36456*10^-11 = The Balmer Constant.

Take the product x...

0* 2 * 1 * 2 * 2 * 3 * 7 * 7 * 31 = 0

SUM (CUT_(n, 3)) for range -1 --> -1 + 31 = 36456
CUT_(-1, 3) = 0

CUT(n,3) = Cake numbers: maximal number of pieces resulting from n planar cuts through a cube (or cake): C(n+1,3)+n+1
1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526, 4992, 5489, 6018, 6580, 7176, 7807, 8474, 9178, 9920, 10701, 11522, 12384, 13288, 14235, 15226
http://oeis.org/A000125

Pretty cool, eh?

- RF

==================================

SUM (1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526)

(1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526)

Mean = 1176 == T_(24*2)
Median = 576 == (24^2)
Standard Deviation for Sample = 1365 == T_90/3 == T_((sigma(p_24))/sigma(p_1))

Totient 90 = 24

Binomial coefficients binomial(n,4).
0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365
http://oeis.org/A000332

1365 = (30/2)*((30/2)-1)*((30/2)-2)*((30/2)-3)/24
30 = 90/3

Second pentagonal numbers: n * (3*n + 1) / 2
0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365
http://oeis.org/A005449

1365 = 30 * (3*30 + 1) / 2
30 = 90/3
==================================
Raphie Frank
There's an error in the above post as well, Jeremy, but it's "somwhat' irrelevant since the error in no way changes the values. Rather, only the range.

- RF

7 +/- 2 Miller 1956
jeremyebert
QUOTE (Raphie Frank+Dec 1 2011, 04:21 AM)
There's an error in the above post as well, Jeremy, but it's "somwhat' irrelevant since the error in no way changes the values. Rather, only the range.

- RF

7 +/- 2 Miller 1956

I'll be catching up on everything here in a few days Raphie. I'm working on a mainframe migration this week. Fun stuff....
Raphie Frank
A nice little graphic...

Hydrogen Density Plot.
http://en.wikipedia.org/wiki/File:Hydrogen_Density_Plots.png

- RF
Raphie Frank
Wish I'd come across this a while back. This paper came out in 2000... (see bolded)

Quantum-like Chaos in Prime Number Distribution and in Turbulent Fluid Flows
http://arxiv.org/html/physics/0005067
by A.M. Selvam

ABSTRACT
Recent studies indicate a close association between the distribution of prime numbers and quantum mechanical laws governing the subatomic dynamics of quantum systems such as the electron or the photon. Number theoretical concepts are intrinsically related to the quantitative description of dynamical systems of all scales ranging from the microscopic subatomic dynamics to macroscale turbulent fluid flows such as the atmospheric flows. It is now recognised that Cantorian fractal spacetime characterise all dynamical systems in nature. A cell dynamical system model developed by the author shows that the continuum dynamics of turbulent fluid flows consist of a broadband continuum spectrum of eddies which follow quantumlike mechanical laws. The model concepts enable to show that the continuum real number field contains unique structures, namely prime numbers which are analogous to the dominant eddies in the eddy continuum in turbulent fluid flows. In this paper it is shown that the prime number frequency spectrum follows quantumlike mechanical laws.

excerpt
---------
Historically, the British mathematician Roger Penrose discovered in 1974 the quasiperiodic Penrose tiling pattern, purely as a mathematical concept. The fundamental investigation of tilings which fill space completely is analogous to investigating the manner in which matter splits up into atoms and natural numbers split up into product of primes. The distiction between periodic and aperiodic tilings is somewhat analogous to the distinction between rational and irrational real numbers, where the latter have decimal expansions that continue forever , without settling into repeating blocks [Devlin, 1997]. Even earlier Kepler saw a fundamental mathematical connection between symmetric patterns and 'space filling geometric figures' such as his own discovery , the rhombic dodecahedron , a figure having 12 identical faces [ Devlin, 1997]. The quasiperiodic Penrose tiling pattern has five-fold symmetry of the dodecahedron. Recent studies[Seife,1998] show that in a strong magnetic field, electrons swirl around magnetic field lines, creating a vortex. Under right conditions, a vortex can couple to an electron, acting as a single unit.
Raphie Frank
Jeremy, check out this triangle...

A077028
The rascal triangle, read by rows: T(n,k) (n >= 0, 0 <= k <= n) = k(n-k)+1.
https://oeis.org/A077028

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 5, 1, 1, 6, 9, 10, 9, 6, 1, 1, 7, 11, 13, 13, 11, 7, 1, 1, 8, 13, 16, 17, 16, 13, 8, 1, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10, 1, 1, 11, 19, 25, 29, 31, 31, 29, 25, 19, 11, 1, 1, 12, 21, 28, 33, 36

The rows sum to cake numbers and it is conjectured that except for rows 0, 1 and 6 every row contains a prime.

I came across it while doing a search for the values one gets for the following formula...

((n-7)*i)^2 + 26 - 2*floor [(n-1)/11]
n = 1 --> 12
= -11, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10, 1, -1

... which is, obviously, a parabola for the first two arguments, one that is related (via the sign function) to the following exploratory formula:

3X + Z^2 + 3Z

X = CUT((n-2), (4 + uv))
Z = -1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, -1
u = 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1 (Single Paradiddle Sequence)
v = -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -1

I think this triangle could possibly be a bridge between your and my explorations.

Best,
Raphie
Raphie Frank
The best part? The "Rascal Triangle" was discovered (or rediscovered) by... get this, Middle Schoolers doing a collaborative project.

- RF
jeremyebert
QUOTE (Raphie Frank+Dec 17 2011, 05:58 PM)
Jeremy, check out this triangle...

A077028
The rascal triangle, read by rows: T(n,k) (n >= 0, 0 <= k <= n) = k(n-k)+1.
https://oeis.org/A077028

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 5, 4, 1, 1, 5, 7, 7, 5, 1, 1, 6, 9, 10, 9, 6, 1, 1, 7, 11, 13, 13, 11, 7, 1, 1, 8, 13, 16, 17, 16, 13, 8, 1, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10, 1, 1, 11, 19, 25, 29, 31, 31, 29, 25, 19, 11, 1, 1, 12, 21, 28, 33, 36

The rows sum to cake numbers and it is conjectured that except for rows 0, 1 and 6 every row contains a prime.

I came across it while doing a search for the values one gets for the following formula...

((n-7)*i)^2 + 26 - 2*floor [(n-1)/11]
n = 1 --> 12
= -11, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10, 1, -1

... which is, obviously, a parabola for the first two arguments, one that is related (via the sign function) to the following exploratory formula:

3X + Z^2 + 3Z

X = CUT((n-2), (4 + uv))
Z = -1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, -1
u = 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1 (Single Paradiddle Sequence)
v = -1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, -1

I think this triangle could possibly be a bridge between your and my explorations.

Best,
Raphie

YOWZERS!!!
"The triangle can be generated by numbers of the form k*(n-k) + 1 for k = 0 to n."

looks very close to the imaginary part of my equation

(((n+1)/2)-k) + i sqrt(k*(n-k+1))

This is awesome Raphie!

Sorry for my lack of anything to say lately. I've been spending a lot of time learning about simultaneous linear equations, matrices and Lie groups these last few weeks. I'm really understanding more about matrix math and how it applies to my stuff. I used a n*3 matrix in my visualizations, but I'm just now starting to understand how the rotations actually work. I'm even starting to understand E8 a little better. Now with these connections back into your work, I think we might be able to find something exceptional.

jeremyebert
QUOTE (jeremyebert+Dec 17 2011, 09:46 PM)
YOWZERS!!!
"The triangle can be generated by numbers of the form k*(n-k) + 1 for k = 0 to n."

looks very close to the imaginary part of my equation

(((n+1)/2)-k) + i sqrt(k*(n-k+1))

This is awesome Raphie!

Sorry for my lack of anything to say lately. I've been spending a lot of time learning about simultaneous linear equations, matrices and Lie groups these last few weeks. I'm really understanding more about matrix math and how it applies to my stuff. I used a n*3 matrix in my visualizations, but I'm just now starting to understand how the rotations actually work. I'm even starting to understand E8 a little better. Now with these connections back into your work, I think we might be able to find something exceptional.

The difference between the triangles is just k-1.

Table[{(k*(n-k+1)), (k*(n-k) + 1) },{n,25},{k,n}]

http://www.wolframalpha.com/input/?i=Table...%7Bk%2Cn%7D%5D+
Raphie Frank
QUOTE (jeremyebert+Dec 18 2011, 02:46 AM)
Sorry for my lack of anything to say lately. I've been spending a lot of time learning about simultaneous linear equations, matrices and Lie groups these last few weeks. I'm really understanding more about matrix math and how it applies to my stuff. I used a n*3 matrix in my visualizations, but I'm just now starting to understand how the rotations actually work. I'm even starting to understand E8 a little better. Now with these connections back into your work, I think we might be able to find something exceptional.

Excellent, Jeremy. As you know, I'm a bit of a "dummy," so I look forward to learning from you.

Best,
Raphie

P.S. The Rascal Triangle/Square can be viewed as first differences of the "Polygonal Number Triangle/Square." The latter, when properly extended in square format has diagonal sums that can easily generate the "Gross Polygonal Sum" of an n-gon (aka "Moser's problem")

In other words, just as one might expect the geometric division of space to have something to do with the geometric division of space, one might also expect that polygonal numbers have something to do with polygons....
Raphie Frank
QUOTE (jeremyebert+Dec 18 2011, 02:46 AM)
YOWZERS!!!
"The triangle can be generated by numbers of the form k*(n-k) + 1 for k = 0 to n."

Thought you might like that part...
Raphie Frank
QUOTE (jeremyebert+Dec 18 2011, 02:46 AM)
"The triangle can be generated by numbers of the form k*(n-k) + 1 for k = 0 to n."

The above gives you the diagonals. If you want to get the rows, then...

A.
WOLFRAM INPUT
((n-(k+1))*i)^2 + (k^2 + 1)

alternate form...
k^2 - n^2 + 12n - 35

e.g.
((n-(5+1))*i)^2 + (5^2 + 1)
parabola

1 | 1
2 | 10
3 | 17
4 | 22
5 | 25
6 | 26
7 | 25
8 | 22
9 | 17
10 | 10
11 | 1

as opposed to...

B.
5*(n-5) + 1
which is a line

In other words, that triangle can be thought of as either an infinite collection of (truncated) lines OR an infinite set of (truncated) parabolas. It's not this. It's not that. It's this AND that.

What you see depends on how you look at it.

Both formulas, by the way, without inserted values, are hyperbolic paraboloids

Via Wikipedia
"The hyperbolic paraboloid (not to be confused with a hyperboloid) is a doubly ruled surface shaped like a saddle. In a suitable coordinate system, a hyperbolic paraboloid can be represented by the equation:

z/c = y^2/b^2 - x^2/a^2
http://en.wikipedia.org/wiki/Paraboloid

- RF
Raphie Frank
Now, thanks to the kids we can recursively define the Polygonal Number array...

A057145 Square array T(n,k) of polygonal numbers T(n,k) = ((n-2)*k^2-(n-4)*k)/2, n >= 2, k >= 1, read by antidiagonals.
via OEIS https://oeis.org/A057145

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, ...
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, ...
1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, ...
1, 6, 15, 28, 45, 66, 91, 120, 153, 190, 231, ...
1, 7, 18, 34, 55, 81, 112, 148, 189, 235, 286, ...
1, 8, 21, 40, 65, 96, 133, 176, 225, 280, 341, ...
1, 9, 24, 46, 75, 111, 154, 204, 261, 325, 396, ...
1, 10, 27, 52, 85, 126, 175, 232, 297, 370, 451, ...
1, 11, 30, 58, 95, 141, 196, 260, 333, 415, 506, ...
1, 12, 33, 64, 105, 156, 217, 288, 369, 460, 561, ...
1, 13, 36, 70, 115, 171, 238, 316, 405, 505, 616, ...
1, 14, 39, 76, 125, 186, 259, 344, 441, 550, 671, ...

e.g.

((9 * 10) + 6)/6 = 16 for 6 = 3rd Triangular Number (3rd column)
((12 * 16) + 6)/9 = 22 for 6 = 3rd Triangular Number (3rd column)

Go to the 4th column and you would add 10, 5 th column, 15, etc. Easy as pie...

In Kid Notation, starting from the 1st Column as "Column 1" (0's not shown):

((NorthEast * SouthWest) + Triangle of NorthWest Column Number)/Northwest
=
SouthEast

P(n,k) = (P(n, k-1)*P(n-1, k) + (k^2 + k)/2)/P(n-1, k-1)

As a little social experiment, let's see how long it takes someone to post this recursive relationship on OEIS since currently you won't find it there.

- RF

P.S. This manner of recursive relationship can almost certainly be extended to additional summations of the Rascal Triangle, but I think it gets a bit complicated pretty fast. I haven't really checked.

P.P.S. Note that in dimensional terms, in order to specify k, you need to know where you are in the triangle column-wise, although row number is not necessary information.
Raphie Frank
QUOTE (Raphie Frank+Dec 19 2011, 06:45 PM)
A057145   Square array T(n,k) of polygonal numbers T(n,k) = ((n-2)*k^2-(n-4)*k)/2, n >= 2, k >= 1, read by antidiagonals.
via OEIS https://oeis.org/A057145

1,  2,  3,  4,   5,   6,   7,   8,   9,  10,  11, ...
1,  3,  6, 10,  15,  21,  28,  36,  45,  55,  66, ...
1,  4,  9, 16,  25,  36,  49,  64,  81, 100, 121, ...
1,  5, 12, 22,  35,  51,  70,  92, 117, 145, 176, ...
1,  6, 15, 28,  45,  66,  91, 120, 153, 190, 231, ...
1,  7, 18, 34,  55,  81, 112, 148, 189, 235, 286, ...
1,  8, 21, 40,  65,  96, 133, 176, 225, 280, 341, ...
1,  9, 24, 46,  75, 111, 154, 204, 261, 325, 396, ...
1, 10, 27, 52,  85, 126, 175, 232, 297, 370, 451, ...
1, 11, 30, 58,  95, 141, 196, 260, 333, 415, 506, ...
1, 12, 33, 64, 105, 156, 217, 288, 369, 460, 561, ...
1, 13, 36, 70, 115, 171, 238, 316, 405, 505, 616, ...
1, 14, 39, 76, 125, 186, 259, 344, 441, 550, 671, ...

Left Down to Up Right Diagonal Sums

0, 1, 3, 7, 15, 30, 56, 98, 162, 255, 385, 561, 793, 1092, 1470, 1940, 2516, 3213, 4047, 5035, 6195, 7546, 9108, 10902, 12950, 15275, 17901, 20853, 24157, 27840, 31930, 36456
https://oeis.org/A055795

In Rascal Triangle terms you would get the very same result if you summed the first 496 terms.

Dimensions of E8 = 248
Vertices of E8 = 240 = totient 496

RELATION OF 496 TO SUPERSTRING THEORY
via Wikipedia

"The number 496 is a very important number in superstring theory. In 1984, Michael Green and John H. Schwarz realized that one of the necessary conditions for a superstring theory to make sense is that the dimension of the gauge group of type I string theory must be 496. The group is therefore SO(32). Their discovery started the first superstring revolution. It was realized in 1985 that the heterotic string can admit another possible gauge group, namely E8 x E8"
http://en.wikipedia.org/wiki/496_(number)

And what was that number, again, of maximal number of electrons per shell of an atom? oh, right, 32.

And what's the Balmer constant again? oh, right, 36456 * 10^-11 meters

And the number of dimensions in M-theory? Let me think.........11?

- RF

The Five Superstring Theories
Here Spin(32)/Z2 is essentially SO(32) (they have the same Lie algebra, but different global properties), and E8 is the rank 8, dimension 248 exceptional (simple) Lie group.

Note that all of these groups have dimension 496. In addition, the first two both have rank 16, and contain SO(16) × SO(16) as a subgroup. The later two have rank 256 and 496 respectively. It turns out that there are string theories corresponding to the Spin(32)/Z2 and E8 × E8 cases (in fact, two for Spin(32)/ Z2), though not for the latter two groups.
http://www.lepp.cornell.edu/~gx26/files/benMar31.pdf
Raphie Frank
WOLFRAM INPUT

A. DIVISION OF 3 SPACE
--------------------------
SUM ((n+1)*(n^2-n+6)/6) for range 0 through 30
= 36456

1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526

The mean is 1176 = (48^2 + 48)/2
The Median is 576 = 24^2
Divisors of 36456 = 48

B. POLYGONAL NUMBER SQUARE DIAGONAL SUM
-----------------------------------------------------
SUM (1/2 (-n^3 + 32n^2 - 29n)) for range 1 through 31
= 36456

1, 31, 87, 166, 265, 381, 511, 652, 801, 955, 1111, 1266, 1417, 1561, 1695, 1816, 1921, 2007, 2071, 2110, 2121, 2101, 2047, 1956, 1825, 1651, 1431, 1162, 841, 465, 31

The mean is 1176 = (48^2 + 48)/2
The Median is 1266
Divisors of 36456 = 48

OBSERVATION # 1
----------------------
The first value of A that is larger than B is 2048. The last value of B that is larger than A is 2047. Together they sum to 4095 which the last of the Ramanujan-Nagell Triangular Numbers. 196560/48 = 4095 = Vertices of the Leech Lattice/48

OBSERVATION # 2
----------------------
First two terms of B sum to 32
Last two terms of B sum to 496

Middle term of B is 1816. It's counterpart in A is 576. The difference between the two is divisible by 31 (40*31 = 2^3 * 5 * 31, all Golden Scale Numbers that are also Mersenne prime exponents)

- RF

P.S. If I were the mystical numerologist people have claimed that I am, but am not, I'd be all excited about the fact that the 11 th term of B, counting both ways are 1111 and 1111+1010 = 2121. Both 1111 and 1010 have significant meaning for New Age Fanatics, Bible Zealouts and the like. Here is one example of that http://globalpsychics.com/enlightening-you...logy/1111.shtml
Raphie Frank
QUOTE (Raphie Frank+Dec 19 2011, 08:21 PM)
WOLFRAM INPUT

A. DIVISION OF 3 SPACE
--------------------------
SUM ((n+1)*(n^2-n+6)/6) for range 0 through 30
= 36456

1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176, 232, 299, 378, 470, 576, 697, 834, 988, 1160, 1351, 1562, 1794, 2048, 2325, 2626, 2952, 3304, 3683, 4090, 4526

The mean is 1176 = (48^2 + 48)/2
The Median is 576 = 24^2
Divisors of 36456 = 48

B. POLYGONAL NUMBER SQUARE DIAGONAL SUM
-----------------------------------------------------
SUM (1/2 (-n^3 + 32n^2 - 29n)) for range 1 through 31
= 36456

1, 31, 87, 166, 265, 381, 511, 652, 801, 955, 1111, 1266, 1417, 1561, 1695, 1816, 1921, 2007, 2071, 2110, 2121, 2101, 2047, 1956, 1825, 1651, 1431, 1162, 841, 465, 31

The mean is 1176 = (48^2 + 48)/2
The Median is 1266
Divisors of 36456 = 48

Subtract B from A

With help from Wolfram Alpha:

n/6(4n^2 - 99n + 95)

n | 1/6 n (4 n^2-99 n+95)
1 | 0
2 | -29
3 | -83
4 | -158
5 | -250
6 | -355
7 | -469
8 | -588
9 | -708
10 | -825
11 | -935
12 | -1034
13 | -1118
14 | -1183
15 | -1225 etc. to n = 31

NOT: 1225 = 35^2 and is the 49th Triangular Number

Now, starting from the middle, 1240, take the first differences working in both directions and here's what you get...

0, 1, 10, 35, 84, 165, 286, 455, 680, 969, 1330, 1771, 2300, 2925, 3654, 4495

And with a little help from OEIS:

A000447
https://oeis.org/A000447
a(n) = 1^2 + 3^2 + 5^2 + 7^2 + ... + (2*n-1)^2 = n*(4*n^2 - 1)/3
0, 1, 10, 35, 84, 165, 286, 455, 680, 969, 1330, 1771, 2300, 2925, 3654, 4495, 5456, 6545, 7770, 9139, 10660, 12341, 14190, 16215, 18424, 20825, 23426, 26235, 29260, 32509, 35990, 39711, 43680, 47905, 52394, 57155, 62196, 67525, 73150, 79079
4 times variance of the area under an n step random walk: e.g. with three steps, area can be 9/2, 7/2, 3/2, 1/2, -1/2, -3/2, -7/2, or -9/2 each with probability 1/8, giving a variance of 35/4 or a(3)/4. - Henry Bottomley, Jul 14 2003

Also a(n)=(1/6)*(8*n^3-2*n), n>0: structured octagonal diamond numbers (vertex structure 9)

REFERENCES
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).

Coincidence? Seems I have a paper I need to look at.

QUOTE (Raphie Frank+Nov 27 2011, 02:03 AM)
...Or that 31 + 4, the "Sophie Germain Constant" just happens to equal 35 and 4*pi^2/sqrt(35) *10^-11 m^3*kg^-1*s^-2 = 6.6730705216 * 10^-11 m^3*kg^-1*s^-2 just happens to give a very nice little first order approximation of Big G that's NOT correct, but is about as correct (a bit more so actually) as what the Google search engine gives you when you do a search for "Gravitational Constant."

Best,
RF
Raphie Frank
QUOTE (Raphie Frank+Dec 19 2011, 10:00 PM)
REFERENCES
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).

Coincidence? Seems I have a paper I need to look at.

http://web.cs.dal.ca/~jborwein/Preprints/P...pers/shells.pdf

And if I wasn't clear:

1, 10, 35, 64, etc. are Tetrahedral Numbers. Every other one (the odd ones).

- RF

Here is an excerpt from page 201
------------------------------------------
A good starting point for the discussion of the arrangement of atoms is the packing of spheres. Once again, it all began with Kepler. Kepler described the arrangement of spheres which we now call face-centered cubic (fee). He mentioned in an off-handed way that this is the densest possible packing of spheres. Mathematicians have been trying to prove this assertion for the past 300 years [2,3]. Why is this so difficult to prove? It is clear that the densest packing of just four spheres is a tetrahedron, and it is clear how this densest packing ofjust four spheres is a tetrahedron, and it is clear how this densest packing should continue. Each newly added sphere should nestle into the hole made by three other spheres, forming a new tetrahedral arrangement. If you were able to continue indefinitely this building scheme and fill up all of space, you would construct a body having a density of 0.78 . . .
------------------------------------------

If there is anything to the relationships I've been coming across, Jeremy, I will tell you right here and now what it implies...

"atomic structure of a crystal"
becomes (also)...
"crystalline structure of an atom"

... and particle/wave duality becomes particle/wave/crystal triality.

A Geometric Theory of Everything (Scientific American)
http://www.scientificamerican.com/article....y-of-everything

- RF
Raphie Frank
Here is the Wolfram Ready Formula:

(32 - n)/6(4(32 - n)^2 - 99(32 - n) + 95) - n/6(4n^2 - 99n + 95)

n | 1/6 (4 (32-n)^2-99 (32-n)+95) (32-n)-1/6 n (4 n^2-99 n+95)
1 | 4495
2 | 3654
3 | 2925
4 | 2300
5 | 1771
6 | 1330
7 | 969
8 | 680
9 | 455
10 | 286
11 | 165
12 | 84
13 | 35
14 | 10
15 | 1

Also, not sure why this works yet, but...

SUM((32 - n)/6(4(32 - n)^2 - 99(32 - n) + 95) - n/6(4n^2 - 99n + 95)) for range 1 through 15
= 19160

= 36456 - (1/6)*(46^3+3*46^2+2*46)
= 36456 - 17296

... where 17296 is the 46th Tetrahedral number.
Raphie Frank
QUOTE (Raphie Frank+Dec 20 2011, 12:16 AM)
Also, not sure why this works yet, but...

SUM((32 - n)/6(4(32 - n)^2 - 99(32 - n) + 95) - n/6(4n^2 - 99n + 95)) for range 1 through 15
= 19160

= 36456 - (1/6)*(46^3+3*46^2+2*46)
= 36456 - 17296

... where 17296 is the 46th Tetrahedral number.

AN OBSERVATION # 1

E.
DODECAHEDRAL GNOMIC #s (sum to EVERY (3n + 1)-th TETRAHEDRAL #)
A093485 (27*n^2 + 9*n + 2)/2.
https://oeis.org/A093485
1, 19, 64, 136, 235, 361, 514, 694, 901, 1135, 1396, 1684, 1999, 2341, 2710, 3106

D.
EVEN TETRAHEDRAL #s
1, 10, 35, 84, 165, 286, 455, 680, 969, 1330, 1771, 2300, 2925, 3654, 4495

Total Values: 31

Sum = 36456

Formula: (2n^4 + 31n^3 + 55n^2 + 32n + 6)/6 = y
for n = 15, then y = 36456

A006566 Dodecahedral numbers: n(3n-1)(3n-2)/2
https://oeis.org/A006566
0, 1, 20, 84, 220, 455, 816, 1330, 2024, 2925, 4060, 5456, 7140, 9139, 11480, 14190, 17296
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral and icosahedral) numbers (cf. A053012). [From Daniel Forgues, May 14 2010]

AN OBSERVATION #2
A057402 Low-temperature magnetization expansion for Kagome net (Potts model, q=4).
https://oeis.org/A057402
1, 0, 0, 0, -4, 0, -24, -32, -120, -336, -768, -2496, -6612, -17296

AN OBSERVATION #3
A085039 Smallest multiple of the n-th prime such that every partial sum is a square.
4, 12, 20, 28, 132, 429, 51, 285, 483, 957, 3224, 6475, 6396, 8729, 17296
https://oeis.org/A085039

Thus 211^2 - 165^2 = 17296 == (46+165)^2 - 165^2; 47 divides 17296 and is the 15-th prime. 17296 also just so happens to be the 15th "amicable" number

See: http://britton.disted.camosun.bc.ca/amicable.html
Raphie Frank
This triangle is related to the Rascal Triangle...

=====================================================
A087401 Triangle of n*r-binomial(r+1,2).
http://oeis.org/A087401
0, 0, 0, 0, 1, 1, 0, 2, 3, 3, 0, 3, 5, 6, 6, 0, 4, 7, 9, 10, 10, 0, 5, 9, 12, 14, 15, 15, 0, 6, 11, 15, 18, 20, 21, 21, 0, 7, 13, 18, 22, 25, 27, 28, 28, 0, 8, 15, 21, 26, 30, 33, 35, 36, 36, 0, 9, 17, 24, 30, 35, 39, 42, 44, 45, 45, 0, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55, 55, 0, 11

There is a curious connection with the character tables of cyclic groups of prime power order. Let G be a cyclic group of order p^n where p is prime and n is nonnegative. Construct an (n+1)x(n+1) matrix A whose rows and columns are indexed by the set 0,1,...,n as follows. The ij entry is obtained by taking any element of order p^(n-j) in G and summing its character values over all characters of order p^i in the dual group of G. Remarkably, all coefficients of the characteristic polynomial of A are powers of p (with alternating signs) and these powers can be read off from the appropriate row of our triangle. For example if n=2 then the characteristic polynomial is X^3 - p^2*X^2 + p^3*X - p^3.
=====================================================

In table format
http://oeis.org/A087401/table

I. First Pentagonal Numbers run right down the middle (even rows)
II. Second Pentagonal Numbers run down the Center Left (odd rows)
III. Term to Center Right values (odd rows) are the indices of Second Pentagonal Numbers that are elements of C(n,4)

Taken collectively, the center terms are the indices associated with Pentagonal Numbers that are elements of C(n,4)

I. U II. are indices of First Pentagonal Numbers
III. are indices of Second Pentagonal Numbers

If you add 2 to each value in the table, then the rows sum to 2*Cake Numbers. At row 30 (496 values), the sum is 2*36456. Subtract out sigma (496) = 992, then you get 2*35960 = 2*binomial(32,4).

NOTE: In general, any time the number of terms in the triangle is an even perfect number, then subtracting out sigma (P_n) will give you twice the Summation of the Tetrahedrals. See: http://oeis.org/A034827

Subtract out the Rascal Triangle and you end up with the PanCake Triangle

PANCAKE TRIANGLE

1
1, 1
1, 1, 2
1, 1, 2, 4
1, 1, 2, 4, 7
1, 1, 2, 4, 7, 11
(not on OEIS)

+

RASCAL TRIANGLE
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 5, 4, 1
1, 5, 7, 7, 5, 1
http://oeis.org/A077028

=

"PENTA-CAKE TRIANGLE"
2
2, 2
2, 3, 3
2, 5, 7, 8, 8
2, 6, 9, 11, 12, 12
(not on OEIS)

&

"PENTA-CAKE TRIANGLE" - 2 = A087401 Triangle of n*r-binomial(r+1,2)
0
0, 0
0, 1, 1
0, 2, 3, 3
0, 3, 5, 6, 6
0, 4, 7, 9, 10, 10
http://oeis.org/A087401

CONJECTURE:

Strong: Excepting the 1-st row, there are at least two primes in every row of the Penta-Cake Triangle.
Weak: If Strong Conjecture is Incorrect, there will be a discernible pattern in relation to where no primes appear.

But, for the record, I have only checked this to row 31 thus far.

Also related...
A141418
A triangular sequence of coefficients of Dynkin diagram weights for the Cartan Groups D_n: t(n,m)=m*(2*n - m - 1)/2.
http://oeis.org/A141418

Same as A087401, but without the 0's

Another way to get to A087401 is just to sum (n(mod n) - k)

e.g. Sum the following left to right...

0
0, 0
0, 1, 0
0, 2, 1, 0
0, 3, 2, 1, 0
0, 4, 3, 2, 1, 0

I am sure you will recognize that triangle, Jeremy. The interior gives the number of divisors of prime powers. "There is a curious connection with the character tables of cyclic groups of prime power order."

"Curious," maybe, but not very surprising, actually...

- RF
Raphie Frank
QUOTE (Raphie Frank+Nov 24 2011, 11:40 PM)
Via Wikipedia

KAGOME LATTICE
"Some minerals, namely jarosites and herbertsmithite, contain layers with kagome lattice arrangement of atoms in their crystal structure. These minerals display novel physical properties connected with geometrically frustrated magnetism."
http://en.wikipedia.org/wiki/Kagome_lattice

GEOMETRICAL FRUSTRATION
"In simple terms, the substance can never be completely frozen, because the structure it forms does not have a single minimal-energy state, so motion on a molecular scale continues even at absolute zero and even without input of energy."

QUOTE (jeremyebert+Nov 25 2011, 02:36 AM)
Is geometrical frustration a source of energy then? Thats just amazing to think about. Very cool stuff Raphie.

QUOTE (Raphie Frank+Nov 25 2011, 02:46 AM)
Personally, I think more in terms of consciousness. I really don't know the relationship to energy, but you can't help but wonder how that works into fantastical notions of perpetual motion machines, which all reasonable evidence has shown to be impossible. Which I don't argue with, by the way, at non-atomic scales.

I was always under the apparently wrong impression that at absolute zero, all movement stops, so it's certainly interesting to me as well

Perpetual motion 'time crystals' may exist
By Natalie Wolchover
updated 2/21/2012 7:22:38 PM ET

excerpt
===================================
From diamonds to snowflakes to salt, crystals are common in nature. The arrangement of their atoms in orderly, repeating patterns extending in all three spatial dimensions doesn't just make them nice to look at; crystals are also the vital components of technologies from electrical transistors to LCD screens.

In groundbreaking new research, Nobel-winning physicist Frank Wilczek contends that “time crystals,” moving structures that repeat periodically in the fourth dimension, exist as well.

A time crystal would be a physical object whose constituent parts move in a repeating pattern. Think of a kaleidoscope, whose sparkly bits swirl on loop forever, or a clock, whose hour hand completes a 360-degree turn every 12 hours. But unlike clocks or other common objects with moving parts, time crystals would run forever under their own steam — perpetual motion devices permitted by the laws of physics.

These bizarre objects have never been conceived of previously, but Wilczek, a theoretical physicist at the Massachusetts Institute of Technology and winner of the 2004 Nobel Prize in Physics for his work on the strong nuclear force, thinks they either already exist in nature or could soon be engineered.

Many types of time crystals are possible. “The simplest realization would be some system whose geometry allows it to move in a circle and come around after a certain time to the same place,” Wilczek told Life’s Little Mysteries. A more complicated configuration could be a collection of atoms moving fluidly in three dimensions but all periodically returning to their starting points.

The characteristic trait of a time crystal, Wilczek says, is that it moves without consuming or shedding any energy. Instead, it is in a stable, minimum-energy state, just as diamonds and other conventional crystals are. Even so, it is also in a state of perpetual motion.
===================================

MORE: http://www.msnbc.msn.com/id/46472940/ns/te...s/#.T1iFWK5bd20

QUOTE (Raphie Frank+Dec 19 2011, 10:24 PM)

If there is anything to the relationships I've been coming across, Jeremy, I will tell you right here and now what it implies...

"atomic structure of a crystal"
becomes (also)...
"crystalline structure of an atom"

... and particle/wave duality becomes particle/wave/crystal [i]triality
.

- RF
jeremyebert
Raphie,
I think this a map of the Klein four-group geometry.
http://dl.dropbox.com/u/13155084/prime-%20...our%20group.png

I've started on the documentation.
http://dl.dropbox.com/u/13155084/Given%20a%20divisor%20k.pdf
jeremyebert
QUOTE (jeremyebert+Apr 30 2012, 09:05 PM)
Raphie Frank
QUOTE (jeremyebert+May 1 2012, 08:00 PM)
More detail.
http://dl.dropbox.com/u/13155084/prime-%20...our%20group.png

I'll have to look into this. As an FYI, the tiling of the hyperbolic plane by heptagons, which maps to Klein's quartic curve, as well as the mathematics of the Fano plane, is 7* a Fibonacci #. Lots of great info about it on the website of John Baez.

- RF
Raphie Frank
QUOTE (Raphie Frank+May 2 2012, 09:48 PM)
I'll have to look into this. As an FYI, the tiling of the hyperbolic plane by heptagons, which maps to Klein's quartic curve, as well as the mathematics of the Fano plane, is 7* a Fibonacci #. Lots of great info about it on the website of John Baez.

- RF

Just to get you thinking ahead...

Via Wikipedia...

"In the construction of finite rings, eight of the eleven rings with four elements have the Klein four-group as their additive substructure."

Look for the 11 of 11, not the 8 of 11.

As an analog...

phi(p) or sigma_(p) divides 360. 14 of the 15 supersingular primes are members of this group. Close but no cigar... To get the 15th of 15 supersingular prime (47+/
- 1 does not divide 360), then phi(p) or sigma_(p) divides 720, not 360.

Probably just a coincidence that 4-dimensional geometry only works with a double, not single, circle...

- RF
Raphie Frank
Galois-theoretic derivation of the quartic formula
http://planetmath.org/GaloisTheoreticDeriv...ticFormula.html

The Klein Four Group (V4) is right there in the middle of the Composition Series.

Great article. One glaring omission, however. It doesn't mention Lucas Sequences.
http://en.wikipedia.org/wiki/Lucas_sequence

In brief:

(sqrt (P^2 - 4Q) + P)/2 = x will give the limit at infinity of successive terms in a recursive sequence where the recursion rule is -Q(a(n)) + P(a(n+1)) = a(n+2).

P = a + b
Q = a * b

Mersenne Numbers, Pell Numbers, Fibonacci Numbers, etc. etc. are all examples of Lucas Sequences.

Now, look at the formula for the roots of the Quartic in the article by David Jao. It's right at the end. It will be blindingly obvious that the exact same mathematics applies to both.

Where P & Q are twin primes, then x will equal P + 1 and P * Q will be one less than a square, or twice the harmonic average of consecutive Triangular Numbers. And then you can bring in totient and sigma functions to your heart's content.

While you're at it, take a look at the role p + 1 plays in relation to supersingular primes. (3, 5), (5, 7), (11, 13), (17, 19), (29, 31) are all supersingular and twins to one another. That's 9 of the 15. Outliers: 41, 59, 71 & 2, 23, 47, although 3 of the 6 are the lower member of a twin prime pair while the other 3 are 2 less than the square of a supersingular prime (2^2, 5^2, 7^2).

- RF

================================
P.S. p^2 - 2 is prime: First values = 2, 3, 5, 7, 13, the divisors of which are 1, 2, 3, 5, 7, 13. This is the complete set of {1 U primes} such that n^2 + 11 divides 360 (or 720).

pi (1, 2, 3, 5, 7, 13) = 0, 1, 2, 3, 4, 6 = n in N such that (360/n^sign(n) radians) is in N == n in N | phi(n) < 3. For n > 0, then these are the positive solutions to the short proof for the Crystallographic Restriction Theorem (aka "allowable n-fold rotatonal symmetries of a Crystal"). Fundamental domain: 360 (degrees)

p_(1, 2, 3, 5, 7, 13) = 2, 3, 5, 11, 17, 41, sometimes referred to as Euler's "Lucky Primes" since they generate primes to n = (p-2) when placed in the equation form: n^2 + n + p.

Kinda makes you wonder if Frampton & Kephart weren't on to something back in 1999:

Mersenne Primes, Polygonal Anomalies and String Theory Classification
Paul H. Frampton, Thomas W. Kephart
(Submitted on 29 Apr 1999)
It is pointed out that the Mersenne primes $M_p=(2^p-1)$ and associated perfect numbers ${\cal M}_p=2^{p-1}M_p$ play a significant role in string theory; this observation may suggest a classification of consistent string theories.
http://arxiv.org/abs/hep-th/9904212

Probably all kinds of magic hiding out in the Eulerian polynomials they analyzed.
================================
Raphie Frank
As I am sure you are well aware, Jeremy, the divisors of 24 are important with regards to the Klein Four Group. Geometrically, they are particularly interesting since any symmetrical n-gon with a number of edges equal to a divisor of 24 will have angles with an integer multiple (ease of "bonding" with other polygons perhaps?). In fact, this holds true for any divisor of 360.

So, anyway, here's a simple multiplication table that generates the divisors, not of 24, but of 360 which is the first integer with 24 divisors...

01 03 06 015
02 06 12 030
03 09 18 045
04 12 24 060
06 18 36 090
12 36 72 180

... with one caveat. You've got to take that table and turn it into the denominator of a quotient table with 360 as the numerator.

360 120 60 24
180 060 30 12
120 040 20 08
090 030 15 06
060 020 10 04
030 010 05 02

1, 2, 3, 4, 6, 12 (left edge of top table) are the divisors of 12. 1, 3, 6 & 15 are the 1st, 2nd, 3rd and 5th Triangular Numbers (T_-2, T_-3, T_-4, T_-6).

There are 48 values in toto in those two tables, 24 of which are unique. All 24 divide 360 and d(360) = 24 (and is highly composite).

The second row is most interesting since all values in both tables are equal to phi(p) for some p in N and are equal either to a Pronic Number or 360 divided by a Pronic number. The last row of the top table also consists of integers 1 less than a prime, while you could completely do away with the 4th row since all are dupicates that can be found in other rows.

All together there are 4 sets of primes p and q such that phi(p)*phi(q) = 360:

(3, 181), (7, 61), (13, 31), (37, 11).
3 = p_(3-1) & 3 = p_2
7 = p_(5-1) & 5 = p_3
13 = p_(7-1) & 7 = p_4
37 = p_(13 -1) & 13 = p_6

Set ala' Lucas Sequence Notation:

P = 3 + 10 = 13
Q = 3 * 10 = 30

Then...
|(sqrt (D) + P)/2| = 10
---------------------------
|(sqrt (D) - P)/2| = 3
sqrt (D) = sqrt (13^2 - 4*30) = sqrt (49) = 7
P = 13
Q + sqrt (D) = 30 + 7 = 37

In general, think in terms of nested Babicka dolls aka "Groups within Groups" or "Communities within Communities"...

e.g.
d(3603600) = 360
d (360) = 24
d(24) = 8
d(8) = 4
d(4) = 3
d(3) = 2

S4 --> A4 --> V4 --> Z(2) --> 1

But what yields S4?

- RF
jeremyebert
QUOTE (Raphie Frank+May 3 2012, 08:59 PM)
As I am sure you are well aware, Jeremy, the divisors of 24 are important with regards to the Klein Four Group. Geometrically, they are particularly interesting since any symmetrical n-gon with a number of edges equal to a divisor of 24 will have angles with an integer multiple (ease of "bonding" with other polygons perhaps?). In fact, this holds true for any divisor of 360.

So, anyway, here's a simple multiplication table that generates the divisors, not of 24, but of 360 which is the first integer with 24 divisors...

01 03 06 015
02 06 12 030
03 09 18 045
04 12 24 060
06 18 36 090
12 36 72 180

... with one caveat. You've got to take that table and turn it into the denominator of a quotient table with 360 as the numerator.

360 120 60 24
180 060 30 12
120 040 20 08
090 030 15 06
060 020 10 04
030 010 05 02

1, 2, 3, 4, 6, 12 (left edge of top table) are the divisors of 12. 1, 3, 6 & 15 are the 1st, 2nd, 3rd and 5th Triangular Numbers (T_-2, T_-3, T_-4, T_-6).

There are 48 values in toto in those two tables, 24 of which are unique. All 24 divide 360 and d(360) = 24 (and is highly composite).

The second row is most interesting since all values in both tables are equal to phi(p) for some p in N and are equal either to a Pronic Number or 360 divided by a Pronic number. The last row of the top table also consists of integers 1 less than a prime, while you could completely do away with the 4th row since all are dupicates that can be found in other rows.

All together there are 4 sets of primes p and q such that phi(p)*phi(q) = 360:

(3, 181), (7, 61), (13, 31), (37, 11).
3 = p_(3-1) & 3 = p_2
7 = p_(5-1) & 5 = p_3
13 = p_(7-1) & 7 = p_4
37 = p_(13 -1) & 13 = p_6

Set ala' Lucas Sequence Notation:

P = 3 + 10 = 13
Q = 3 * 10 = 30

Then...
|(sqrt (D) + P)/2| = 10
---------------------------
|(sqrt (D) - P)/2| = 3
sqrt (D) = sqrt (13^2 - 4*30) = sqrt (49) = 7
P = 13
Q + sqrt (D) = 30 + 7 = 37

In general, think in terms of nested Babicka dolls aka "Groups within Groups" or "Communities within Communities"...

e.g.
d(3603600) = 360
d (360) = 24
d(24) = 8
d(8) = 4
d(4) = 3
d(3) = 2

S4 --> A4 --> V4 --> Z(2) --> 1

But what yields S4?

- RF

Your right, its something more than just the Klein Four-Group:

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."

I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = n

Z/2Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,-1}

Z/3Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,2}

Z/4Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,3}

Z/6Z
(s+k)^2 = {1,0}
s^2 = {0,1}
n = {1,5}

Z/8Z
(s+k)^2 = {1,4,1,0}
s^2 = {0,1,4,1}
n = {1,3,5,7}

Z/12Z
(s+k)^2 = {1,9,4,0}
s^2 = {0,4,9,1}
n = {1,5,7,11}

Z/24Z
(s+k)^2 = {01,09,16,12,01,09,04,00}
s^2 = {00,04,09,01,12,16,09,01}
n = {01,05,07,11,13,17,19,23}

(s+k)^2 + s^2 = 1 (mod 12) ** only modulo 12. ???? additive inversion of the identity element to its multiplication group Z/12Z*????
((s+k)^2 - s^2)^2 = 1 (mod 12)

To provide a possible answer to the S4 question:
In the Klein Four-Group the quadratic residues of (s+k)^2 and s^2 are permutations of the symmetric group of order 4 (S4).

Raphie Frank
QUOTE (jeremyebert+May 4 2012, 08:00 PM)
Your right, its something more than just the Klein Four-Group:

Wiki:
"The divisors of 24 — namely, {1, 2, 3, 4, 6, 8, 12, 24} — are exactly those n for which every invertible element x of the commutative ring Z/nZ satisfies x^2 = 1.
Thus the multiplicative group (Z/24Z)× = {±1, ±5, ±7, ±11} is isomorphic to the additive group (Z/2Z)3. This fact plays a role in monstrous moonshine."

I checked. Only the divisors of 24 produce the specific symmetry such that:

s=(n-k^2)/(2k)
s+k =(n+k^2)/(2k)
(s+k)^2 - s^2 = n

Just as an abstraction and flight of fancy, Jeremy, imagine there were such a thing as a...

KOLOLOKO GROUP
(Kolo means "wheel" or "bicycle" in Czech)

aka "Crazy Wheel"
aka "Mobius Strip"

... the p'rime divisors of which were all p in N such that phi(p') or sigma(q') divides 720. (p' or q' = n in N | d(n) < 3)

In such an imaginary group there would be 26 primes + 1, for a total of 27, decomposed into 16 KOA p'rimes phi(p')|720 and 19 LA p'rimes sigma(p')|720

{KOA P'RIMES} U {LA P'RIMES} = {KOALA PRIMES}

In such an imaginary group, assuming multiplicity, then the Monster Group would be a subset of this larger group and "Monster Moonshine" would become, just for the heck of it, let's call it "Koala Sunshine."

Koala's are a lot nicer than monsters, so I root for this imaginary group on humanistic grounds alone. :-)

- RF
jeremyebert
QUOTE (Raphie Frank+May 5 2012, 04:52 AM)

Just as an abstraction and flight of fancy, Jeremy, imagine there were such a thing as a...

KOLOLOKO GROUP
(Kolo means "wheel" or "bicycle" in Czech)

aka "Crazy Wheel"
aka "Mobius Strip"

... the p'rime divisors of which were all p in N such that phi(p') or sigma(q') divides 720. (p' or q' = n in N | d(n) < 3)

In such an imaginary group there would be 26 primes + 1, for a total of 27, decomposed into 16 KOA p'rimes phi(p')|720 and 19 LA p'rimes sigma(p')|720

{KOA P'RIMES} U {LA P'RIMES} = {KOALA PRIMES}

In such an imaginary group, assuming multiplicity, then the Monster Group would be a subset of this larger group and "Monster Moonshine" would become, just for the heck of it, let's call it "Koala Sunshine."

Koala's are a lot nicer than monsters, so I root for this imaginary group on humanistic grounds alone. :-)

- RF

ha! perfect.
Raphie Frank
QUOTE (Raphie Frank+May 3 2012, 09:47 PM)
P.S. p^2 - 2 is prime: First values = 2, 3, 5, 7, 13, the divisors of which are 1, 2, 3, 5, 7, 13. This is the complete set of {1 U primes} such that n^2 + 11 divides 360 (or 720).

Only just occurred to me, Jeremy...

Start from the divisors of 12 mod (12). Alternatively, the solutions to the Crystallographic Restriction Theorem for n > 0 (0 can be conceived of as the "Do-Nothing" State). Or, alternatively, the set of n in N such that totient(n) < 3
= 0, 1, 2, 3, 4, 6

A ) Insert as prime # indices calling 1 p'_0
= 1, 2, 3, 5, 7, 13

p'^2 - 2 is prime and p'^2 + 11 divides 360.

Not only is this a unique set (as well as the complete set of divisors of the unique prime divisors of 196560, the number of vertices of the Leech Lattice, as well as the divisors of the primes identified by Frampton & Kephart in 1999 as being associated with anomaly cancellations in 26 Dimensional String Theory...) but also, there can be no other primes such that p^2 + 11 is a multiple of 360. This follows since all p^2 - 1 are congruent to 0(mod 24) for p > 3.

B ) Insert 1, 2, 3, 5, 7, 13 as prime indices and now you get Euler's Lucky primes, which are famous for their role in prime generating polynomials [k^2 - k + p is prime to p-1]:
= 2, 3, 5, 11, 17, 41

Not only is this also a unique set, but it maps to 6 of the 9 Heegner numbers by the rule 4x - 1

Heegner Numbers:
1, 2, 3, 7, 11, 19, 43, 67, 163 = -d

For j denotes the j invariant, then j*(1 + sqrt (-d))/2 is an integer. See: http://en.wikipedia.org/wiki/Heegner_number

and...
j-invariant and Monster Moonshine
http://en.wikipedia.org/wiki/J-invariant#T...n_and_moonshine

C) Insert the divisors of 12 = 1, 2, 3, 4, 6, 12 as prime indices:
= 2, 3, 5, 7, 13, 37

Euler's Convenient primes. Integers not constructible in following manner: (a^1b) * (b^1c) * (c^1a). Change the 1's to 3's and what do you get?

Klein's Quartic Curve

- RF

P.S. There's a simple way to get a general grasp of the j-invariant without having to understand all the notation, which is pretty complicated. Look at the algebraic definition and you'll see how it all fits together. http://en.wikipedia.org/wiki/J-invariant#A...raic_definition

P.P.S. What did Frampton & Kephart analyze back in 1999 to come up with their observations? The Eulerian polynomials.
jeremyebert
Mekigal
you would like my paintings . I call them paint by numbers . They separate the sieve in an unbelievable orderly fashion . Yeah a layering that would blow your mind . One day . I am looking at em right now and they still blow my mind . Cris crossing by pairs .
-1,+1
5,7
11,13
17,19,

It is all about -1 and the +1. You could put it all into very specific angels by very specific layering one on top of each other out to ............... if you know what I mean . Each pairing would create its own Pythagorean triplet motion not on a curve but on a straight line . Yeah . Or you could do stair stepping of the sieve by the pairing also .

Course you do know that all primes fall in the six n line I started above by the pairings already I imagine
Mekigal
You know I really want people to succeed . I am so misunderstood in this thing we call existence. I want the human race to succeed . You peoples of the world stand in my way from the conception of my birth . Why is that ? Well they say " Becuae you Charly Brown Mikel .
O.K. tricky dickies ( Learn about watergate and you will know what that means in its fullness) Water Gate . That is funny now that I think about it . You know the day I quit collage cause the world saw me as carpenter only ( and not a human being , except for people like Bridget O"conner and her sisters. plus her Mom was a ******** st.

I am getting off track. Oh yeah it was the day Squeeky From pulled a gun on president Ford. 22 pistol with no rounds. It was not loaded . Crazy bitch . I think I was married to her sister once . Holy **** they may be fun , but they redefine trouble.

O.K. lets see if we can straighten out your brains so you can see there is no mystery to prime numbers . Alpha said it best . They are just a set inside of a set. They are multiples of 1 and the ones that remain prime are the ones not used up yet. They are targeted to be used up by multiplication. He didn't say it like that cause his command of language is weak. Course he is a mathematician and that is his native tongue so there you go.

Lets pick on the Muslims favorite slaughter food " The Lamb"

My new expression I woke up with do to my Vanity affair

Me Evil Jesus

Now you look at that and what do you think . You think " That crazy *** person thinks they are Jesus.
See that is not it : You thinking is haywired

Now think of Me as my personal identification . Personal name and Jesus as a broader aspect . A clone so to speak . You see what the difference is . All those Christains that run around the globe are what I call Jesus clones . They aspire to be la bunch of little jesuses . They do it too. It really that hard to be a jesus . It is depressing bout not that hard . You can see why people get anxiety from that behavior .

That is how you got to look at the prime number set . It is a reflection of what occurred. What occurred is in very specific repetition. It is the mirror reflection of that layering by pairing . Like a photograph and its negative. Lets say I want to make an impression of something so I could copy it . So I mold some clay over it .Now you have a reverse image. That is like the prime number set.
There is no mystery beyond that . The prime creates first but the future of the prime is destined to be broken by the multiples creeping up behind the primes .

The Future is dependent on the past and the past is dependent on the future .

Same old

The crystal stuff , Crystals in time ? You guys are way out there with that one . No wonder people run around rooting in the dirt looking for shiny rocks .

I would never call them time crystals. You need better terminology than that . It seems to leave a false impression . Time is like water friends . I do suppose you could figure away to freeze time though . Are the appearance of frozen time . Going the speed of light and all compared to the slow moving waves we live in

Mekigal
now for the rotation by its code . The pathways that repeat by numbers
The next set in the series of prime by pairing is what ?
23,25 except the past has caught up with 25 so it has been knocked out of existence of being prime time set. It is still a multiple of one though . Just not prime anymore. Its pairing is still prime so we find the non prime still useful by the pairing . So the roll over by coding is following the law of 9s . This is where on my painting they are represented by the same color as the first pairing that does not have a negative which is 5,7. So programers . ( as I believe only a young minded programer can understand what i am saying in its fullest conception cause old fooggys are rocks set in gridlocks of the mind ) you see how it is coded in the repeat. by the 5,7 . Now all this repeats by triplets and that is it too......................... if you know what I mean . We are done . That is it in a nut shell.

Some one with a brain needs to see my paintings . I am a rock in computer world and the people around me see no value in my paintings , but I am telling you right now the mystery is solved in my paintings . The grid is in perfect harmony . I call it the periodic table of numbers . I know there is someone out there that can interpret the data better than I can . I am just saying " I got a snap shot of the answer" No body seems to care enough to check it out . My sons math teacher in middle school was more bewildered than amazed . Mesmerized is more like it . Like it through his brain into a Disney land state. ( I gave him a simplified version of the paint series.

Arithmeticians don't look back is my conclusion. They go on to bigger better things so my paintings have the appearance of child's play . That is why no one takes them serious. They are serious and if some one with a brain looked a them they would see it . How serious they are. I would load them up for you to see but **** hole I am a old rock lost by displacement of the information age . True my son is a wizard programer changing the world by writing program for a nonprofit bringing internet stuff to Africa ( one world united by communication, Man that is powerful , You all so lucky to have such a good programer among you, He will blow your mind . I don't even compare to that boys imagination . 50 generation beyond me) He won't take the time . The demands put on him by the world supersede helping Me post a painting . You *******. That is fine no big deal . I know you all need him more than I do . Good thing he loves Me or I would really be out of his life all together .
Mekigal
Lets see if I can simplify . The growth pattern does not change . The grid just enlarges by specific growth patterns . Quantum jumps ? Step by step in congruent intervals . Predicted equal jumps of distance . The angel changes on the grid . But the growth is equivalent stacked like cards in a real specific patterning. Predictable patterning by stepwise growth . It could all be reduced do degrees of growth is the thing . By slope .

First we back drop which is vertical lines of 6
123456
789.....
...........

now the layering beyond that by pairing. Each layer lays on top of that back drop . The first set of 5,7 is at a 45% , But once you get to 35 in the interruption of the back drop field you are moving by 7 vertical to 1 horizontal. That set will continue out to ....................................... at 7 to 1 threw the back drop fields That would be the negative side of the set . That set moves backwards threw the field . The positive side of the pair still moves at 45 degrees. I don't know if that is the case always with each layering as the relationship may be different in the pairing but the point is the stepwise motion of 35 is moving threw the field in that fashion . Meaning every interruption of prime by 35 will and only follow that path . It will take five stepwise passes threw the field before it interrupts the prime number line by a 7 to 1 ratio.

Now lets take 143 cause that is where the pairing first starts . See it don't start with sq. Sq. is like a tail to the past so we take 143 and then the next negative line passing threw the back drop field is a 9 to 5 ratio as it passes threw the field .

Now I bet you dollars to donuts if you continue the field out further you will see subsequent sets of grow pattern in interruption by even more stepwise fashions. I need the third set of pairs to know the growth . But dollars to donuts you will have your growth pattern by the ratio of the third set and then all will follow that growth enlargement to .............. They will follow the ratio of growth . Could be wrong about that but I doubt that I am . The fields are to much in perfect harmony
Mekigal
does anybody understand that . You take the 35 for example . It is at a 7 to 1 ratio if you count 35 as one from 0 it will take five 7 to 1 ratios to interrupt the next prime in the back drop field . Lets check and see.
0,35
35,70
70,105
105,140
140,175

175 being the interruption of the prime number line field .

We are not finished though as the next interruption happens in 2 more intervals still moving by the 7 to 1 ratio .

so we got
175,210
210,245

245 being the next interruption of the prime number set. Then it all starts over again .

it will go 5 sets of a 7 to 1 ratio to the next prime interruption then 2 more 7 to 1 to interrupt the next prime . It is true in the simplest form they are all still a set of the originated 5,7 sets . Can you see how the 35 is more elevate jumps in the field . See the thing is you take any number to make bigger jumps by this law if you know the ratio of the jumps threw the fields then unravel backwards from there once you make the jumps . The growth should be predictable. The jump ratios that is . They have to follow the logic of the back drop fields of rotation

I might be off one in the whole concept . It might be 6 times . Maybe not . If I leave out the 0,35 and go with four 7 to 1 ratios then 2 7 to 1 ratios I would be back at the start I think . That is like , is there really 9 numbers or 10. Still have trouble with that one . I guess it depends on if 0 is a real number or not . Empty place holder more than likely and that is where the confusion comes in I believe
Mekigal
i made a mistake . it is 7 to 2 not 7 to 1 , one over on rows but if you count 7 down you got to count the 7 row as one and then the next row as 2 so the slope would be 7 to 2 . Seven vertical and 2 horizontal. My bad. What is that slope ? what is the degrees . See on the cart each set has its own slope . I was not looking at quite right before . 5,7 is a 45% . I am to lazy to do the conversion . To distracting because of all the other thoughts I carry in my head . Like Mary trying to control my guitar playing . I am crying with joy this morning cause I got to be me last night at our gig at the loft down town Missoula . I got my chance to show that I am a great jazz musician . You should of seen the bar tender . Blew his mind no doubt . His knees were wobbling. So go So go . World I am a guitar player .

O.K. O.K. don't get all pissy, back to the subject

All factoring by the chart can be reduced to a slope . A predictable slope

First set of 5,7 <2 to 2
2nd set of 11,13 < 3 to 2
3rd set of 17,19 < 4 to 2
4th set of 23,25< 5 to 2
5th set of 29,31,< 6 to 2
6th set of 35,37 < 7 to 2
7th set of 41,43< 8 to 2
8th set of 47,49 < 9 to 2

and so on out to .................................................... like stair stepping.

Now you might be able to see how the Pythagorean theorem can play into the equation. Except you might need a little trig to get the slope . A little bit of tangent work and boom you got the flow chart by degrees .

Now you got to keep in mind if you just shoot a straight line instead of crisscrossing the events the 5 will only reflect 5s or what call the red side of 5,11,17,23 which is the negative one in the series that contain half of the multiples of one in a 6n line prime number set . Now if you notice by the law of 9 all of the negative set are the Tesla 5,2,8 and all the positive are 7,4,1. So you that don't know prime numbers by the law of nine can only be of those forms and they repeat over and over again by that patten in the 6n line of reason . So now you got another tool to help you understand . We are getting into new oratory for me so feel free to interject .

Lets see if patterning hold up in any form by isolating by laws of nine .
Lets take all the 5s of the negative line of 5,11,17,23,29.........................
all the 5s would be 5 23 41............................................Lets lot the ratios now

2 to 2
5 to 2
8 to 2

Lets see what happens when we take all the 2s
11, 29,47

3 to 2
6 to 2
9 to 2

Now for the last set
of 17,35,53

4 to 2
7 to 2
10to 2 and if you reduce the 10 you can see it is a one 1+0=1
So all nine numbers are accounted for on the next level growth . All contained by slopes of growth due to the 6n numbering chart . The chart shows it all simple like for your feeble minds to grasp . Periodic number chart of prime patterning pops out like a sore thumb . Anyone who can't see that is a bonehead . You don't need to go all curvy down the road like the more worldly charts all flipping around the back and all that turning into a spherical night mare . Slope works fine and you can see by the slope how it never intersects. Like a river . That point of reaching the entire whole of infinity . Why it can't be done . Your always behind the 8 ball so to speak

you see the growth patten in that . straight forward i would say
Mekigal
What the chart does is it isolates all factors in perspective rows .

What you do if you want to try it at home is map out all whole numbers by repeat in columns moving by 6 . I call it the 666 cause thats just the way my evil mind works .

you can go negative too to get the full grasp of the hour glass effet

Like this . This is my favorite sampling of order in whole number sets

-7-6-5-4-3-2-1 0 1 2 3 4 5 6 7
-1 0 1 2................................
............................................
............................................
...........................
...............
..........

now the back drop is all multiples of 1,2,3and last but not least 6s

all vertical lines. All vertical lines contain all multiples of 1s,2s and 3s then the 6s are just the conclusion of 2s and 3s mixture. That is the back drop . You can see why too. I make the two vertical line green my self as i looked for complimentary color coding to make it all easy as possible to look at . My wife the color guru help me with that. Then I make the 4 line green to . It probably should be a little different shade but I didn't see any need to seeing how it is the sq. of 2 . Then I make the 3 vertical line yellow . Then the 6 vertical line yellow green just to show it is a mixture of green and yellow . The prime lines I make white with black numbering so they pop out of the chart better . Now it is all about red and blue . I screwed up on my best painting being all dyslexic and all . It shows the first 2 crisscrosses . 2 sets the pairing of 5,7 and 11,13 and well i should probably made the 11 red and the 13 blue but I didn't . I switched them around and there is no way I am going to go back and change it . That paintings done. My other painting that shows no crisscross I got right with the color coding where as all the negatives falling to the left as you face the canvas and all the positives falling to the right as you look at the canvas. That one shows how it continues in the pattern to infinity . It shows the repeat in reduction value and repeats the triplet patterning .

Give it a try . It will end the idea that prime numbers are random for you and you can cross that off your list of new things to discover. Watch out cause when you start seeing the depth it will blow you mind . You taste for the infinite will be scratched real good
...
note: the 6 line is just a repeat of the 0 line so it is not really an introduction to a new line like 1s 2s and 3s . It is the 0 line which acts like a trunk of a tree in the painting that shows negatives dropping to the left and the positives dropping to the right as you face the canvas. It is all about point of reference. Think of the 0 line as point of reference of 0s . Like 6 is 0 between 5 and 7 . Can you see how that work ? It is pretty crazy if you ask me . I mean it is logical that numbering would be that way by the definition of -1+1=0 but to see if in physical space like that with you brain comprehending that vastness of it is a whole different thing than just conceptualizing it .

So lets take 6 for example . You got 5 and seven on each side . Past and future or Negative and positive . The 6 will always be half of the sum of the equal distance out from 6 . 5+7=12, 4+8=12, 9+3=12 You see that ? That is with all numbering . You can see why the 0 line is like a tree trunk in that respect . Although it works with any number system by simple math , but it don't pop out the one lines that contain all the primes straddled on either side of the 0 line like this chart shows
Mekigal
cross roads is a better terminology than crystal. The places where resonance occurrences happen. They can stack up big . Like ideas of perfect storms ( George Klunny you the man ) Good movie aye?
Raphie Frank
QUOTE (Mekigal+May 25 2012, 05:45 PM)
cross roads is a better terminology than crystal. The places where resonance occurrences happen.

You'll have to take that up with Nobel Laureate Fred Wilczek, Mekigal, not me and Jeremy. He came up with the term "time crystals." Funny, though that Jeremy and I were discussing the idea of perpetual motion in relation to Kagome Lattices just this past December.
Mekigal
QUOTE (Raphie Frank+May 30 2012, 12:33 AM)
You'll have to take that up with Nobel Laureate Fred Wilczek, Mekigal, not me and Jeremy. He came up with the term "time crystals." Funny, though that Jeremy and I were discussing the idea of perpetual motion in relation to Kagome Lattices just this past December.

Well that is a lot like the number chart I am talking about . The cross of under over webbing by every other . You take the grouping of:
14,15
20,21 and then the resonance of 35 . Those are web crossings in the matrix . 7s crossing 5s . If you think of of 35 being a resonance form of 5,7 factors . 35,70,105.....
and the no resonance crossings as the under version of web . You can see that in the number chart real easy . Still cross roads like where every other resonates as a whole number .

Did you try the lattice I suggested and if so did you see the web . How squared primes can only fall in the positive one line moving by six . Example
1
7
13
19
25
31
37
43
49

All sq. primes are in that line and that line only . You can see the start in the sampling by the 25 and the 49 being the first sq. of the matrix . You will also see the web move as it flows down the chart horizontally. Why it pierces the prime line where it does . It is like an optical motion in its self as resonance drifts threw the vertical lines as you move down the chart .

O.K. Fred Wikczek ? I will check him out . Is he a crank ?
Raphie Frank
QUOTE (Mekigal+May 30 2012, 09:48 PM)
O.K.  Fred Wikczek ? I will check him out . Is he a crank ?

Actually, it's Frank, not Fred. You can decide for yourself if he's a "crank."

FRANK WILCZEK
Herman Feshbach Professor of Physics
2004 Nobel Laureate
(at MIT)
http://web.mit.edu/physics/people/faculty/wilczek_frank.html
Mekigal
QUOTE (Raphie Frank+May 31 2012, 02:16 AM)
Actually, it's Frank, not Fred. You can decide for yourself if he's a "crank."

FRANK WILCZEK
Herman Feshbach Professor of Physics
2004 Nobel Laureate
(at MIT)
http://web.mit.edu/physics/people/faculty/wilczek_frank.html

dynamical degrees of freedom is a mouthful . He seems to have a handle on something with quotes like that . That makes me think of my whole number chart organizing by the flip flop of whole numbers . Limited degrees of freedom sound more realistic

Consider this . The chart works in negative by mirror imagery so in the negative the -1 becomes -7 -13 -19 . That may not seem like much but the crisscross lines up perfect with the positive relative. The seven positive line of 7,14,21.................. follows right in its perfect harmony with out skipping a beat . It creates the hour glass effect of the prime number groupings . It looks funny in a painting I never finished for if you turn the painting up side down it looks identical to the positive side and with out designation of its negative aspect by putting a - sign you can not tell the difference when you flip the painting . Course I started from zero in the middle of the canvas and only used 7 lines as all is repeat after that in reality . Copies of the same thing . I should finish that painting and see what else emerges . So the numbering is up side down from each other. The - verses the positive in the painting not finished and that is why the mirror effect happens like it does in that painting .

I want to look at that book of harmony him and his wife wrote . It was called beautiful written . That intrigued Me .

Later . My wife is after Me to go to the store . I like the guy and his time in California more than likely makes him naturally cool
Raphie Frank
QUOTE (Mekigal+May 30 2012, 09:48 PM)
20,21 and then the resonance of 35 . Those are web crossings in the matrix . 7s crossing 5s . If you think of of 35 being a resonance form of 5,7 factors . 35,70,105.....

Not sure how this figures in to your thinking Mekigal, but the integers modulo 7 do in fact seem to play a special role in theoretical physics...

================================================
117) John Baez, Klein's quartic curve, http://math.ucr.edu/home/baez/klein.html

You may think I'm digressing, but the relation between Klein's quartic curve and the Fano plane underlies what I want to talk about today. Greg Egan and I realized that this relation is just part of a bigger picture involving special relativity in 3-dimensional spacetime... over the integers mod 7.
http://math.ucr.edu/home/baez/week219.html
================================================

35, by the way, is the last integer congruent to 5 (mod 10) that is semiprime, 1 less than a square and 2 less than a prime, 37, which is the last of Euler's Convenient primes.

Heuristically speaking, it so happens that (without respect to units) 4*pi^2/sqrt 35 *(10^-pi(35)) = 6.67307... * 10^-11 is also a good way to remember the, more or less, average measurement of the Gravitational Constant going back to 1929 or so. Perhaps why Google chooses this value, more or less, over the current 2010 Co-Data Value for it which is 6.67384 * 10^-11 (if memory serves me correctly)

- RF
Raphie Frank
Jeremy, this is absolutely unbelievable...

p_(1, 2, 3, 4, 6) = 2, 3, 5, 7 13
Now, think about summing the primes after those primes...

Y =
3 = +(3) START: p_(1 + 1) TERMS: (2*1) - 1
- 12 = - (5 + 7) START: p_(2 + 1) TERMS: (2*2) - 2
31 = + (7 + 11 + 13) START: p_(3 + 1) TERMS: (2*3) - 3
- 60 = - (11 + 13 + 17 + 19) START: p_(4 + 1) TERMS: (2*4) - 4
- 88 = - (17 + 19 + 23 + 29) START: p_(6 + 1) TERMS: (2*5) - 6

If odd, then +. If even, then -

Now insert those terms for the Y variable into the following equation we discussed back in November...

sqrt ((4*pi^2*(9.10938291*10^-31)*sqrt (67092478 - Y))/(67092478 + Y)/10^34)

And compare to the values here:
http://en.wikipedia.org/wiki/Planck_constant

Essentially a 100% match. With the added bonus...
3, 13, 31, 61, 89, the 2nd, 5th, 8th, 9th and 10th Mersenne Primes Exponents.

- RF
Mekigal
QUOTE (Raphie Frank+Jun 4 2012, 01:10 AM)
Not sure how this figures in to your thinking Mekigal, but the integers modulo 7 do in fact seem to play a special role in theoretical physics...

================================================
117) John Baez, Klein's quartic curve, http://math.ucr.edu/home/baez/klein.html

You may think I'm digressing, but the relation between Klein's quartic curve and the Fano plane underlies what I want to talk about today. Greg Egan and I realized that this relation is just part of a bigger picture involving special relativity in 3-dimensional spacetime... over the integers mod 7.
http://math.ucr.edu/home/baez/week219.html
================================================

35, by the way, is the last integer congruent to 5 (mod 10) that is semiprime, 1 less than a square and 2 less than a prime, 37, which is the last of Euler's Convenient primes.

Heuristically speaking, it so happens that (without respect to units) 4*pi^2/sqrt 35 *(10^-pi(35)) = 6.67307... * 10^-11 is also a good way to remember the, more or less, average measurement of the Gravitational Constant going back to 1929 or so. Perhaps why Google chooses this value, more or less, over the current 2010 Co-Data Value for it which is 6.67384 * 10^-11 (if memory serves me correctly)

- RF

all forms of 8 go as a result of 5,7 harmonies count backwards . This is another tool in the cabinet . In the pairing also .
Series : 35,70,105,140,175.......................................=8,7,6,5,4............................

so all pairings in the set that contains all primes can be reduced to 8 series line of thinking.
5x7
11x13= 35,70,105........
17x19=323,646,969..............
23x25=575.....................
29x31=.......................
.........
.........

see how they all move by 8,7,6,5,4,3,2,1,
You can see why too

11x13 is equivalent to 2x4
17x19=8x1
23x25=5x7
29x31=2x4
then all is repeat to ............................................
852s x 174s all =8s
It is easy to get the code . That is a bunch more painting charts specific to each factoring . What it means to be 2s

12
34
56
78
that is where it all starts . See the separation by sets . This is the start of even odd. Now 11 will follow the same code identical code, so will all reduction numerology style stuff ( not to be confused terminology as with mysticism) Meaning if you reduce all elevens by adding up each number value you will see 11 follows the same path way as 2s by code.

11=2
22=4
33=6
44=8
55=1
66=3
77 =5
88=7
you see that ?
123
456
789
......
......
......

now there is a mix for you to think about. That is the key to the prime set. Now flip flopping occurrences with each increase . I forget but I think it goes like this
3,6,12,24
Those will be all the sets that will reflect the 147,258,369 maps originally created by division of 3 groups . They will all follow the same coding

Each set has there own code . Meaning all numbers in the set change to suit the set .

Like the 11 series does the 2 series . More than that .it is like all numbers to the set in the same direction another words the code in secession is no different when following lets say the 5 pattern of 5 series
5,10,15,20,25..............
the code is no different for 1,6,11,16,21,26,31,.....................
They are grouped in tandem. All numbers in the pattern

So you can get the code easy enough by a 5 series like this
12345
6789..
........
.....
they will all be 5,1,6,2,7,3,8,4,9,5
that is the code in tandem. Any combo of 14,23,32,41 will go strictly by the law of 5s or code of 5

The 9 is also the key as it makes all numbering in unison ( I don't know another terminology to describe it)
Unison as I call it happens at 9,18,27,....................... flip flopping like rotation
AKAIS@cox.net
Could you send me a link to your equation. I would love to check it out. Please LMK Thanks, Andrew 1 401 709 2357 AKAIS@cox.ne
Mekigal
Well that John Baez sure hit on the lee lines pretty heavy . Fans what Have I been saying ? Aye Lie lines in string theory O.M.G. and the Persian creation of Algebra
Surprise ! Surprise! Blues power. Albert King ! Quote ! Blues Power!
Unbelievable is right
Mekigal
QUOTE (jeremyebert+Jun 13 2011, 09:35 PM)
Thanks Raphie. Good to hear from you. Another connection I found when I was looking for the area of the 2 (˝ vertex) parabolas

(n-1)/2 = h (triangle height) (x)
2*sqrt(n)= b (triangle base) (+-iy)
(b*h)/2 = a (triangle area)

2*(a^2) = Pentagonal pyramidal number

if (a*(4/3))^2 is an integer then n is a number having a digital root of 1, 4, 7 or 9.

1, 4, 7, 9, 10, 13, 16, 18, 19, 22, 25, 27, 28, 31, 34, 36, 37, 40, 43, 45, 46, 49, 52, 54......

I've never really had much use for digital roots though they are intriguing.

I’m in the process of redoing that pdf with as much detail as I can muster.

is that the terminology "digital root "

O.K. I am trying to communicate
digital roots is where it is at , partially anyway . More cross-reference than anything as far as I can tell. Maybe more than that though . I suspect more .

Lets say you reduce all multiples to there digital root . The funny thing is the sets of multiples . I have thrown away most of my stuff on this , but lets see if I can recreate a set .

No No lets see if I can explain it with the 3 chart

123
456
789
.....
......
......
......

o.k. EVERY THING IS CHANNELED IN THE 3 lines . All whole numbers to infinity . Now they all follow the basic rules of addition . First line a 2nd line b and 3rd line c

First line is 147 by digital roots 2nd. line is all 258 by digital roots . #rd line 369

a+b=c always . a+ b can only be in the c line . You can see by the sampling 1+2=3 and 4+5=9. Now you can get an idea of the skip pattern . It goes deeper than that . That is a scratch on the surface . Now if you take the code of multiples it all holds true to the code .

It is like 2 functions and it pops out a new equation in the perfect match . It may seem small , but its not . If you consider supplementary angles a small thing that you take for granted then I can see how you would think it is a small thing .

. I didn't make my self clear yesterday about the 3,6,3,6
The mixing as the sets get broader
Lets take 12 for example
that will be a further subdivision of the 3 and the 6 . I think it creates 4 prime lines ?Instead of the 2 lines in the 6 = 123456.
Might be wrong as it has been a while sense I thought about it
123456789101112

so
1
13
25
37 yeah prime line of 147

5
17
29
41 yeah 582

7
19
31
43 yeah a double line or more subdivision of the 147

and now for the 11
11
23
35 yeah the 258 again
so now we have I think what you call the quads . Instead of pairs we have the quads .
The other thing this set does is it stacks the multiples of 5 in the number chart you can see that in the sampling how the 5s line up .
example .
43
53
73
83 now compare it to

55
65
85
95 all good and the same . When does that occur again I don't know cause the next set is fragmented by 7x19 and 11x13 so we get
103
113
133
143 compared to all fives of

115
125
145
165

if you follow this line of thought there is bigger imprint . I look at it as cards . The repeat of the cards as they grow .

It is what I call the 35th card where as the first card ends on 89 2nd card ends on 99 third card ends on 109 that is horizontal out to seven rows
Then vertical it goes
89, 179,269,359 out to 5 cards vertical . now you move out to the card that starts with 421 and ends on 509. That 509 card is the same card as the card that starts with 1 and ends in 89 . The repeat pattern if you only had 5x7 to consider .
So you can see it too. The relationship of theses 2 sets
421
431
441

is the same as
1
11
21
even so much in this first bigger card that 1 and 11 are prime and 421 and 431 are still prime not to mention that 441 and 21 are only multiples of 3 and 7 at there core

That is where the card repeats by this number pattern
Mekigal
so now you can feel for the continued subdivision of the prime line that ocure at the 3,6,3,6,pattern or at 3,6,12,24,30,42

so lets look at it
41
83
125
167
209
now see how it is still a 528 pattern but it is flip flopped where as in the 3 line
123
456
789 the pattern is 147 then the 6 =123456
it is 174 then the
12=123456789101112
is 714
and 24 is 174
and 42 is 174

i would think this would be the turn around at 60 . Never been there before .
nope just checked it it is also 174 . 120 would be a 147 .

There must be a pattern and I am not seeing it yet . Anyway you can see it has to end in a even number to subdivide the prime number lines and it needs to add up to a 3 or a 6 also by digit reduction to the digital root
Mekigal
Note : I realize I didn't depict the quads like 11,13,17,19 and 191,193,197,199
the 5s are a cross section to those quads vertical to the horizontal quad motion

11 13 15 17 19 and the 15 is vertical , perpendicular to that quad except in this case 0 would be the start of the cross 0,5,15,25,35

then for 191, 193 195 197 199 the comparable cross section would be 175,185,195,205 and 215. I suspect for all quads it would hold true . I don't know that for sure . I bet a dollar it does
Mekigal
you can see what happens when you convert the prime lines to movement by 12 the center line changes to 3,15,27,39 becomes the center line instead of 6,12,18 and it creates a knot ( lack of better description in terms) that moves in bigger jumps . So the more you subdivide the line the more the center line becomes manageable. I don't know that for sure but in the case of 6 to 12 it is true and the only barrier between the line is 3,15,27.................................. And now you have quads . where as with the 6 you have pairings

Anybody got anything to say . This is typically were I get ridiculed by a mathematician that knows better than me and I take my toys and go play some where else or get banned if i don't . Go ahead I can take it . I actually like it when someone shows me the fallacies of my math, be it in there head or mine . Sock it to me
Mekigal
nope i was wrong about all that . There are 2 barriers , the 101,103 is split by the second barrier of 105 then the rest of the set 107,109 . So that makes sense 4 prime lines and 2 barrier lines to maintain the 1 to 2 relationship , 4 to 2 . I imagine as the line subdivides so do the barriers as to maintain the 1 to 2 ratio . So it is suspect that it would grow to 8 to 4 . I don't know that for sure . That be a guess , but I would bet it would be reveling to run it out and see

It does appear to have a 5 in the middle in every case , but I still don't know that for sure
Mekigal
O.K. you might think that the group 11,13,17,19 might be the same as 101,103,107,109 but to me this grouping is just as accurate because of its relationship with the 5x7 grouping
210 would be the same as 0 to
5,7
11,13
17,19
as 210 is to
215,217
221,223
227,229

I know the number of primes are different but the relationship to the 5,7 crossing is the same where 5 is equivalent to the 215 and the 217 both being divisible by nothing less than 5,7 . I don't know what you call that 43x5 and 31x7 course you got the 221 messing with your brain being introduced to the line by the 13x17 but still that is the repeat . 210 is the repeat marker of the grouping . So all 210 from that point will be the same in that respect in the 6=123456 number chart . So it should occur at the same formation at 420 if my math is right . Don't know if or how many primes there are at that point but the 420 holds true for penetration . We can assume that at 420 that 425 and 427 will be 5s and 427 will be 7s. Yep as we can see 5x85 is and7x61

It is the same crossing , but a little more complex with 85 bottoming out at 17 as the prime creator 17x5x5. Now that makes me wonder what 630 looks like ? Don't know got to go

Warning ::: Don't get lost . It is easy to do . Get lost in your own mind that is . Thank your lucky stars you don't have the paintings . Then you could end up where I am . Got to go . Gonna be late
AKAIS@cox.net
QUOTE (Mekigal+Jun 9 2012, 01:22 AM)
O.K. you might think that the group 11,13,17,19 might be the same as 101,103,107,109 but to me this grouping is just as accurate because of its relationship with the 5x7 grouping
210 would be the same as 0 to
5,7
11,13
17,19
as 210 is to
215,217
221,223
227,229

I know the number of primes are different but the relationship to the 5,7 crossing is the same where 5 is equivalent to the 215 and the 217 both being divisible by nothing less than 5,7 . I don't know what you call that 43x5 and 31x7 course you got the 221 messing with your brain being introduced to the line by the 13x17 but still that is the repeat . 210 is the repeat marker of the grouping . So all 210 from that point will be the same in that respect in the 6=123456 number chart . So it should occur at the same formation at 420 if my math is right . Don't know if or how many primes there are at that point but the 420 holds true for penetration . We can assume that at 420 that 425 and 427 will be 5s and 427 will be 7s. Yep as we can see 5x85 is and7x61

It is the same crossing , but a little more complex with 85 bottoming out at 17 as the prime creator 17x5x5. Now that makes me wonder what 630 looks like ? Don't know got to go

Warning ::: Don't get lost . It is easy to do . Get lost in your own mind that is . Thank your lucky stars you don't have the paintings . Then you could end up where I am . Got to go . Gonna be late

Hey there, You responded to my posts aswell. This is exactly what I am talking about. If one considers the spiral p^3, I am using p but I think there should be anouther varialble because it might be a differnt axis, then it expands in a 3^(1/2) and 2^(1/2) kind of way and I would think that the realtions in numbers you have here can be found on that spiral if one looks at it correctly, not that I am saying I know what that it is, but sort of like at oneself with a mirror., Change in angle, chang in x, change in y, change in, z, corresponds to the distance from origin to center of sphere at a particular rate..... Am I mkaing sense? I can see it. I think you need to look at Mayan calander and your numbers will make more sens, I know its silly but 12, 21st theere is a alignment with univers that only happens in 10,000 years almost on the clock. I think you are on the clock if you know what I mean. Andrew 1 401 709 2357
Raphie Frank
Please think about starting another thread to continue this exploration. This is Jeremy Ebert's thread (and to a lesser extent mine since I've been collaborating with / guiding Jeremy to the best of my abilities.). And this is all going far, far off-topic. Don't get me wrong. I wouldn't mind continuing to discuss your ideas Mekigal, but not here on this thread.

Raphie

P.S. Mekigal, You might want to look in to Cunningham chains. 89, 179,269,359 are all prime elements of a Cunningham chain of the first kind (2p + 1 is prime aka "Sophie Germain primes"). You could continue the chain: 719, 1439 and 2879 are also prime numbers.

The longest known Cunningham chains have 17 elements.
Raphie Frank
QUOTE (Raphie Frank+Jun 9 2012, 04:13 AM)
P.S. Mekigal, You might want to look in to Cunningham chains. 89, 179,269,359 are all prime elements of a Cunningham chain of the first kind (2p + 1 is prime aka "Sophie Germain primes"). You could continue the chain: 719, 1439 and 2879 are also prime numbers.

The longest known Cunningham chains have 17 elements.

ERROR CORRECTION: 269 does not belong in that chain.

89, 179, 359, 719, 1439, 2879...

... are all prime and follow the rule 2p+1. START: 89

- RF

See: A075712: Rearrangement of primes into Germain groups.
http://oeis.org/A075712
Raphie Frank
Here's the expansion, Jeremy, for the geometrically based model for various determinations of Planck's Constant I mentioned a few days back (placed in to a discrete formula). When I stated previously, "essentially a 100% match", I wasn't suggesting that in a loose way...

0 < n < 6
(n + floor [n/5]) = 1, 2, 3, 4, 6
aka "The proper divisors of 12"
aka "Integers with a totient of 1 or 2"
aka "Allowable n-fold rotational symmetries of a crystal by the Crystallographic Restriction Theorem"

((2(n - 1))^2 - 1) = - 1, 3, 15, 35, 63
((2n + 1) - (n + floor [n/5]))^2 = 4, 9, 16, 25, 25

(((2(n - 1))^2 - 1) + ((2n + 1) - (n + floor [n/5]))^2)*((-1)^(n + floor [n/5]))
which reduces to...
(((2n - 2)^2 - 1) + (n + 1 - floor [n/5])^2)*((-1)^(n + floor [n/5]))
= - 3, 12, -31, 60, 88

3 = 3
12 = 5 + 7
31 = 7 + 11 + 13
60 = 11 + 13 + 17 + 19
88 = 17 + 19 + 23 + 29

Without respect to units (which are Joule-Seconds)...

sqrt (((4*pi^2*(9.10938291*10^-31))/(2*10^34))*((2*sqrt (67092478 - 3))/(67092478 + 3)))
= 6.62606779 * 10^-34
vs. 6.6260678 * 10^-34 --> Josephson Constant Determination of Planck's Constant

sqrt (((4*pi^2*(9.10938291*10^-31))/(2*10^34))*((2*sqrt (67092478 + 12))/(67092478 - 12)))
= 6.626068899 * 10^-34
vs. 6.62606889 * 10^-34 --> Watt Balance Determination of Planck's Constant

sqrt (((4*pi^2*(9.10938291*10^-31))/(2*10^34))*((2*sqrt (67092478 - 31))/(67092478 + 31)))
= 6.626065710 * 10^-34
vs. 6.6260657 * 10^-34 --> Faraday Constant Determinationion of Planck's Constant

sqrt (((4*pi^2*(9.10938291*10^-31))/(2*10^34))*((2*sqrt (67092478 + 60))/(67092478 - 60)))
= 6.62607245 * 10^-34
vs. 6.6260724 * 10^-34 --> Magnetic Resonance Determination of Planck's Constant

sqrt (((4*pi^2*(9.10938291*10^-31))/(2*10^34))*((2*sqrt (67092478 + 88))/(67092478 - 88)))
= 6.62607453 * 10^-34
vs. 6.6260745 * 10^-34 --> x-Ray Crystal Density Determination of Planck's Constant
See: http://en.wikipedia.org/wiki/Planck_constant

9.10938291(40)*10^-31 kg = Mass of an Electron (CODATA 2010)
2* 10^34 kg = Estimated Mass of a Giant Molecular Cloud

Quite possibly related...
Effective Mass
http://en.wikipedia.org/wiki/Effective_mas...-state_physics)

Induced Gravity
http://en.wikipedia.org/wiki/Induced_gravity
Induced gravity (or emergent gravity) is an idea in quantum gravity that space-time background emerges as a mean field approximation of underlying microscopic degrees of freedom, similar to the fluid mechanics approximation of Bose–Einstein condensates. The concept was originally proposed by Andrei Sakharov in 1967

And almost certainly related...
Crystallographic Restriction Theorem
http://en.wikipedia.org/wiki/Crystallograp...riction_theorem

What this predicts in principle: Planck's Constant, the "Fundamental" quantum of action, is itself composed of quantized energy states, and those quantized states have something to do with the heterotic interaction between the n-fold periodic rotational symmetries of a crystal and the prime number distribution, not to not mention sphere packing...

1*2 = 2 = K_1
2*3 = 6 = K_2
3*4 = 12 = K_3
4*6 = 24 = K_4
(K_1 + 6)*5 = 40 = K_5
(K_2 + 6)*6 = 72 = K_6
(K_3 + 6)*7 = 126 = K_7
(K_4 + 6)*8 = 240 = K_8

A002336 Maximal kissing number of n-dimensional laminated lattice.
http://oeis.org/A002336

... and Heegner Numbers.

4*p_(p_1) - 1 = 11
4*p_(p_2) - 1 = 19
4*p_(p_3) - 1 = 43
4*p_(p_4) - 1 = 67
4*p_(p_6) - 1 = 163

A003173 Heegner numbers: imaginary quadratic fields with unique factorization (or class number 1).
http://oeis.org/A003173

Meaning it all also has something to do with KLEIN'S j-INVARIANT
http://en.wikipedia.org/wiki/J-invariant

- RF

QUOTE (Raphie Frank+May 16 2012, 11:14 PM)
Only just occurred to me, Jeremy...

Start from the divisors of 12 mod (12). Alternatively, the solutions to the Crystallographic Restriction Theorem for n > 0 (0 can be conceived of as the "Do-Nothing" State). Or, alternatively, the set of n in N such that totient(n) < 3
= 0, 1, 2, 3, 4, 6

A ) Insert as prime # indices calling 1 p'_0
= 1, 2, 3, 5, 7, 13

p'^2 - 2  is prime and p'^2 + 11 divides 360.

Not only is this a unique set (as well as the complete set of divisors of the unique prime divisors of 196560, the number of vertices of the Leech Lattice, as well as the divisors of the primes identified by Frampton & Kephart in 1999 as being associated with anomaly cancellations in 26 Dimensional String Theory...) but also, there can be no other primes such that p^2 + 11 is a multiple of 360. This follows since all p^2 - 1 are congruent to 0(mod 24) for p > 3.

B ) Insert 1, 2, 3, 5, 7, 13 as prime indices and now you get Euler's Lucky primes, which are famous for their role in prime generating polynomials [k^2 - k + p is prime to p-1]:
= 2, 3, 5, 11, 17, 41

Not only is this also a unique set, but it maps to 6 of the 9 Heegner numbers by the rule 4x - 1

Heegner Numbers:
1, 2, 3, 7, 11, 19, 43, 67, 163 = -d

For j denotes the j invariant, then j*(1 + sqrt (-d))/2 is an integer. See: http://en.wikipedia.org/wiki/Heegner_number

and...
j-invariant and Monster Moonshine
http://en.wikipedia.org/wiki/J-invariant#T...n_and_moonshine

C) Insert the divisors of 12 = 1, 2, 3, 4, 6, 12 as prime indices:
= 2, 3, 5, 7, 13, 37

Euler's Convenient primes. Integers not constructible in following manner: (a^1b) * (b^1c) * (c^1a). Change the 1's to 3's and what do you get?

Klein's Quartic Curve

- RF

P.S. There's a simple way to get a general grasp of the j-invariant without having to understand all the notation, which is pretty complicated. Look at the algebraic definition and you'll see how it all fits together. http://en.wikipedia.org/wiki/J-invariant#A...raic_definition

P.P.S. What did Frampton & Kephart analyze back in 1999 to come up with their observations? The Eulerian polynomials.
Mekigal
QUOTE (Raphie Frank+Jun 10 2012, 08:27 AM)
ERROR CORRECTION: 269 does not belong in that chain.

89, 179, 359, 719, 1439, 2879...

... are all prime and follow the rule 2p+1. START: 89

- RF

See: A075712: Rearrangement of primes into Germain groups.
http://oeis.org/A075712

Good I got that . Thanks ! That don't reveal any pattern . It don't show the rotation . It is more of an anomaly of the rotation . After the fact . Thanks my man . You get a golden star . I got that finally . What I didn't under stand was what the 2 meant . Dyslexia is a bitch . So Yeah I get it now . That will change every thing in my thinking . Double gold star for your assistance my man. So to be clear (89x2+1=179 and (179x2+1=359

Mekigal
That is not bilateral thinking . What I am talking about is jamming 3/4 into 4/4 time . They have the same metered beat also . So how many 3/4is until they match up with 4/4 if they are metered the same . The spot where they hit the same down beat over and over again . That would be it ion a much bigger scale . You have to think 4/4 and 3/4 at the same time . Mostly we learn it as one movement . Yet it is not . Like banging on a gong . You hear a predominate note but in reality it is a cluster of notes .

I was intrigued by your guys thread because of the frequency diagrams you showed at the beginning .
I saw that one that went
1/8 2/7 3/6 5/4 6/3 7/2 8/1

So just saying . That is the key to music and the language we use to create harmony ( resonance
We take the 1 3 5 to make the major chord . Those notes do special things together . They will vibrate each other naturally. Resonance . Same thing that causes bridges and sky scrapers to potentially fall . Long term effects it makes every thing fail . Just like cracks in your house.
Decks are notorious when in heavy use to fall off a structure it is attached to.
So those are inverted harmonies you got there in that equation .
The division of time has been my brass ring . Born with no rhythm do to listen to Debussy . The king of no metering by constant measures. So yeah like where is the down beat . Well I figured it out when I turned eighteen . Oh yeah 1 2 3 4 there it is and 1&2&3&4& is half then there first Mo Fos . Well on my way . It starts to get complicated after that and by the time you throw 3/4 and 4/4 on top of each other
not to mention cut time over the both of them . Well get the picture
Mekigal
QUOTE (AKAIS@cox.net+Jun 9 2012, 02:53 AM)
I think you are on the clock if you know what I mean. Andrew 1 401 709 2357

Yeah I been trying to get a witness . Did you see my miracles ? Unbelievable. I still can't believe it my self . It is not Me . It is the Dog or the bitch dog lizard woman . She loves Me for some reason . She let me know when I turned 21 . Almost drove me over the edge then. After that she kind of left me to my own doings except by making me fail if i didn't comply . Like kicking a wall. Now she is making up to me for all her meanness .
She says it is for my own good and still teases the hell out of me in very vindictive ways . Anything for a laugh kind of thing . Oh look at Mikey squirm . But know I am way past the squeamishness. Yeah now it is like " Can you believe that ? Unbelievable ? Some one else needs to see this . That whole Japan thing had to take the cake all though it was not until after I did the talk with music that she said I did enough and she would let me have more freedom to think for my self . What ever that means cause I am still following the script so I am thinking that too was a tease . All like the carrot and horse
Raphie Frank
QUOTE (Mekigal+Jun 15 2012, 12:26 AM)
Did you see my miracles ? Unbelievable. I still can't believe it my self . It is not Me . It is the Dog or the bitch dog lizard woman . She loves Me for some reason .

Mekigal, if the "bitch dog lizard woman" loved me too, would you still think yourself so special?

Metaphorically speaking, "She" loves us all.

Get off your F***in' high horse, okay?
Mekigal
QUOTE (Raphie Frank+Jun 15 2012, 08:23 AM)
Mekigal, if the "bitch dog lizard woman" loved me too, would you still think yourself so special?

Metaphorically speaking, "She" loves us all.

Get off your F***in' high horse, okay?

can't . Wish I could . Not sure if she loves you . You got cancer ? She kills you with cancer if she don't love you . Not to say all people with cancer are not loved . Good chance if you are a healthy individual she loves you .
High hoo silver . Anyway I am trying to teach you something about the subject you breach . Parabolic Coordinates and the Prime Root . I wanted to show you the particle aspect of it all and why there are brakes in the chains .
You don't seem all that interested in what I have to teach you though ?
I doubt you even gave credit to my great accomplishments . Did you even do the chart . I think not . You dismissed it didn't you ?You dismissed it as crankery right out of the gate didn't you ? Do the chart . Map out all the factors . Then read all that I said from the beginning of Me entering this thread

Maybe just a commubication brake down > I don't know
I was going to give you the cell version . But now I see I gots to give it to someone else more deserving and not so incline to be so judgmental

Boo Who
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