To add comments or start new threads please go to the full version of: Parabolic Coordinates And Prime Roots
PhysForum Science, Physics and Technology Discussion Forums > Relativity, Quantum Mechanics and New Theories > Relativity, Quantum Mechanics, New Theories
Pages: 1, 2, 3, 4, 5, 6, 7

Raphie Frank
I don't know anything about stress tensors, Jeremy, but what I can tell you is that there are those who have been a bit more receptive to my ways of thinking than you have personally witnessed who have suggested that I investigate them.

Insofar as what you and I have been individually exploring might be related (you think they're related?duh...), then I would simply pass on that advice.

- RF
jeremyebert
The Triangular Number / Divisor Summatory function ratio seem to be directly related to the Prime Counting function. possibly converging to 2 or 1/2...

D(1000) = 7069
T(1000) = 500500
ratio = 70.80209364832368
PrimePi(1000) = 168/ ratio = 2.3728111888111883659252210726337436010054682482293726145183


D(100000) = 1166750
T(100000) = 5000050000
ratio = 4285.4510392114848939361474180415684593957574458967216627383
PrimePi(100000) = 9592/ ratio = 2.2382708172918270817291827081729182708172918270817291827082...


D(200000) = 2472113
T(200000) = 20000100000
ratio = 8090.2855168837346836491697588257494701900762626951114289678...
PrimePi(200000) = 17984/ ratio = 2.2229128950355248223758881205593972030139849300753496232518...



D(300000) = 3829833
T(300000) = 45000150000
ratio = 11749.898755376539917014658341499485747811980313501920318718...
PrimePi(300000) = 25997/ ratio = 2.2125297027009909966966776777410741964193452688491038363207




D(350000) = 4522065
T(350000) = 61250175000
ratio = 13544.735646214727121348322060828404722178916048309787674436
PrimePi(350000) = 29977/ ratio = 2.2131845746563173084811594415852689400479263936796915274120



D(390000) = 5081120
T(390000) = 76050195000
ratio = 14967.210969235129262839688887489372421828258336744654721793
PrimePi(390000) = 29977/ ratio = 2.2092960450660251429993046040184380855302211914118037435670

jeremyebert
QUOTE (jeremyebert+Oct 24 2011, 03:02 PM)
The Triangular Number / Divisor Summatory function ratio seem to be directly related to the Prime Counting function. possibly converging to 2 or 1/2...

D(1000) = 7069
T(1000) = 500500
ratio = 70.80209364832368
PrimePi(1000) = 168/ ratio = 2.3728111888111883659252210726337436010054682482293726145183


D(100000) = 1166750
T(100000) = 5000050000
ratio = 4285.4510392114848939361474180415684593957574458967216627383
PrimePi(100000) = 9592/ ratio = 2.2382708172918270817291827081729182708172918270817291827082...


D(200000) = 2472113
T(200000) = 20000100000
ratio = 8090.2855168837346836491697588257494701900762626951114289678...
PrimePi(200000) = 17984/ ratio = 2.2229128950355248223758881205593972030139849300753496232518...



D(300000) = 3829833
T(300000) = 45000150000
ratio = 11749.898755376539917014658341499485747811980313501920318718...
PrimePi(300000) = 25997/ ratio = 2.2125297027009909966966776777410741964193452688491038363207




D(350000) = 4522065
T(350000) = 61250175000
ratio = 13544.735646214727121348322060828404722178916048309787674436
PrimePi(350000) = 29977/ ratio = 2.2131845746563173084811594415852689400479263936796915274120



D(390000) = 5081120
T(390000) = 76050195000
ratio = 14967.210969235129262839688887489372421828258336744654721793
PrimePi(390000) = 29977/ ratio = 2.2092960450660251429993046040184380855302211914118037435670

I guess it follows then that n/LOG(n) ~ 2*(T(n)/D(n))
jeremyebert
QUOTE (Raphie Frank+Oct 24 2011, 04:22 AM)
I don't know anything about stress tensors, Jeremy, but what I can tell you is that there are those who have been a bit more receptive to my ways of thinking than you have personally witnessed who have suggested that I investigate them.

Insofar as what you and I have been individually exploring might be related (you think they're related?duh...), then I would simply pass on that advice.

- RF

Very related, I'm working on a lexicon for importing the different "multiplication tables" into my 3D rendering.
jeremyebert
QUOTE (jeremyebert+Oct 24 2011, 05:22 PM)
I guess it follows then that n/LOG(n) ~ 2*(T(n)/D(n))

It make since that T(n)/D(n) is proportionately less than PrimePi(n) because not all divisors are prime. I guess that’s a good question. How many divisors of n are prime?
T(n) = nth Triangular Number = total of divisors and non-divisors
D(n) = Sum of divisor counts from 1 to n
PrimePi(n) = Count of primes less than n

http://oeis.org/A161664/a161664.pdf
jeremyebert
QUOTE (jeremyebert+Oct 25 2011, 05:18 PM)
It make since that T(n)/D(n) is proportionately less than PrimePi(n) because not all divisors are prime. I guess that’s a good question. How many divisors of n are prime?
T(n) = nth Triangular Number = total of divisors and non-divisors
D(n) = Sum of divisor counts from 1 to n
PrimePi(n) = Count of primes less than n

http://oeis.org/A161664/a161664.pdf

With my geometries in 3d as I have posted, where n>1, n is a circle of a sphere

http://en.wikipedia.org/wiki/Circle_of_a_sphere

when n=Infinity, n is a Great circle

http://en.wikipedia.org/wiki/Great_circle

when n=1, n is a point of a sphere.

Do you follow that Raphie?

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png


http://dl.dropbox.com/u/13155084/PL3D2SPHE...ttice_3D_2.html
Raphie Frank
QUOTE (jeremyebert+Oct 26 2011, 01:44 AM)
hen n=1, n is a point of a sphere.

Do you follow that Raphie?

Take it as a given that I understand, if nothing else, this one point...

But. such understanding, however derived, however, also constitutes legitimitate basis for others to "prove" you 'insane,' 'delusional,' 'stupid' a 'liar' or many of the empirically 'justifiable' descriptors which have been appended to me, yours truly, inclusive of by IVY LEAGUE trained scholars upon this forum such as RP****r.

I only suggest this -- and with a bit of irritation (truth be told) because, in essence, you are simply making (aka "reinforcing") an obvious point -- and making obvious points -- points I made upon this forum 3 or 4 years ago... Hopefully the ENLIGHTENED NON-moderator, Dr. ROBERT PENNER, will emerge from hiding to put the Owellian amd Dictatorial RP***er in his place. Dr. ROBERT PENNER is a brilliant man (with a laudable social conscience (IMHO)), but his intellectual counterpart, R***er is an intellectual twig with priggish and Pharisiree-tic tendencies..

SUGGESTED READING: "The Grand Inquisitor" by Dostoeyevsky

- RF
Raphie Frank
Sorry, Jeremy. That last post was out of place. Not quite as out of place as an utterly inexplicable 40 day suspension, but out of place all the same.

Here is a potentially important set of numbers, one that can be generated recursively as you will see below...

2, 1, 2, 3, 7, 31, 8191

Call that progression 'v' (The odd numbers are all 1 or Mersenne Primes, associated with the 1- and 2-Perfect Numbers and 1, 2, 3, 7,and 31 are all indices of Mersenne Prime Exponents 2, 3, 5, 17, 127)

0*(2-1) = K_0
0*(1-1) = K_0
0*(2-1) = K_0
1*(3-1) = K_1
4*(7-1) = K_4
8*(31-1) = K_8
24*(8191-1) = K_24

These are the number of vertices of the important Unimodular Lattices (e.g. 196560 = vertices of the Leech Lattice)

|C(n-1,6)| = v'
1, 0, 0, 0, 0, 0 1

Binomial Coefficient C(n,6) Respect to modulo 7:
0, 0, 0, 0, 0 1 -1, 0, 0, 0, 0, 0 1 -1, 0, 0, 0, 0, 0 1 -1, 0, 0, 0, 0, 0 1 repeating

--------------------------------------------------------
1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']

b = n (mod 2)
1 == Co-Totient (11)
10 == Totient (11)

... returns the following formula, based on idealized values in accord with empirical reality, assuming the Josephson Constant Derivation of Planck's Constant and Mp = 1*10^12, Mg = 2*10^34, B = 3.6456*10^-7 (The respective best guess minimum masses of a Primordial Black Hole and a Giant Molecular Cloud + the Ballmer Constant [units = meters])

2/B * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

Me~ 9.10938215 * 10^-31 kg
h-bar ~ 6.62606776 * 10^-34 J*s
m = 1 kg
2s = 2*pi * sqrt (1 meter/pi^2 meters/second^2)
GM/4pi^2 is (Kepler's Constant of Planetary Proportionality)^-1 m^3/s^2
--------------------------------------------------------

2, 1, 2, 3, 7, 31, 8191 = p_1, p_0, p_1, p_2, p_4, p_11, p_1028

-1 = |2^(T_-1) - 2|
0 = |2^(T_-1) - 1|
1 = |2^(T_-1) + 0|
2 = |2^(T_0) + 1|
4 = |2^(T_1) + 2|
11 = |2^(T_2) + 3|
1028 = |2^(T_4) + 4|

2^|-1| + 2*(-1) = 0
2^|-1| + 2*(-1) = 0
2^|-1| + 2*(-1) = 0
2^|0| + 2*(0) = 1
2^|1| + 2*(1) = 4
2^|2| + 2*(2) = 8
2^|4| + 2*(4) = 24

Pi_n(12) = 12, 5, 3, 2, 1, 0, 0, 0
Pi_n(12) - 1 = 11, 4, 2, 1, 0, -1, -1, -1
--------------------------------------------------------

As you see, the formula:

1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']

... can be generated entirely from recursion, no less so than the Electric Constant (based on one idealized value, pi) can be generated.

The Planetary Positioning ratio at n = 10, can be estimated, on the basis of best data as of 10/22/2011 either by the following:

SUM
p_0 * ceiling [e^0/0^sgn(0)]/10^(pi(11))
p_11 * ceiling [e^11/11^sgn(11)]/10^(pi(11))

... or the following:

Sum the Main Diagonals of the Square Number Spiral, evaluated at a "radius" of 31 = p_11, divided by 10^pi(11)

1.68764 + .00001 = 1.68765

Best,
Raphie

(To get R ~ 4/B [but not exactly] requires the introduction of no more than two variables. phi^31 and 35 + 5*sqrt 35; 35 == p_3 * ceiling [e^3/3^sgn(3)] and Co-Totient(35) = 11))

P.S. I'm typing this all from memory, so it's quite possible there's a "mind typo" somewhere...

SUGGESTED READING: "The Grand Design" by Mlodinow and Hawking (2010)
Raphie Frank
Firstly... h-bar ~ 6.62606776 * 10^-34 J*s/(2*pi). It is the reduced Planck's Constant. Thus, h-bar^2 =h/(4*pi^2)

Secondly, Jeremy, as an FYI...

Because of the geometric nature of the square number spiral, many of our observations related to divisibility by primes also apply.

e.g. 168765 -1 is divisible by 31

Just as...

1/4*(CUT(31,4) + CUT(-31,4)) = .5*SUM(CUT(n, 3)) for the range 0 --> 30 = 36456/2 is also divisble by 31, where CUT(n,k) denotes The Division of Space Square aka "The Bernoulli Square" aka "The Whitney Square" aka "The Knight's Move Pascal Triangle" Yes, one set of numbers that goes by many names...

Also FYI...

1, 2, 3, 7 and 31 all follow the form |ceiling [n^n + n + 1]| as well as n^2 + n + 1

0^2 + 0^1 + 0^0 = 1
phi^2 + phi^0 + phi^0 = 2
1^2 + 1^1 + 1^0 = 3
2^2 + 2^1 + 2^0 = 7
5^2 + 5^1 + 5^0 = 31
90^2 + 90^1 + 90^0 = 8191

A set of numbers you may well recognize because the odd terms, where n is in N are all associated with the Ramanujan-Nagell Equation.

2^x - 1 = y
-------------------
2^13 - 1 = 8191
2^5 - 1 = 31
2^3 - 1 = 7
2^2 - 1 = 3
2^1 - 1 = 1

sqrt (4y - 3) is in N

- RF
Raphie Frank
QUOTE (Raphie Frank+Oct 26 2011, 05:47 PM)

P.S. I'm typing this all from memory, so it's quite possible there's a "mind typo" somewhere...

Here's at least one "mind typo"...

2/B * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

should have read...

B/2 * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

Turned around, you get B/4 which approximates R^-1, or 1/Rydberg Constant (but only approximately).

The 2's cancel. The 4*pi^2's cancel and one Mg cancels, as well as all the units and the scalars of 10. Thus, simplified...

B*(Me*m)/(h^2*Mg) * GM ~ 1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']

1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']
= .1*36456*(67092481 - 2)/(2*8191)
= .1*36456*(8191^2 - sqrt(2)^2)/(2*8191)
= .1*36456/2*Carol_13/Mersenne_13

Carol Numbers follow the form (2^n - 2)
36456 is the summation of the first 30 "Birthday Cake" Numbers aka CUT(n,3) related to both Polygons and the Geometric Division of Space (as well as much more).
67092481 is also equivalent to 1 + 1*2*4*8*15*26*42*64 where 1,2,4,8,15,26,42,64 are the 0-th through 7-th Birthday Cake Numbers.
2, of course, is the product of the 0-th through 1-st Birthday Cake Number.

Alternatively, 2*8191 is equal to the following: ((103682*289154)+(1860498*16114))/10^2 - (2*299792458) = 16382
= ((L_24*G_24)+(L_30*G_18))/10^2 - (2*299792458)
where (3*10^8 - 2*(103682 + 89) = (3*10^8 - p_1*(L_24 + p_24)) = (3*10^8 - L_0*(L_24 + F_11)) = 299792458

... which is relevant only insofar as this observation begat the others since 299792458 m/s is the speed of light in a vacuum.

L, G and F denote Lucas, Golden and Fibonacci Numbers respectively. Oh, and...

((103682*289154)-(1860498*16114)) = 256 = 2*127 + 2 == (p_1 + p_1*p_31) = 2*sigma(p_31)
Raphie Frank
QUOTE (Raphie Frank+Oct 26 2011, 05:47 PM)
Here is a potentially important set of numbers, one that can be generated recursively as you will see below...

2, 1, 2, 3, 7, 31, 8191

Call that progression 'v' (The odd numbers are all 1 or Mersenne Primes, associated with the 1- and 2-Perfect Numbers and 1, 2, 3, 7,and 31 are all indices of Mersenne Prime Exponents 2, 3, 5, 17, 127)

0*(2-1) = K_0
0*(1-1) = K_0
0*(2-1) = K_0
1*(3-1) = K_1
4*(7-1) = K_4
8*(31-1) = K_8
24*(8191-1) = K_24

pi (K_0 + 2(-1)^0) = pi (2) = 1
pi (K_1 + 3(-1)^0) = pi (5) = 3
pi (K_4 - 5(-1)^1) = pi (19) = 8
pi (K_8 + 17(-1)^0) = pi (257) = 55
pi (K_24 + 127(-1)^0) = pi (196687) = 17711

1 = F_2 = F_(2*1)
3= F_4 = F_(2*2)
8= F_6 = F_(2*3)
55 = F_10 = F_(2*5)
17711 = F_22 = F_(2*11)

As well as being the first iterated primes = pi_n(11) = pi_n+1(31) = pi_n+2(127)...
11 = 12-1
5 = 6-1
3 = 4-1
2 = 3-1
1 = 2-1
0 = 1-1
... where 1, 2, 3, 4, 6, 12 are the divisors of 12.

0, 0, 1, 0, 0 is equal either to T_(n-1)mod 2 or (1 - F_(n+1))mod 2

p_1 = p_(E_0) = 2 = E_1
p_2 = p_(E_1) = 3 = E_2
p_3 = p_(E_2) = 5 = E_3
p_7 = p_(E_4) = 17 = E_6
p_31 = p_(E_8) = 127 = E_12

for E_n denotes the nth Mersenne Prime Exponent, inclusive of -1, 0, 1, indexed from -1.

0, 1, 2, 4, 8, coincidentally, are the -1-st through 3-rd Birthday Cake Numbers, while 1, 2, 3, 6, 12 all make an appearance in our p^n*q^k-n triangle and have the property that totient(n) divides sigma(n). Alternatively, {floor [1,2,4,6,12,24]}/2 = (0, 1, 2, 3, 6, 12), the first six Highly Composite Numbers...

See the recursion at work?

0*(2-1)
|0*(1-1) - p_2| = p_2
|0*(2-1) + p_1| = p_1
|1*(3-1) + p_2| = p_3
|4*(7-1) - p_3| = p_8
|8*(31-1) + p_7| = p_55
|24*(8191-1) + p_31| = p_17711

Pretty unbelievable, if you ask me. And even more unbelievable how simply one can combine these numbers to generate what I showed you above.

- RF

Note: 2, of course, is also a Fibonacci Number = F_3
Raphie Frank
Here is the (+155) : PI(n,k) Square, an exploratory construct, which is where most of the relationships I am showing you come together.

155 36 12 6 4 3 2 1
035 11 05 3 2 2 1 0
011 05 03 2 1 1 0 0
005 03 02 1 0 0 0 0
003 02 01 0 0 0 0 0
002 01 00 0 0 0 0 0
001 00 00 0 0 0 0 0
000 00 00 0 0 0 0 0

Top Edge = Left Edge + 1
Rows = Iterated Prime Counting Function (Descending)

Left Edge = SUM[|n-2|!] - 2

-2 + (2 + 1 + 1 + 1 + 2 + 6 + 24 + 120)

---------------------
(+155) : PI(n,k) - 1
---------------------

154 35 11 05 03 2 1 0
034 10 04 02 01 1 0 -1
010 04 02 01 00 0 0 -1
004 02 01 00 -1 -1 -1 -1
002 01 00 -1 -1 -1 -1 -1
001 00 -1 -1 -1 -1 -1 -1
000 -1 -1 -1 -1 -1 -1 -1
x-1 -1 -1 -1 -1 -1 -1 -1

Note, in particular, the values in the 3rd column (11, 4, 2, 1, 0, -1, -1, -1), and also in the top row (11 05 03 2 1 0)

Based on the (+36) : PI(n,k) square, of which a copy is contained in the (+155) : PI(n,k) square, I have an empirically derived formula for laminated lattice Kissing Numbers to D = 24.

The Key terms for that formula are 1, 2, 3, 4, 6, 12, 36.

- RF
Raphie Frank
QUOTE (Raphie Frank+Oct 26 2011, 07:42 PM)
Here's at least one "mind typo"...

2/B * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

should have read...

B/2 * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

Turned around, you get B/4 which approximates R^-1, or 1/Rydberg Constant (but only approximately).

The 2's cancel. The 4*pi^2's cancel and one Mg cancels, as well as all the units and the scalars of 10. Thus, simplified...

B*(Me*m)/(h^2*Mg) * GM ~ 1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']

And here's a second "mind typo." As I said I'm typing from memory and balancing a number of variables...

B/2 * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

should read...

B/2 * (2*Me*Mp*m)/(h-bar^2*Mg) * GM/4pi^2

Thus...

BMp*(Me*m)/(h^2*Mg) * GM ~ 1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']

v = 2, 1, 2, 3, 7, 31, 8191
v' = 1, 0, 0, 0, 0, 0, 1

All units should cancel and the scalars should work out.

- RF
jeremyebert
QUOTE (jeremyebert+Oct 25 2011, 08:44 PM)
With my geometries in 3d as I have posted, where n>1, n is a circle of a sphere

http://en.wikipedia.org/wiki/Circle_of_a_sphere

when n=Infinity, n is a Great circle

http://en.wikipedia.org/wiki/Great_circle

when n=1, n is a point of a sphere.

Do you follow that Raphie?

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png


http://dl.dropbox.com/u/13155084/PL3D2SPHE...ttice_3D_2.html

Wolfram view of my Divisor summatory function:

D(y) = ((y(y+1))/2) - (((y-x^2)/(2x)) - (((y-x^2)/(2x)) mod 0.5))

http://www.wolframalpha.com/input/?i=((y(y...mod+0.5))&cdf=1
jeremyebert
QUOTE (Raphie Frank+Oct 26 2011, 05:24 PM)
And here's a second "mind typo." As I said I'm typing from memory and balancing a number of variables...

B/2 * (2*Me*Mg*m)/(h-bar*Mg)^2) * GM/4pi^2

should read...

B/2 * (2*Me*Mp*m)/(h-bar^2*Mg) * GM/4pi^2

Thus...

BMp*(Me*m)/(h^2*Mg) * GM ~ 1/10*PRODUCT [(v^(2^ b ) - 2v')/v^v']

v = 2, 1, 2, 3, 7, 31, 8191
v' = 1, 0, 0, 0, 0, 0, 1

All units should cancel and the scalars should work out.

- RF

Raphie,
I'm really looking forward to overlaying these vertices in my model; they seem to have very interesting properties, especially with the exploratory links into physics.
P.S. Your memory is much better than mine even with the "mind typos".
jeremyebert
QUOTE (jeremyebert+Oct 27 2011, 09:05 PM)
Wolfram view of my Divisor summatory function:

D(y) = ((y(y+1))/2) - (((y-x^2)/(2x)) - (((y-x^2)/(2x)) mod 0.5))

http://www.wolframalpha.com/input/?i=((y(y...mod+0.5))&cdf=1

OOPS!! I forgot the summation part, disregard.
Raphie Frank
Partition- and Fibonacci-based recursions of numbers associated with the Ramanujan-Nagell Equation...

Second Differences of Partition Numbers
1, 0, 1, 0, 2, 0, 3, 1, 4, 2, 7, 3, 10, 7...
https://oeis.org/A053445 (associated with "Verma modules" which I know nothing about...)

Less 1
0, -1, 0, -1, 1, -1, 2, 0, 3, 1, 6, 2, 9, 6...

Take as p'rime indices [inclusive of (-1, 0 --> 0, 1)
1, 0, 1, 0, 2, 0, 3, 1, 5, 2, 13, 3, 23, 13...

Break down into odd and even...

1, 1, 2, 3, 5, 13, 23
0, 0, 0, 1, 2, 3, 13

2^1 - 1 = 1
2^1 - 1 = 1
2^2 - 1 = 3
2^3 - 1 = 7
2^5 - 1 = 31
2^13 - 1 = 8191
2^23 - 1 = 8388607

2*(T_0) - 1 = -1
2*(T_0) - 1 = -1
2*(T_0) - 1 = -1
2*(T_1) + 1 = 3
2*(T_2) - 1 = 5
2*(T_3) - 1 = 11
2*(T_13) - 1 = 181

---------------------------
Alternatively...

2*(F_1 - 1) + 1 = 1
2*(F_1 - 1) + 1 = 1
2*(F_1 - 1) + 1 = 1
2*(F_1 - 0) + 1 = 3
2*(F_3 + 1) + 1 = 5
2*(F_5 + 1) + 1 = 11
2*(F_11 + 1) + 1 = 181

a(n) = 2*(F_a(n-1) + sgn(n-3)) + 1
---------------------------

((-1)^2 + 3)/4 = 1
((-1)^2 + 3)/4 = 1
((-1)^2 + 3)/4 = 1
(3^2 + 3)/4 = 3
(5^2 + 3)/4 = 7
(11^2 + 3)/4 = 31
(181^2 + 3)/4 = 8191

(1^2 + 3)/4 = 1
(1^2 + 3)/4 = 1
(1^2 + 3)/4 = 1
(3^2 + 3)/4 = 3
(5^2 + 3)/4 = 7
(11^2 + 3)/4 = 31
(181^2 + 3)/4 = 8191

Alternatively...
((-1)^2*sgn(-1^-1) + 3)/4 = 1/2
((-1)^2*sgn(-1^0) + 3)/4 = 1
((-1)^2*sgn(-1^1) + 3)/4 = 1/2
((+3)^2*sgn(3^2) + 3)/4 = 3
((+5)^2*sgn(5^3) + 3)/4 = 7
((+11)^2*sgn(11^4) + 3)/4 = 31
((+181)^2*sgn(181^5) + 3)/4 = 8191

(a(n)^2*sgn(a(n)^sgn(n-1)) + 3)/4

Multiply this latter sequence by 2, 1, 2, 3, 7, 31, 8191 and you get 1, 1, 1, 3, 7, 31, 8191. Divide it and you get... 2, 1, 2, 1, 1, 1, 1, a nice little "transform" multiplier between the two sequences.

Thus...
2, 1, 2, 3, 7, 31, 8191 (Prime-based Recursion and more...)
.5, 1, .5, 3, 7, 31, 8191 (Fibonacci-based Recursion)
1, 1, 1, 3, 7, 31, 8191 (Partition/Prime Number-based Recursion and more...)

Read more about the sequence 1, 3, 5, 11, 181 on the website of Robert Munafo under the entry for "91"
http://mrob.com/pub/math/numbers-9.html

- RF
Raphie Frank
QUOTE (jeremyebert+Oct 28 2011, 12:14 PM)
OOPS!! I forgot the summation part, disregard.

Pretty cool, Jeremy.
Raphie Frank
QUOTE (Raphie Frank+Oct 28 2011, 03:57 PM)

Thus...
2, 1, 2, 3, 7, 31, 8191 (Prime-based Recursion and more...)
.5, 1, .5, 3, 7, 31, 8191  (Fibonacci-based Recursion)
1, 1, 1, 3, 7, 31, 8191  (Partition/Prime Number-based Recursion and more...)


The above should have read...

2, 1, 2, 3, 7, 31, 8191 (Prime-based Recursion and more...)
.5, 1, .5, 3, 7, 31, 8191 (Partition/Prime Number-based Recursion and more...)
1, 1, 1, 3, 7, 31, 8191 (Fibonacci-based Recursion and more...)

And, obviously, squares, triangles and powers of two are all involved...

Note: See Column 4 of the (+155) : PI(n,k) Square for the p'rime indices (0, 0, 0, 1, 2, 3, 6) associated with 1, 1, 1, 2, 3, 5, 13 and thereby 1, 1, 1, 3, 7, 31, 8191.

pi(6) = 3
pi(3) = 2
pi(2) = 1
pi(1) = 0
pi(0) = 0
pi(0) = 0

- RF

More "mind typos"

CORRECTION
---------------
Alternatively...

2*(F_1 - 1) + 1 = 1
2*(F_1 - 1) + 1 = 1
2*(F_1 - 1) + 1 = 1
2*(F_1 - 0) + 1 = 3
2*(F_3 + 0) + 1 = 5
2*(F_5 + 0) + 1 = 11
2*(F_11 + 1) + 1 = 181

a(n) = 2*(F_a(n-1) + floor [(n-3)/3]) + 1

Next number in the sequence?
2*(F_181 + 1) + 1
= 2*(18130010821454963453907530667147829489881+1) + 1
= 36260021642909926907815061334295658979765

(36260021642909926907815061334295658979765 ^2 + 3)/4
=
328697292386074078726213123201950465498346822281223126008677523511333920169863807 (about 3.3 times the number of atoms in the visible universe according to Wolfram Alpha)

This number is not prime since the digits sum to 336.
Raphie Frank
Here's the crazy thing, Jeremy. Consider the following form...

(p'_n*p'_2n)*(M_n - |T_n - 1|)
--------------------------------
T_(M_n) - T_(|T_n - 1|)

p' --> 0, 1 UNION primes
M --> Mersenne #
T --> Triangular Number

The n solutions that are in N?

0, 1, 2, 3, 13

-->

-1/-1 = 1
6/1 = 6
21/3 = 7
130/13 = 10
33546241/33546241 = 1

33546241 + 33546241 + -1 + -1 = 67092479 = 8191^2 - 2 = Carol_13

!

For the special case of 0 and 13, then p_2n = par(n)

e.g.
p'_(2*0) = p'_0 = par(0) = 1
p'_(2*13) = p'_26 = par(13) = 101

- RF

P.S. The "uniquety" of the solutions to the above equation were suggested by me and confirmed by CRGreathouse on another forum.
Raphie Frank
As per Robert Munafo's website...
http://mrob.com/pub/math/numbers-9.html

2*z^2(z^2 - 1) = 3*(y^2 -1)

(181 + 1)/2 = 91
(11 + 1)/2 = 6
(5 + 1)/2 = 3
(3 + 1)/2 = 2
(1 + 1)/2 = 1

Then...

sqrt ((2*0^2(0^2 - 1))/3 + 1) = 0 = p'_-1
----------------------------------------------------------------------
sqrt ((2*1^2(1^2 - 1))/3 + 1) = 1 = p'_0 = L_1 = F_5 - 4
sqrt ((2*2^2(2^2 - 1))/3 + 1) = 3 = p'_2 = L_2 - 0
sqrt ((2*3^2(3^2 - 1))/3 + 1) = 7 = p'_4 = L_4 - 0
sqrt ((2*6^2(6^2 - 1))/3 + 1) = 29 = p'_10 = L_7 - 0
sqrt ((2*91^2(91^2 - 1))/3 + 1) = 6761 = p'_870 = F_20 - 4
----------------------------------------------------------------------

These numbers are all p'rime as well, and with an interesting connection, it seems, to Iterated Fibonacci numbers.

1 = 2 - 1 = F_3 - 1 = F_(1+2) - 1
2 = 3 - 1 = F_4 - 1 = F_(2+2) - 1
4 = 5 - 1 = F_5 - 1 = F_(3+2) - 1
7 = 8 - 1 = F_6 - 1 = F_(4+2) - 1
20 = 21 - 1 = F_8 - 1 = F_(6+2) - 1

As opposed to...
F_(1+1) - 0 = 1 = log_2(1 + 1)
F_(2+1) - 0 = 2 = log_2(3 + 1)
F_(3+1) - 0 = 3 = log_2(7 + 1)
F_(4+1) - 0 = 5 = log_2(31 + 1)
F_(6+1) - 0 = 13 = log_2(8191 + 1)

And as for the prime indices a down and dirty quick look shows...

2*floor [n^2/pi]
----------------------------------------
2*floor [1^2/pi] = 0; 0 = F_-4 + 3
2*floor [2^2/pi] = 2; 2 = F_-2 + 3
2*floor [3^2/pi] = 4; 3 = F_0 + 3
2*floor [4^2/pi] = 10; 4 = F_1 + 3
2*floor [37^2/pi] = 870; 37 = F_9 + 3

2*000 = 000; 000 = T_1 - 1; 1 = L_1
2*001 = 002; 001 = T_1 - 0; 1 = L_1
2*002 = 004; 002 = T_2 - 1; 2 = L_0
2*005 = 010; 005 = T_3 - 1; 3 = L_2
2*435 = 870; 435 = T_29 - 0; 29 = L_7

1, 0, 1, 1, 0 is the Lucas Series mod 2, indexed from -1
1 divides 2 = 2
5 divides 435 = 87
2 + 87 = 89 = F_11 = p_24

- RF
Raphie Frank
This is neat...

sqrt ((6*1^2) - 5) = 1
sqrt ((6*3^2) - 5) = 7
sqrt ((6*7^2) - 5) = 17
sqrt ((6*29^2) - 5) = 71
sqrt ((6*6761^2) - 5) = 16561

1 = p_0
3 = p'_2
7 = p'_4
29 = p'_10
6761 = p'_870

1 = p_1
7 = p'_4
17 = p'_7
71 = p'_20
16561 = p'_1917

A080806
Positive integer values of n such that 6n^2-5 is a square.
https://oeis.org/A080806

floor [10^01/9^01] = 1
floor [10^10/9^10] = 2
-------------------------
floor [10^((10+1)*1!)/9^((10+1)*1!)] = 3
floor [10^((10+1)*2!)/9^((10+1)*2!)] = 10
floor [10^((10+1)*3!)/9^((10+1)*3!)] = 1047

0 + 1 = 1
2 + 2 = 4
4 + 3 = 7
10 + 10 = 20
870 + 1047 = 1917

One possible next iteration would be...

sqrt ((6*1^2) + 30) = 6 = 6*1
sqrt ((6*7^2) + 30) = 18 = 6 * 3
sqrt ((6*17^2) + 30) = 42 = 6* 7
sqrt ((6*71^2) + 30) = 174 = 6*29
sqrt ((6*16561^2) + 30) = 40566 = 6*6761

Another one...
sqrt ((24*1^2) + 120) = 12 = 12*1
sqrt ((24*7^2) + 120) = 36 = 12*3
sqrt ((24*17^2) + 120) = 84 = 12*7
sqrt ((24*71^2) + 120) = 348 = 12*29
sqrt ((24*16561^2) + 120) = 81832 = 12*6761

- RF

And another "mind typo" to fix...
QUOTE (Raphie Frank+Oct 28 2011, 05:17 PM)
33546241 + 33546241 + -1 + -1 = 67092479 = 8191^2 - 2 = Carol_13


33546241 + 33546241 + -1 + -1 = 67092480 = 8191^2 - 1^2
= x-bar ((Carol_13)^1, (Mersenne_13)^2)

(p'_13*p'_(2*13))(M_13 - |T_13 - 1|) = T_(M_13) - T_(|T_13 - 1|)
(41*101)*(8191 - 90) = (33550336 - 4095)
(4141)*(8101) = 33546241

(p'_0*p'_(2*0))(M_0 - |T_0 - 1|) = T_(M_0) - T_(|T_0 - 1|)
(1*1)*(0 - 1) = (0 - 1)
(1)*(-1) = -1

(4141)*(8101) + 33546241 + (1)*(-1) + -1
= 67092480
= 8191^2 - 1^1
= M_13^d(13) - M_0^d(0)

p'_(2*13) = par(13)
p'_(2*00) = par(00)
jeremyebert
Raphie,
I want to know more about the CUT(n,k) function. I want to be able to code it so I can work with it in my equations.
jeremyebert
A little reducing...

Summatory version of the factorial n! is a triangular number T(n).


Sum of natural numbers from 1 to n = T(n)


Sum of successful divisions from 1 to n = D(n)


Sum of failed divisions from 1 to n = C(n) ***Cicada Function


C(n)+D(n)=T(n)


D(n)= SUM ((2*floor[(n - k^2)/k]) + 1) where 1<=k<=floor(sqrt(n))

C(n)= Sum(-floor[(n-k^2)/k] where 1<=k<=n

***Reducing the function makes my geometries twice as large.

****Also it obscures the connection to the "Coupon Collector's Problem" which I feel is valid.


H(n) = nth Harmonic Number

n*H(n) - a(n) = D(n)

http://dl.dropbox.com/u/13155084/chart.pdf

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png

http://en.wikipedia.org/wiki/Coupon_collector%27s_problem
jeremyebert
QUOTE (jeremyebert+Oct 25 2011, 05:18 PM)
It make since that T(n)/D(n) is proportionately less than PrimePi(n) because not all divisors are prime. I guess that’s a good question. How many divisors of n are prime?
T(n) = nth Triangular Number = total of divisors and non-divisors
D(n) = Sum of divisor counts from 1 to n
PrimePi(n) = Count of primes less than n

http://oeis.org/A161664/a161664.pdf

I guess thats also called the Big Omega function.
http://en.wikipedia.org/wiki/Prime_factor
Raphie Frank
QUOTE (jeremyebert+Oct 29 2011, 05:33 PM)
Raphie,
I want to know more about the CUT(n,k) function. I want to be able to code it so I can work with it in my equations.

It's just the sideways summation of Pascal's Triangle.

You can generate it by the recursion rule

A + B =
xxxxxx C

Starting from a left edge of all 1's and a top edge of all 1's


1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 11 1 1 1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1

Just fill in the blanks.

n denotes row number; START (0) --> "Cuts"
k denotes column number; START (0) --> "Dimensions"

For instance, CUT (n,3) = C(n,0) + C(n,1) + C(n,2) + C(n, 3)

... where C(n,K) denotes Binomial Coefficient C(n,k). These are the "birthday cake" numbers. CUT (n,4) = C(n,0) + C(n,1) + C(n,2) + C(n, 3) + C(n, 4), which pertains to Moser's problem and the division of a 4-D space. etc.

All kinds of patterns emerge if you fill in those blanks up above, including that you get the summation of the Fibonacci Numbers for the left to right main (down) diagonals and powers of two for the right to left (down) main diagonals. (With a little observation you can see how plainly the two are related.)

You can also easily obtain the "Dying Rabbits" Sequence, not quite as interesting as the "Don't want to die" Cicada Sequence, but interesting all the same.

But perhaps most interesting is that no one seems to know how easily one can derive numbers associated with high upon high mathematics with the mathematics accessible to a pre-teen. Hopefully this will one day change.

Also interesting is that one can "discover" this triangle (or square) without even knowing it is (directly) related to Pascal's triangle. A point I can vouch for since it took me quite a while to notice it myself.

I originally came across it by starting a Pascal "Trapezoid" with a top row of...

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

and 1's down the left side. The first differences of that triangle (or square) are actually much more interesting in many respects than CUT(n,k), at least observationally speaking (since you can pretty much map the entire solar system on to it, including many planetary orbits at Aphelion and Perihelion).

Best,
Raphie
Raphie Frank
Here. I'll get you started...

1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 2 2 2 2 2 2 2 2 2 2 2 2 2
1 3 4 4 4 4 4 4 4 4 4 4 4 4
1 4 7 8 8 8 8 8 8 8 8 8 8 8
1 5 11
1 6
1 7
1 8
1 9

The first differences (left to right) of this triangle, as per the mathematical description I gave above, is Pascal's Triangle (or "Square", allowing for the shift of perspective...)

Easy as pie...

A + B =
xxxxxx C

The "Pythagorean Theorem" equivalent for Statisticians (in relation to Variance).

- RF
Raphie Frank
QUOTE (jeremyebert+Oct 29 2011, 11:23 PM)
Sum of failed divisions from 1 to n = C(n) ***Cicada Function


C(n)+D(n)=T(n)

This is very interesting, Jeremy. The Cicada sequence, in essence, and philosophically speaking, can be thought of as a manner "multiplicative failure" that conveniently works to sustains life. Safe harbor in a sense...

- RF
Raphie Frank
Jeremy, this is pretty strange...

Old value for Electron Mass sans scalar:

9.10938215

New Value as of this summer (2010 Codata)

9.10938291

Josephson Constant Derivation of Planck's Reduced constant sans scalar...

(6.6260678/(2*pi))^2

OLD VALUE
1000*((9.10938215 + 9.10938215)/2)/(6.6260678/(2*pi))^2
= 8190.999653481

NEW VALUE
1000*((9.10938291 + 9.10938291)/2)/(6.6260678/(2*pi))^2
= 8191.000336860

AVERAGE VALUE
1000*((9.10938291 + 9.10938215)/2)/(6.6260678/(2*pi))^2
= 8190.999995170

(K_24 + 24)/24 = 8191
(G_24 + L_24)/200 + (G_(24-6) + L_(24+6)/200 - 299792458 = 8191
(289154*103682)/200 + (16114 *1860498)/200 - 299792458 = 8191

Even stranger is that I have two exploratory constructs for the Fine Structure Constant, one callibrated to pi and one "Lewis and Carroll Looking Glass" type one based on Lattices and Planetary Orbits. They average to 137.035999055^-1. If you have a little time check out the last two measurements for alpha^-1 on the website of Robert Munafo (who, incidentally, has seen those derivations I came up with last year BEFORE Codata revised the official value)

- RF
Raphie Frank
This time with scalars, incorporating the new value for electron mass...

(p_13*par(13))*(M_13 - |T_13 - 1|) = T_(M_13) - T_|T_13 - 1|
(p_00*par(00))*(M_00 - |T_00 - 1|) = T_(M_00) - T_|T_00 - 1|

(p_13*par(13))*(M_13 - |T_13 - 1|) + T_(M_13) - T_|T_13 - 1| = 67092482
(p_00*par(00))*(M_00 - |T_00 - 1|) + T_(M_00) - T_|T_00 - 1| = -2

-----------------------------------------------------------------
100*(67092482 - -2)/8191 = 819100.03662556464
8191 = M_0 + M_13

(4*(9.10938291*10^-31*10^(3*12))/((2*10^34)*(6.6260678 * 10^-34)/(2*pi))^2
= 4*meters^2*(Me*Mp^3)/(h-bar*Mg)^2 * (meters/second)^2
= 819100.0336860

Units: Dimensionless, if multiplied by...

A ) (299792458 meters/second)^2 = c^2, instead of (1 meter/1 second)^2
B ) (3.6456*10^-7 meters)^2 = B^2, instead of (4 meters^2), although then you've got a 4 in there to be accounted for...
-----------------------------------------------------------------

based on the following values...

9.10938291*10^-31 kg = Electron Mass
10^12 kg ~ Primordial Black Hole Mass (min)
2*10^34 kg ~ Giant Molecular Cloud Mass (min)
(6.6260678 * 10^-34)/(2*pi) Joule-Seconds --> Planck's Reduced Constant (Josephson Derivation)
4*m^2 --> ((r + h)^2 with r + h = 2

Kind of makes you go hmmmm....

((103682*289154)/2+(1860498*16114)/2) - 100*299792458 = 819100
((100*196560) + (100*24))/24 = 819100

- RF
Lasand
Have you guys figured all this out yet?

http://www.bordalierinstitute.com/target1.html
Raphie Frank
QUOTE (Lasand+Nov 1 2011, 10:29 PM)
Have you guys figured all this out yet?

http://www.bordalierinstitute.com/target1.html

It's a process Lasand, but Jeremy has made some real discoveries. Particularly impressive because he is self-taught.

- RF
Mazulu
QUOTE (Lasand+Nov 1 2011, 10:29 PM)
Have you guys figured all this out yet?

http://www.bordalierinstitute.com/target1.html

Nice job.
Raphie Frank
Mazulu and Lasand.

You both might want to read "The Grand Design" by Mlodinow and Hawking (2010). In that book the authors make the case for A) Model Dependent Realism, B ) the strong anthropic principle (in non-mystical manner), and C) Natural Selection as applicable to dynamic systems of inorganic as well as organic origin. Specifically, they argue that natural selection is applicable to survival of entire universes.

And they spend most of the last Chapter exploring Conway's "Game of Life" in order to demonstrate how immense complexity can result from very simple rules.

And if you want a hands-on example of such complexity, it's right here on this thread. Recursively, given enough time, we can very easily identify whether a number is prime or not just by subtracting A from B. If the result is 2, then the number is prime. But good luck predicting the next time that will be the next result. And good luck obtaining the computational power to make the need for such prediction essentially a moot point.

Right, Jeremy?

- RF
jeremyebert
QUOTE (Raphie Frank+Nov 2 2011, 04:08 AM)
Mazulu and Lasand.

You both might want to read "The Grand Design" by Mlodinow and Hawking (2010). In that book the authors make the case for A) Model Dependent Realism, B ) the strong anthropic principle (in non-mystical manner), and C) Natural Selection as applicable to dynamic systems of inorganic as well as organic origin. Specifically, they argue that natural selection is applicable to survival of entire universes.

And they spend most of the last Chapter exploring Conway's "Game of Life" in order to demonstrate how immense complexity can result from very simple rules.

And if you want a hands-on example of such complexity, it's right here on this thread. Recursively, given enough time, we can very easily identify whether a number is prime or not just by subtracting A from B. If the result is 2, then the number is prime. But good luck predicting the next time that will be the next result. And good luck obtaining the computational power to make the need for such prediction essentially a moot point.

Right, Jeremy?

- RF

Raphie summed it up nicely.

My model shows the symmetry of a geometric expansion outlining the divisors of a number,
really nothing more than the Pythagorean theorem at work. With direction and support
from Raphie, analysis of my model led to a rediscovery of the Divisor Summatory Function,
and an alternative view to the standard hyperbolic simplex method. The geometry
present in my model shows that just as there are varying levels of composite numbers
( highly composite numbers ), there are varying levels of prime numbers, hence the unsolved
infimum of the Dirichlet divisor problem. I believe these geometries can be exploited to gain
a deeper understanding of primality. This alternative view, along with points Raphie
has been making, dare I say it, seem to have connections to areas well beyond the divisibility
of a number.
jeremyebert
QUOTE (jeremyebert+Oct 29 2011, 06:23 PM)
A little reducing...

Summatory version of the factorial n! is a triangular number T(n).


Sum of natural numbers from 1 to n = T(n)


Sum of successful divisions from 1 to n = D(n)


Sum of failed divisions from 1 to n = C(n) ***Cicada Function


C(n)+D(n)=T(n)


D(n)= SUM ((2*floor[(n - k^2)/k]) + 1) where 1<=k<=floor(sqrt(n))

C(n)= Sum(-floor[(n-k^2)/k] where 1<=k<=n

***Reducing the function makes my geometries twice as large.

****Also it obscures the connection to the "Coupon Collector's Problem" which I feel is valid.


H(n) = nth Harmonic Number

n*H(n) - a(n) = D(n)

http://dl.dropbox.com/u/13155084/chart.pdf

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png

http://en.wikipedia.org/wiki/Coupon_collector%27s_problem


Some more generalizing...

T(n) = sum(k) where 1<=k<=n

D(n) = sum(k+floor[(n-k^2)/k]) where 1<=k<=n

C(n) = sum(-floor[(n-k^2)/k]) where 1<=k<=n

C(n)+D(n)=T(n)


jeremyebert
QUOTE (jeremyebert+Nov 3 2011, 06:28 PM)

Some more generalizing...

T(n) = sum(k) where 1<=k<=n

D(n) = sum(k+floor[(n-k^2)/k]) where 1<=k<=n

C(n) = sum(-floor[(n-k^2)/k]) where 1<=k<=n

C(n)+D(n)=T(n)
Raphie Frank
AN ASIDE #1

Jeremy, below is an old post I came across. My thinking may have evolved since then, but I yet find it difficult to disagree today, except on a trivial level, with anything I wrote upon this forum about 4 years ago...

- RF

QUOTE (Raphie Frank+Oct 18 2007, 01:03 AM)
Dear RPenner,

Here is a part of that post I wrote last night, but deemed too long too post...

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

You make the point, RPenner, and I thank you sincerely for the time you took to respond to my earlier post, that one must understand QED in order to derive formulas of possible interest to physicists and I just can't say as I am convinced that this is the case. Such knowledge certainly improves ones odds, and I have been working at frantic pace to educate myself in that area, but is such knowledge either a necessary or sufficient condition to generating such formulas? IF, in fact, the laws of physics apply to patterns of human organization and perception, as Jung, for instance posited, and great Physicist/Thinkers from Pauli to Einstein to Penrose would seem to agree or have agreed, then it stands to reason at least to postulate that for each and every physical phenomeneon there is a mental correlative; that the mental and the material realms can be thought of as bijective mappings of one another.

IF, in fact, this is the case, then, for instance, the concept of the holographic mind, in which every part contains the imprint of the entirety, as a holographic plate divided can yet project the entire image, may not be so farfetched after all. Whether we are speaking of quarks, leptons or bosons, or physics, biology and sociology or cells from the brain, nose or posterior region, the question becomes far more one of perspective and orientation and emphasis than one of objective validity. In other words, one multi-dimensional "crystal," many facets.

And what would be one of the implications IF this were to be the case? And this, as before, is all a question, not an answer, but I believe it to be a worthwhile question to ask... Well, at least in theory, we could study social systems or the psyche or even language to generate hypotheses regarding the workings of the cosmos and  vice-versa. But there's a problem, an obvious one to me, and it's something I call "Babelization." Even as knowledge and information explodes in exponential upon exponential fashion according to what Ray Kurzweil termed "The Law of Accelerating Returns" (and long gone are the days when one man or woman can know "everything"), we simply don't have the common argot to effectively communicate across disciplines even when we're speaking the same language.

Biology uses one set of symbols and terms and paradigms. Chemistry another. Physics another. And these are practically sibling disciplines. Now bring in the philosophers and the psychologists and the politicians and sociologists and artists and on and on and you've got a multi-threaded system chalk full of co-evolving strands of knowledge and experience diverging in logarithmic fashion even as we are facing global dangers of unprecedented magnitude: Nuclear proliferation to rogue nations and terrorist groups, Global Warming, dwindling supplies of renewable energy, increasing concentration of wealth in violation of the "Tinbergen Norm"  (Tinbergen developed the "Gravity Model of Economics)  etc. etc.

An ironic thought, I know, in an age of globalization and connectedness on a scale also unprecedented, and I do not wish to sound alarmist because i have great hope and optimism for the future of humankind also, but the danger is that, as Yeats might have put it, the center cannot hold. All systems may be self-organizing, which I believe to be the case, but unfortunately such self-organization oft times takes violent form and that is something obviously I believe we can agree that most of us want to avoid.

The bottom line is this:

I believe we need to start figuring out some shortcuts to better communication across disciplines (not to mention across race, color, creed, class, etc.) so that we might more effectively "tie it all together"  (social string theory, anyone?) before we blow it all apart. And in order to do that, my take is that we need better "algorithms" for understanding and communicating with one another and ought to figure out ways to pool our resources in distributed computing, parallel-processing fashion.

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

All that said, let me make one point clearly: I do not believe that understanding can come from pure thought alone. But what I do believe is that reason, intuition, imagination, experience, observation, testing etc. are all together a part of the overall equation.

Furthermore, as pertains to my comments regarding a pooling of resources, I believe we must figure out ways to bridge disciplines. Not empiricism, not poetic social theories, but the two in tandem. Not positivism, not phenomenology, but both in tandem. "Hard" and "soft," reason and intuition in tandem. My goal is not to be a mathematician or a physicist, but to learn enough to be able to work with mathematician and physicists.

Another point I'd like to make, you mention the level of mathematics involved, and point it out with a certain amount of judgement associated, but I feel, as you feel with Guest00, that you are missing the thrust of the basic point I am trying to make (although perhaps I have not made it well...). In Vicoish, back to basics manner, I don't believe that one NECESSARILY requires higher levels of complex mathematics to intuit certain knowledge, so much as one  requires complex and interdependent and relational patterns of thought.

For instance, one of the beauties of number theory is that one can let the embedded algorithm of the series "do the work" for you. As one need not understand a computer's assembly code in order to use it, one also need not understand prime numbers in order to use public key code, and so on.

Fib numbers, for instance, related obviously to Lucas numbers, and therefore to the powers of phi in compound form (phi^2n + phi^-2n = L^n; phi^2n-1 - phi^-2n-1 = L^n) can be thought, relative to one another, as a flowing series of whole number Pythagorean triples (i.e. a "moving equilibrium" in the vein of some of the premises of functionalist sociology) with each F(2n) equal to F(n-1)^2 + F^(n+1)^2.

Does any of this have application in functionalist sociology? Or with respect to string theory or bit string physics? Or IA? Perhaps, yes. Perhaps no. The answer is that I DON'T know, but, to reiterate a point I have made repeatedly, I believe the inquiry to be more than worthwhile.

We need new approaches, because the old ones are no longer sufficient as Peter Woit, I believe, made amply clear by negation in his book "Not Even Wrong," a critical response to string theory.

In closing, let me simply make reference to the words if Stanford Professor of Physics Emeritus H. Pierre Noyes...

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

A Short Introduction to BIT-STRING PHYSICS"
http://arxiv.org/pdf/hep-th/9707020.pdf

When I first met Ted Bastin in 1972 and heard of the Combinatorial Hierarchy (hereinafter CH), my immediate reaction was that it must be dangerous nonsense. Nonsense, because the two numbers computed to reasonable accuracy — 137 ?  hc/e 2 and 2127 + 136 ?  hc/Gm 2p — are empirically determined, according to conventional wisdom. Dangerous, because the idea that one can gain insight into the physical world by “pure thought” without empirical input struck me then (and still strikes me) as subversive of the fundamental Enlightenment rationality which was so hard won, and which is proving to be all too fragile in the “new age” environment that the approach to the end of the millennium seems to encourage [84, 86]...

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

I largely agree with Noyes. For whatever that is worth.

Kindest Regards,
Raphie
Raphie Frank
AN ASIDE #2

Jeremy, I told you some time back that I am not a mathematician -- tho' I may work with numbers -- but, on a fundamental level, a social theorist. Here's a bit more context for you. As with the above post, you will see familiar themes from our conversations this past better part of a year.

FWIW, let's just say it was not at all surprising to me when Ken Ono and Jerome Malenfant made their respective discoveries this year.

Nor will it be surprising to me in the future when some "brilliant" scholar discovers that prime numbers, at a baseline level, are guided by the same mathematics that describes fractals.

(Periodic and Aperiodic together.)

- RF

QUOTE (Raphie Frank+Oct 19 2007, 10:16 PM)
Dear Dr. Penner,

Tell you what, because I see your point and actually AGREE with you, by and large. Will you PLEASE just keep an open mind, rather than assuming there are no guidelines or "triangulation" points of reference behind what I am doing?

I am not in school YET, and need the peer (or professorial, as the case may be... I would welcome your advice regarding a text book on the Standard Model) review in order to refine and discern that which I don't know and need to learn, particularly in terms of nomenclature and semiotic notation, although I suspect we will never see eye to eye because my approach is structural and prescriptive in nature and begins with QUALITATIVE observation and those observations guide the language in which I learn to express those observations, as does the audience to whom I am expressing it, as does the feedback I receive in dialectical manner.

As for your constant reference to the infinite number of ways we can express any numeric quantity, that's  a lemma, and even if it were an opinion, it's one I agree with, but that's PRECISELY my point. In an infinite universe with an infinite number of ways to express anything, or in an infinite system of infinite order ala Ramsey theory, order that MUST BE IMPOSED BY THE OBSERVER, there are then infinite ways to CONSTRUCT our conceptions of "reality," based upon whatever points of references we agree to culturally.

But here's the thing, once you get started, there's a path dependent component, just as the computer keypad yet has the vowels placed on the outside against all common sense, Italian society resisted the Arabic (actually Indian in origin) numeric system for quite some time when Fibonacci attempted to introduce it. They were fine with all those letters because it was a common language and commerce depended upon it, meaning there was RESISTANCE, economic and otherwise, to an idea we now generally agree was the better idea.

As for me, to contextualize this, my reference is lifelong observation and thought and reflection, and self-education borne of a liberal arts background, and a constant challenging of pre-convention and societal norms leading to "data points" of experience most don't have.  And I have offered up those challenges because I have been committed from young adulthood, not to a "good enough" world, but to a "better" one, and it's a choice motivated and guided by personal experience, experience I will not delve into herein, but suffice it to say that I understand the value of belief in oneself and others more than most  as "factors of production."

I am interested in the hidden variable, the potential energy that we leave untapped and never find because we are not looking for it, and I speak socially here, not mathematically. I am interested in "hidden populations" and "statistical discrimination." It's what drives me because I see far, far too many good people for whom the current "system" does not work, many of whom are driven to self-abuse (I speak specifically of artists I know).

Why do we have the "No Child Left Behind Act" Dr, Penner? Does this not imply a consensus social conviction that many children were left behind before? What are we doing for those people, now adults? What heuristics or theoretical paradigms do we have in place to identifty, for instance, the "adult prodigy," because, no, Dr. Penner, it may be sufficient, but it is certainly not necessary that giftedness manifests itself at an early age.

Our definitions "rule" us, and limit the measure of the man (and woman) far too often and for far too many of us.

Necessity is the mother of invention. I do what I do for people like Helen Buyninski, booted from Columbia with straight A's and 1600 SAT's because she was a little too different (name used by permission) and, in a general sense, for those so desperate in neglected corners of our global society that they graffiti the walls of buildings with the word WATER.

The butterfly effect, Dr. Penner. That's what I am getting at here. I BELIEVE the laws of physics can be applied on an anthropomorphized level. I believe we are all connected, all of us, of that I am quite convinced -- it's all a matter of degree --  and it seems to me that studying fractals and patterns of information and material/immaterial correspondences (ala Actor-Network Theory) that propagate and spider out across all disciplines, as well as the relationships between them, are as good a place as any to start figuring out how to throw out some metaphorical life preservers to those amongst us lacking "access."

Because those lacking access simply cannot do it alone. They are (mixed metaphorical analogy here continued...) cast adrift from the ocean liner of state and society, and let me tell you, it's a whole lot easier to climb up the side of a ship when there's a ladder, meaning those with financial, intellectual and cultural capital to spare have, IN MY VIEW, a moral obligation to make sure those ladders are offered down the side of the ship. But ala the "just world hypothesis," we look down at those in the water and think "Ah they deserve to be there! Let 'em drown..."

The bottom line is this: It's not so much if I am right or wrong. The issue is "Would we be better off as a world society if I were more right than wrong regarding some of my more unconventional approaches and ideas (many of which can be found online, and many of which I will yet present...)? Obviously I believe so or I wouldn't be doing this, setting myself up knowingly in such manner and risking the appendage of "crackpot" to my name in people's minds.

In time, I hope I can convince you that at least my approach is worthwhile. I HOPE I can, but deem it unlikely. Barring that, I simply request the open mind and a little patience as I learn an unfamiliar language, because your opinion carries much "weight" in these here parts, and to be quite frank (no pun intended), it is rather terrifying to disagree with you in such vigorous manner.

Kindest Regards and great respect,
Raphie Frank

Raphie Frank: Self Portrait of a Poet of the Possible
http://politinotions.blogspot.com/2007/01/...of-poet-of.html


(And, by the way, in regards to the above, I have since taken, and am currently taking, classes and a not-so-bad school.)

P.S, You and I both, would behoove ourselves to learn a bit more about the specific maths involved with fractals. They are the key, IMHO, to you taking your work to the next level.
Raphie Frank
QUOTE (jeremyebert+Nov 4 2011, 02:10 AM)
interesting view
http://www.wolframalpha.com/input/?i=k%2Bf...k%5E2%29%2Fk%5D

An interesting cleavage. Apparently (at least approximately...) symmetrical. with a 180 flip.

- RF
Raphie Frank
QUOTE (jeremyebert+Nov 3 2011, 11:28 PM)

Some more generalizing...

T(n) = sum(k) where 1<=k<=n

D(n) = sum(k+floor[(n-k^2)/k]) where 1<=k<=n

C(n) = sum(-floor[(n-k^2)/k]) where 1<=k<=n

C(n)+D(n)=T(n)

Would it be an accurate description of a painting if one were merely to describe the places upon the canvas with applique'?

Positive and negative space together comprise the entirety.

Just as a thought: Figure out a way to describe the "negative space" and the "positive space" will be defined by implication...

- RF

P.S. Easier said than done, of course...
Raphie Frank
See the following in relation to recent discussions...

Making the Schrödinger Equation Relativistic
----------------------------------------------------
http://en.wikipedia.org/wiki/Dirac_equatio...on_Relativistic

m^2*c^2
----------- * Wave Function
h-bar^2

Dirac's Coup
--------------
http://en.wikipedia.org/wiki/Dirac_equation#Dirac.27s_Coup

E = c*sqrt (p^2 + m^2c^2)

Or restated...

E = sqrt (c^2*p^2 + m^2*c^4)

Set c = 1, as is often the custom in the interests of simplification, and you are left with E' = sqrt (p^2 + m^2), or E'^2 = p^2 + m^2

--- When does the sum of squares equal another square? (e.g. 6^2 + 8^2 = 10^2)
--- What happens if you set p = m in value?
--- What happens if you set one or the other or both to a prime number? To the same prime number?

e.g.
sqrt (8191^2 + 2^2) = 8191.000244170427
sqrt (8191^2 + 8191^2) = sqrt (2) * 8191

- RF

P.S. And a fun little equivalency...

For 0 the summation of CUT(n,3) from -1 to -1, and -1 to 30 respectively...

totient (0)*0^2 = 0 = 0 + d(0) = 0 + 0
totient (39)*39^2 = 36504 = 36456 + d(36456) = 36456 + 48

0/2*(totient (totient (0)*0^2 - 2*totient(0))) = 0 = K_0
4/2*(totient (totient (4)*4^2 - 2*totient(4))) = 24 = K_4
39/2*(totient (totient (39)*39^2 - 2*totient(39))) = 196560 = K_24

7/2*(totient (totient (7)*7^2 - 2*totient(7))) = 322
p'_(12*L_12) = p'_(12*322) = 36457
totient (36457) = 36456

for L_n denotes a Lucas Number

(4096*196560)/log_2(4096) = 67092480 = 8191^d(8191) - 0^d(0)
Raphie Frank
QUOTE (Raphie Frank+Nov 4 2011, 04:45 PM)
For 0 the summation of CUT(n,3) from -1 to -1, and -1 to 30 respectively...

totient (0)*0^2 = 0 =  0 + d(0) = 0 + 0
totient (39)*39^2 = 36504 = 36456 + d(36456) = 36456 + 48

0/2*(totient (totient (0)*0^2 - 2*totient(0))) = 0 = K_0
4/2*(totient (totient (4)*4^2 - 2*totient(4))) = 24 = K_4
39/2*(totient (totient (39)*39^2 - 2*totient(39))) = 196560 = K_24

7/2*(totient (totient (7)*7^2 - 2*totient(7))) = 322
p'_(12*L_12) = p'_(12*322) = 36457
totient (36457) = 36456

for L_n denotes a Lucas Number

(4096*196560)/log_2(4096) = 67092480 = 8191^d(8191) - 0^d(0)

Here's the output from Wolfram for the first several values...

0 | 0

n | 1/2 (n) phi((n)^2 phi(n)-2 phi(n))
1 | 0.5
2 | 1
3 | 9
4 | 24
5 | 110
6 | 96
7 | 322
8 | 480
9 | 702
10 | 840
11 | 2112
12 | 1680
13 | 4316
14 | 2688
15 | 6660
16 | 8064

39 | 196560

For the special case of 0 and 39, then totient (n) is equal to Dimension # (= 0, 24)

I don't know if this is unique, but up to n = 39, I didn't notice any other laminated lattice Kissing Numbers.

The observation that led to the above?

(G_48 - sigma(48 + 0) - G_0)/100 - 8191 = 299792458
(29980065026 - 124 - 2)/100 - 8191 = 299792458
for G_n denotes the Golden Scale

This follows from the fact that L_n*G_n +/- 2 is a Golden Scale Number and (L_24*G_24 - 128)/100 - 8191 = 299792458

Thus, by substitution, since 48 is a Highly Composite Number with 10 divisors...

(G_d(36456) - sigma(d(36456) + d(0)) - G_d(0))/(d(d(36456) + d(d(0))^2 - 8191 = 299792458

I've now shown you at least 4 different ways to get to the number 299800649 (=299792458 + 8191) from Fibonacci-Type Numbers. The others relate to primes in their various ways. This is the 1st one that directly seems to tie in to the division of 3 (and 4) dimensional space without losing the relationship with the primes (as you can see below).

The current set of numbers I am working with...

2, 2, 2, 1, 2, 3, 7, 31, 8191 = V_k
4, 4, 4, 2, 4, 10, 50, 962, 67092482 = R_k = V_k^2 + V_k(mod 2)

V_k is entirely determined via recursion...

48, coincidentally, is the numbers of roots in the exceptional Lie Group, F_4

F4 (mathematics)
http://en.wikipedia.org/wiki/F4_(mathematics)

It also happens to be the sum of 2 + 2 + 1 + 2 + 3 + 7 + 31, the "inside" of Progression V_k up there above.

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


2^1 + 2^1 + 1^1 + 2^1 + 3^1 + 7^1 + 31^1
*
2^0 * 2^0 * 1^0 * 2^0 * 3^0 * 7^0 * 31^0
=
48 + 0


36456 + 48 = totient(39)*39^2; totient 39 = 24

3.6456*10^-7 --> The Balmer Constant
http://en.wikipedia.org/wiki/Johann_Jakob_Balmer

The divisor chain for (0 + 36456) --> 48 --> 10 --> 4 --> 2 --> 2...

...where 0 and 36456 are the summation of CUT(n,3) from -1 to -1, and -1 to 30 respectively...

0 + 36456 = 36456

If you find this all a bit "freaky," Jeremy, well... you're not alone.

- RF
Raphie Frank
Let y = 36456
Let z = 299800649 = (G_d(36456) - sigma(d(36456) + d(0)) - G_d(0))/(d(d(36456) + d(d(0))^2 - 8191

(10x)*(z - 8191)
=
-81910 y + 10*y*z

Numerically, without units, assuming the values...

B= 3.6456*10^-7
Mp= 10^12
c = 299792458

... this equation is a simple equivalent statement to (B*Mp*c).

Square this value to get the numerator * B^2 of the formula I showed you two posts previously.

Since 10 = d(d(36456)), then just 3 (including 0, the additive identity) geometrically based numbers, along with a couple iterated elementary arithmetic functions, a few binary operations and one number progression associated with optimal division of the octave (The Golden Scale), are required: 0,36456 and 8191. Although one could stretch it...

8191 --> SUM 2^n for the range 0 through 12 [= 0 through (48/4) = d(d(d(d36456))))^d(36456)/d(d(d36456)))]. To state the obvious, 2^n is associated with the doubling of frequency associated with one octave.

(B*Mp*c) = d(d(36456))*36456* ((G_d(36456) - sigma(d(36456) + d(0)) - G_d(0))/(d(d(36456) + d(d(0))^2 - 8191)

(B*Mp*c) = d(d(36456))*36456* ((G_d(36456) - sigma(d(36456) + d(0)) - G_d(0))/(d(d(36456) + d(d(0))^2 - SUM(d(d(d(d36456))))^x)

x-range: 0 --> d(36456)/d(d(d36456))))

And that error term, 8191, of course, is precisely the key term that gives us the denominator of our exploratory formulation of the equation two posts previous. It also happens to be the last element of Progression V_k

- RF

P.S. Or, in shorthand notation, for d_n(y) denotes the iterated divisor function and y = 36456 and 0' = 0 (as a variable)...

(B*Mp*c)
=
(d_2(y)*d_0(y) + d_2(0')*d_0(0'))
*
((G_d_1(y) - sigma(d_1(y) + d_1(0')) - G_d_1(0'))
------------------------------------------------------------
(d_2(y) + d_2(0'))^2
-
SUM(d_4(y) + d_4(0')^x))

x-range: 0 --> d_1(y)/d_3(y) - d_1(0')/d_3(0')

Note the symmetry?
Raphie Frank
And here is how Pogression R_k fits in (I tried to get "fancy" with the plus and minus terms, but simpler worked best in the end)...

= ((R_4 - R_-4) + R_0(-1)^1)^(1/(1+0)) / ((R_4 - R_-4) + R_0(-1)^0)^(1/(1+1))

= ((67092482 - 4) - 4)/(sqrt ((67092482 - 4) + 4)

= (67092474)/(sqrt (67092482)

= 8190.99908436089
Round = 8190.9990

Josephson Constant Value of h = 6.6260678(27) * 10^-34 Joule-Seconds
------------------------------------------------------------------------------
(Me*Mg)^1/(h-bar*Mg)^2
=
(2*(9.10938291*10^-31)*(2*10^34))/((1/(2*pi))*(2*10^34)*(6.6260678)*10^-34))^2
= 8191.00033686027


Watt balance Value of h = 6.62606889(23) * 10^-34 Joule-Seconds
------------------------------------------------------------------------------
(2*(9.10938291*10^-31)*(2*10^34))/((1/(2*pi))*(2*10^34)*(6.62606889)*10^-34))^2
= 8190.99764199254

(8191.00033686027 + 8190.99764199254)/2
= 8190.998989426405
Round = 8190.9990

Until Co-Data changed it's value for the Electron Mass, my general model (pre-progression V_k and R_k) only worked with the Josephson Constant Derivation. Now, with a litle help from Progression V_k and R_k, it more or less gives a perfect match for the average of the two-mid range values for h.

If one were to have a hankering, one could simply substitute the values in this post wholesale into the equation in the last post.

The problem being, of course, that this would have c work out to a bit less than 299792458.001 meters/second and c^2 would be "wildly" off. In meter terms that would be almost 549,000 meters!

Although in percentage terms, it wouldn't be so bad...

10^12*((299800649 - 8190.99908436089)^2/299792458^2) = 1.0000000000061084866251

2006 Co-data Value for Fine Structure Constant^-1/ 2010 Co-data Value

137.035999679/137.035999074
= 1.0000000044148983

Compared to...

(299800649 - 8190.99908436089)/299792458
1.000000000003054243

- RF

QUOTE (Raphie Frank+Nov 4 2011, 05:57 PM)

The current set of numbers I am working with...

2, 2, 2, 1, 2, 3, 7, 31, 8191 = V_k
4, 4, 4, 2, 4, 10, 50, 962, 67092482 = R_k = V_k^2 + V_k(mod 2)

V_k is entirely determined via recursion...

48, coincidentally, is the numbers of roots in the exceptional Lie Group, F_4

F4 (mathematics)
http://en.wikipedia.org/wiki/F4_(mathematics)

It also happens to be the sum of  2 + 2 + 1 + 2 + 3 + 7 + 31, the "inside" of Progression V_k up there above.

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


2^1 + 2^1 + 1^1 + 2^1 + 3^1 + 7^1 + 31^1
*
2^0 * 2^0 * 1^0 * 2^0 * 3^0 * 7^0 * 31^0
=
48 + 0



- RF
Raphie Frank
Jeremy, have you come across this number progression?

3*n*(3*n-2)

0, 3, 8, 24, 63 etc.

It's 3 * the octagonal numbers. I know you came across octagonal numbers somewhere...

Also, FYI... Here's the error relative to high and low values for h

-----------------------------------------------
X-ray crystal density h (High Value for h)
(2*(9.10938291*10^-31*2*10^(34))/((2*10^34)*(6.6260745 * 10^-34)/(2*pi))^2
= 8190.983772097983

((67092482 - 4) - 88)/(sqrt ((67092482 - 4) + 88)
= 8190.983702 (a little bit less)
((67092482 - 4) - 87)/(sqrt ((67092482 - 4) + 87)
= 8190.98388476324

Farraday Constant h (Low Value for h)
(2*(9.10938291*10^-31*2*10^(34))/((2*10^34)*(6.6260657 * 10^-34)/(2*pi))^2
= 8191.005529

((67092482 - 4) + 32)/(sqrt ((67092482 - 4) - 32)
= 8191.005677 (a little bit more)
((67092482 - 4) + 31)/(sqrt ((67092482 - 4) - 31)
= 8191.005494
-----------------------------------------------

32 == sigma(p'_11)
88 == totient(p'_24)

4, 32 & 88 show up in a sequence related to "Kagome" Lattices.

Kagome lattice
http://en.wikipedia.org/wiki/Kagome_lattice

OEIS Search
4, 12, 32, 88, 240

Geometrical frustration
http://en.wikipedia.org/wiki/Geometrically_frustrated_magnet

- RF
jeremyebert
QUOTE (jeremyebert+Aug 18 2011, 02:29 PM)
RPenner/Raphie,

Another link into Eulers Generalized Pentagonal Numbers and the divisor function d(n):



For our Divisor summatory function we have:

D(n) = SUM(d(n)) :

for k = 0 --> floor [sqrt n]
SUM (d(n)) = SUM ((2*floor[(n - k^2)/k]) + 1)


The notable difference in the equation from the published version is the:

(n - k^2)/k (congruence of squares)

which is derived from the

z = (n - k^2)/2k + i n^(1/2)

forming a parabolic coordinate system.


The function (n - k^2)/2k forms a divisor symmetry centered on the square-root of n.

Example:

k = divisors of n {1,2,3,4,6,9,12,18,36}
n = 36


+17.5, +8.0, +4.5, +2.5, 00.0, -2.5, -4.5, -8.0, -17.5




key results:
sqrt(n) = 0
Sum Terms = 0

Offsetting by -((n-1)/2) = -17.5 and taking the absolute values gives us:


0, 9.5, 13, 15, 17.5, 20, 22, 25.5, 35

Key results:
sqrt(n) = (n-1)/2;




Another way to generate these terms is:

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

The key ratio here being the (k-1)/2k function.

reducing this ratio sequence we get:

01/04, 01/03, 03/08, 02/05, 05/12, 03/07, 07/06, 04/09, 09/20, 05/11, 11/24, 06/13, 13/28, 07/15, 15/32, 08/17, 17/36

or

01 01 03 02 05 03 07 04 09 05 11 06 13 07 15 08 17
04 03 08 05 12 07 16 09 20 11 24 13 28 15 32 17 36


Showing a direct connection to Eulers Generalized Pentagonal Numbers and the divisor function d(n)



**

A026741 ( n if n odd, n/2 if n even. ) = xx, 00, 01, 01, 03, 02, 05, 03, 07, 04, 09, 05, 11, 06, 13, 07, 15, 08, 17
A022998 ( If n is odd then n else 2*n. ) = 00, 01, 04, 03, 08, 05, 12, 07, 16, 09, 20, 11, 24, 13, 28, 15, 32, 17, 36

A026741 = Partial sums give Generalized Pentagonal Numbers A001318 = 00, 01, 02, 05, 07, 12, 15, 22, 26, 35, 40, 51, 57
A022998 = Partial sums give Generalized Octagonal Numbers A001082 = 00, 01, 05, 08, 16, 21, 33, 40, 56, 65, 85, 96, 120

Raphie,
I came across Generalized Pentagonal Number/ Generalized Octagonal Number ratio while working on the tensor idea.
Raphie Frank
QUOTE (Raphie Frank+Nov 7 2011, 05:20 PM)
4, 32 & 88 show up in a sequence related to "Kagome" Lattices.

Kagome lattice
http://en.wikipedia.org/wiki/Kagome_lattice

OEIS Search
4, 12, 32, 88, 240

Geometrical frustration
http://en.wikipedia.org/wiki/Geometrically_frustrated_magnet

- RF

((67092482 - 4) - 12)/(sqrt ((67092482 - 4) + 12)
= 8190.997619338476

Watt balance Value of h = 6.62606889(23) * 10^-34 Joule-Seconds
------------------------------------------------------------------------------
(2*(9.10938291*10^-31)*(2*10^34))/((1/(2*pi))*(2*10^34)*(6.62606889)*10^-34))^2
= 8190.99764199254

Hmmm....

((67092482 - 4) - 88)/(sqrt ((67092482 - 4) + 88) --> X-ray crystal density Derivation of h
((67092482 - 4) + 32)/(sqrt ((67092482 - 4) - 32) --> Farraday Constant Derivation of h
((67092482 - 4) - 12)/(sqrt ((67092482 - 4) + 12) --> Watt Balance Derivation of h
((67092482 - 4) + 4)/(sqrt ((67092482 - 4) - 4) --> Josephson Derivation of h
((67092482 - 4) - 1)/(sqrt ((67092482 - 4) + 1)
DATA SOURCE: http://en.wikipedia.org/wiki/Planck_constant

RELATED SEQUENCE FOR RANGE OF INTEREST
002605 a(n)=2*(a(n-1)+a(n-2)), a(0)=0, a(1)= 1
0, 1, 2, 6, 16, 44, 120...
"Individually, both this sequence and A028859 are convergents to 1+sqrt(3)"
http://oeis.org/A002605

Love the literal translation of Kagome, btw. It means, literally, "eyes (or holes) in a basket"

- RF

QUOTE (Raphie Frank+Nov 4 2011, 05:57 PM)
If you find this all a bit "freaky," Jeremy, well... you're not alone.

- RF


Go to the following page on the site of Robert Munafo and you'll see that my interest in 1 + sqrt 3 is not "new":
http://mrob.com/pub/num/n-b137_035.html
Raphie Frank
Cake Numbers...

1, 2, 4, 8, 15, 26, 42, 64, 93, 130, 176
http://oeis.org/A000125

Cake Numbers/2

.5, 1, 2, 4, 7.5, 13, 21, 32, 46.5, 65, 88

Cake Numbers/2 - (Cake Numbers/2)(mod 2)

0, 0, 2, 4, 7, 12, 20, 32, 46, 64, 88

1 * 2 * 4 * 8 * 15 * 26 * 42 * 64 = 67092480
1 * 2 = 2

Delta(67092480, 2) = 67092478 = 67092482 - 4

- RF
Raphie Frank
So, Jeremy, I've come up with a name for our joint exploratory efforts. We're investigating "SpacePrime." :-)
jeremyebert
QUOTE (Raphie Frank+Nov 7 2011, 02:19 PM)
So, Jeremy, I've come up with a name for our joint exploratory efforts. We're investigating "SpacePrime." :-)

Perfect!

Reminds me of a friends website. He likes to think of primes as space as well.

http://divisorplot.com/

Jeffery works with fractals a lot as well.
jeremyebert
QUOTE (jeremyebert+Nov 7 2011, 02:47 PM)
Perfect!

Reminds me of a friends website. He likes to think of primes as space as well.

http://divisorplot.com/

Jeffery works with fractals a lot as well.

If you take a look at his page you will notice his "Square Root Parabolas" are my concentric circles.
jeremyebert
QUOTE (jeremyebert+Nov 8 2011, 07:26 AM)
If you take a look at his page you will notice his "Square Root Parabolas" are my concentric circles.

Interesting link to my n*H(n) Coupon Problem connection...

pi(n) * H(pi(n)) ~ n

http://divisorplot.com/HarmonicFrequencyOfPrimes.pdf
Raphie Frank
QUOTE (jeremyebert+Nov 7 2011, 07:47 PM)
Perfect!

Reminds me of a friends website. He likes to think of primes as space as well.

http://divisorplot.com/

Jeffery works with fractals a lot as well.

"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
Raphie Frank
0 - 0 + 4 - 16 + 48 - 120
= 0, -0, 4, -12, +32, - 88
= (1^2 - 1) - (1^2 - 1) + (2^2-0) - (4^2 - 0) + (7^2 - 1) - (11^2 - 1)

------------------------------------------
SUM (x^2 - x(mod 2))(-1)^n(mod 2)
------------------------------------------

And what does x equal?

Here's a hint...

QUOTE (Raphie Frank+Oct 30 2011, 09:22 AM)
Here. I'll get you started...

1 1 1 1 1 1 1 1 1 1 1 1 1 1
1 2 2 2 2 2 2 2 2 2 2 2 2 2
1 3 4 4 4 4 4 4 4 4 4 4 4 4
1 4 7 8 8 8 8 8 8 8 8 8 8 8
1 5 11
1 6
1 7
1 8
1 9

The first differences (left to right) of this triangle, as per the mathematical description I gave above, is Pascal's Triangle (or "Square", allowing for the shift of perspective...)

Easy as pie...

A + B =
xxxxxx C

The "Pythagorean Theorem" equivalent for Statisticians (in relation to Variance).

- RF


CUT(n,2) = Triangular Numbers + 1

Next Number in sequence = 168 = (13^2 - 1) = -88 + 256

Wolfram-ready input...
(Click on "more terms" to see the shape)
---------------------------------------------------------------
SUM ((((n^2 + n + 2)/2)^2 - ((n^2 + n + 2)/2)^2(modulo 2))(-1)^(n modulo 2))
---------------------------------------------------------------

0, -0, 4, -12, +32, - 88, 168
(-1 + 1), (1 - 1) , (3 + 1), -(13 - 1), +(31 + 1), - (89 - 1), (167+1)

p'_-2, p'_0, p'_2, p'_6, p'_11, p_24', p'_39

- RF
jeremyebert
QUOTE (Raphie Frank+Nov 8 2011, 12:53 PM)
0 - 0 + 4 - 16 + 48 - 120
= 0, -0, 4, -12, +32, - 88
= (1^2 - 1) - (1^2 - 1) + (2^2-0) - (4^2 - 0) + (7^2 - 1) - (11^2 - 1)

------------------------------------------
SUM (x^2 - x(mod 2))(-1)^n(mod 2)
------------------------------------------

And what does x equal?

Here's a hint...



CUT(n,2) = Triangular Numbers + 1

Next Number in sequence = 168 = (13^2 - 1) = -88 + 256

Wolfram-ready input...
(Click on "more terms" to see the shape)
---------------------------------------------------------------
SUM ((((n^2 + n + 2)/2)^2 - ((n^2 + n + 2)/2)^2(modulo 2))(-1)^(n modulo 2))
---------------------------------------------------------------

0, -0, 4, -12, +32, - 88, 168
(-1 + 1), (1 - 1) , (3 + 1), -(13 - 1), +(31 + 1), - (89 - 1), (167+1)

p'_-2, p'_0, p'_2, p'_6, p'_11, p_24', p'_39

- RF

Sweet! The Wolfram-ready input makes things so much easier for me to grasp.

CUT(n,2) gives me much to think about!

"maximal number of pieces formed when slicing a pancake with n cuts" : http://oeis.org/A000124

pancakes, unit circles, roots of unity and cyclic groups...

Pascal, that rascal.

Thanks Raphie!
jeremyebert
Reduced equation for the lattice points in my model. (locus of points on concentric circles with radius = (n+1)/2)

1<=k<=n

(n+1)/2 * e^(i acos(1-k(2/(n+1))))
jeremyebert
QUOTE (Raphie Frank+Nov 8 2011, 12:39 PM)

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

with electron flow, sine and cosine are "eli the ice men in disguise"

e=voltage
l=inductance
c=capacitance
i=current

current comes before voltage by 90 degree's in an capacitive circuit
ice
voltage comes before current by 90 degree's in an inductive circuit
eli

My vocational electronics class. Ha!

http://www.electronicstheory.com/html/e101-31.htm
jeremyebert
QUOTE (jeremyebert+Nov 8 2011, 10:03 PM)
with electron flow, sine and cosine are "eli the ice men in disguise"

e=voltage
l=inductance
c=capacitance
i=current

current comes before voltage by 90 degree's in an capacitive circuit
ice
voltage comes before current by 90 degree's in an inductive circuit
eli

My vocational electronics class. Ha!

http://www.electronicstheory.com/html/e101-31.htm

more eli the ice men

e^ix = cos(x) + i sin(x)

with relation to n

1<=k<=n

time (moments) = t = acos(1-k(2/(n+1)))/ acos((n-1)/(n+1))

frequency = w = acos( (n-1)/(n+1) )

amplitude = a = n^((n+1)/2)


a*e^iwt


http://en.wikipedia.org/wiki/Angular_frequency

http://en.wikipedia.org/wiki/Sine_wave
Raphie Frank
QUOTE (jeremyebert+Nov 9 2011, 03:32 AM)
more eli the ice men

?

- RF
jeremyebert
QUOTE (Raphie Frank+Nov 9 2011, 05:02 AM)
?

- RF

Sorry Raphie, just some electronics humor from vocational school, like the resistor color code:
Bad Beer Rots Our Young Guts But Vodka Goes Willingly
Black Brown Red Orange Yellow Green Blue Violet Grey White
0 1 2 3 4 5 6 7 8 9

eli the ice man was how I first learned about pythagorean theorem, sine, cosine, phase, reactance and resonance. My teacher didn't use imaginary numbers. I wish he would have though, it simplifies things a lot. I wonder if my model has any application in electronics theory... hmm.
jeremyebert
QUOTE (jeremyebert+Nov 8 2011, 10:32 PM)
more eli the ice men

e^ix = cos(x) + i sin(x)

with relation to n

1<=k<=n

time (moments) = t = acos(1-k(2/(n+1)))/ acos((n-1)/(n+1))

frequency = w = acos( (n-1)/(n+1) )

amplitude = a = n^((n+1)/2)


a*e^iwt


http://en.wikipedia.org/wiki/Angular_frequency

http://en.wikipedia.org/wiki/Sine_wave

Oops!!

****amplitude = a = n^((n+1)/2)

should read:
amplitude = a = (n+1)/2
Raphie Frank
QUOTE (jeremyebert+Nov 9 2011, 03:32 AM)
e^ix = cos(x) + i sin(x)

As an FYI...

2*(cos(30 degrees) + i sin(30 degrees)) = 1 + sqrt 3 (real part)
2*(cos(45 degrees) + i sin(45 degrees)) = sqrt 2 (real part)
2*(cos(60 degrees) + i sin(60 degrees)) = 1 (real part)
2*(cos(90 degrees) + i sin(90 degrees)) = 0 (real part)
2*(cos(120 degrees) + i sin(120 degrees)) = -1 (real part)
2*(cos(180 degrees) + i sin(180 degrees)) = -2
2*(cos(360 degrees) + i sin(360 degrees)) = 2

De Moivre not at all in disguise.

That's the unit circle divided by 1, 2, 3, 4, 6, 8, 12, the proper divisors of 24.

- RF
Raphie Frank
Also, as an FYI...

Sum of all terms on the two principal diagonals of a 2n+1 X 2n+1 square spiral
a(n) = 1 + 10*n^2 + [(16n^3 + 26n)/3]
http://oeis.org/A114254

For n = 31
1 + 10*31^2 + floor [(16*31^3 + 26*31)/3] = 168765
1 + 10*31^2 + floor [(16*31^3 + 26*31)/3]/10^5 = 1.68765
For n = 0
1 + 10*0^2 + floor [(16*0^3 + 26*0)/3] = 1
1 + 10*0^2 + floor [(16*0^3 + 26*0)/3]/10^0 = 1

Compared to...

31*ceiling [e^11/11] = 168764
31*ceiling [e^11/11]/10^5 = 1.68764
1*ceiling [e^0/1] = 1
1*ceiling [e^0/1]/10^0 = 1

31 = p_(p_5), 11 = p'_5 and 11 = 11^sgn(11), 5 = 5
1 = p'_(p'_-1), 0 = p'_-1 & 1 = 0^sgn(0), -1 = -1

If we borrow the data from the OEIS page...

21..22..23..24..25
20..7...8...9...10
19..6...1...2...11
18..5...4...3...12
17..16..15..14..13

... it's easy to see, for instance, that square numbers are at 45 degrees, Pronics + 1 at 135 and 315 degrees, etc.

Note that the Pronic Diagonal is consistent, but the Square Diagonal alternates between n^2 and n^2 + 1.

A formula such as this...
1 + 10*n^2 + [(16n^3 + 26n)/3]
... obscures just how simple that sum really is.

But hopefully you can intuit at least, why 31 (a prime), for instance, evenly divides 168764 = 168765 - 1

168764/31 = 5444

Ceiling(e^n/n) = 3, 4, 7, 14, 30, 68, 157, 373, 901, 2203, 5444...
http://oeis.org/A132407

- RF

P.S. Note, socially speaking, that the entry for A132407 uses parentheses, not brackets, to denote the Ceiling function. Strictly speaking, that is "wrong," but it is still understood, and such usage is most certainly not "proof" that one is "stupid" or "incompetent," mathematically speaking. Rather, it is the manner of mistake one who is self-taught is quite liable to make.
Raphie Frank
And one (of many) fun little equivalencies from which the above post in part stems...

2^2 + 3^3 = 4 + 27 == 31
2^0 + 3^1 = 1 + 3 == 4

(Golden Ratio^4 + Golden Ratio^-4)/10^1 = .7
(Golden Ratio^31 + Golden Ratio^-31)/10^5 = 30.10349

Venus @ Semimajor = 0.723332 AU
http://en.wikipedia.org/wiki/Venus

Neptune @ Semimajor = 30.10366151 AU
http://en.wikipedia.org/wiki/Neptune

floor [Golden Ratio^4 + Golden Ratio^-4]*10^0 = 7
floor [Golden Ratio^31 + Golden Ratio^-31]/10^4 = 301
floor [Venus @ Semimajor]*10^1 = 7 AU*10
floor [Neptune @ Semimajor]*10^1 = 301 AU*10

Apply the construction rule 3X + 4 (the long discredited Titius-Bode Rule, formerly "Law"), but instead of x being a power of two, make x a Division of Space [CUT(n,k)] number, say CUT(4 - 4, 4) and CUT(4+4, 4) = 1, 99 (both Kaprekar Numbers)

(3 * 1) + 4 = 7
(3 * 99) + 4 = 301

The two planets with least eccentricity are Neptune and Venus.

Note:
7 + 233/10^3 = 7.233
301.0349 + 17711/10^7 = 301.036671

p_17711 - p_31 = 196687 - 127 = 196560 = K_24
p_233 - p_4 = 1471 - 7 = 1464 = (arbitrary number so far as I know)
17711 = F_22 and 233 = F_13

31*ceiling[e^11/11]/10^5 gives the floor (to 6 digits) of the (empirical) Planetary Positioning ratio at n = 10 based on best data available as of 10/22/2011
1 + 10*31^2 + floor [(16*31^3 + 26*31)/3]/10^5 = 1.68765 gives the ceiling.*

1*ceiling [e^0/1]/10^0 = 1
1 + 10*0^2 + floor [(16*0^3 + 26*0)/3]/10^0 = 1
... is the ratio of Mercury to itself (n = 1)*

But, of course, those measurements are subject to change higher or lower so this is just an approximation.

- RF

* Asteroid Belt set at 2.816 AU, which gives the minimum possible ratio.
Raphie Frank
QUOTE (Raphie Frank+Nov 9 2011, 05:56 PM)
And one (of many) fun little equivalencies from which the above post in part stems...

2^2 + 3^3 = 4 + 27 == 31
2^0 + 3^1 = 1 + 3 == 4

(Golden Ratio^4 + Golden Ratio^-4)/10^1 = .7
(Golden Ratio^31 + Golden Ratio^-31)/10^5 = 30.10349

And, of course, as you know...

CUT(31,4) = 36457
CUT(-31,4) = 36455

... the average of which is the long since not discredited (in contrast to the Titius-Bode "Law") value of the Balmer constant properly scaled.

31 == CUT(5, 4)
4 == CUT(2, 4)

2, 5 and 31 are the 0-th, 1-st and 5-th Golden Scale Numbers, or, if in a Mersenne frame of mind, the 1-st, 3-rd and 8-th Mersenne Prime Exponents.

(2 - 2(mod2))*1 = 2 = K_1 = 2*T_1 = 2*T_(T_1)
(5 - 5(mod2))*3 = 12 = K_3 = 2*T_3 = 2*T_(T_2)
(31 - 31(mod 2))*8 = 240 = K_8 = 2*T_15 = 2*T_(T_5)

K = Laminated Lattice Kissing Number
T = Triangular Number

As for 31 & 4...

2*31 + (2*5) = 72 = K_6 = 2*T_8
2*04 + (2*2) = 12 = K_3 = 2*T_3

... for which is there is a Geometric Interpretation related to the maximal number of regions obtainable by inscribing a polygon inside a circle.

- RF

With personal references removed, and one line in particular bolded, here was my thinking as of November, 2009...

QUOTE (Raphie Frank+Nov 1 2009, 07:46 PM)

Imagine if Johann Balmer had presented upon this forum his formula 3.6564*10^-7 * (n^2 - 4) = B*(n^2 - 4), which was based upon but four visible light wavelength values related to the emission spectrum of the hydrogen atom.

He would have been badgered and ridiculed without end and you, I take it, would have no problem with this. In fact, you might well redline his post [EDIT] as unscientific and preposterous speculation.

Meanwhile, Neils Bohr's model of the atom only came about because of the insights he came to after happening upon this formula. The Balmer formula, generalized by the Rydberg Formula, as I am confident you are aware, is key to predicting energy release (the "quantum jump") when electrons jump between shells.

Thank goodness the Balmer formula did not have to make it past you first, [NAME DELETED], and thank goodness Balmer made his formula known to the world, something he might not have done were his only two choices to remain silent or to post upon this forum.

To personalize the above, one can only imagine if Balmer himself even realized that 36456 is the difference in the maximal number of regions created by inscribing a 1 and a 32-sided polygon in a circle and connecting all the dots (this also gives the number of volumes in a 4-space divided by p-1 hyperplanes), a formula given by:

=====================================================
x = 1/24 (k^4 - 6k^3 + 23K^2 - 18k + 24)
=====================================================

... and one you will find on page 98 of Marcus du Sautoy's book "Symmetry" (2008) http://www.harpercollins.com/books/9780060...etry/index.aspx . The progression begins 1, 2, 4, 8, 16, but the next number is not 32, as one might guess, rather it is 31. du Sautoy mentions that formula as a "warning" to himself to never assume that he knows what number comes next.

Science moves forward when people make connections that have never be made before; when people think things like:

=====================================================
"Wow! 32! That's the maximal number of electrons in an orbit and you can derive the Balmer Constant exactly (scaled down by 10^11) by subtracting the first five values of row 0 (including 0's) from row 31 of Pascal's triangle." Shrodinger substituted waves for particles. What if the average of each of those waves could be viewed as the edge of an n-sided polygon equal to 2*n^2 up to n =4? What might then follow?
=====================================================

Sautoy's formula is one I know well, [NAME DELETED], because it is also related to the maximal number of points, intersections, areas and connections of an n-sided polygon (2x -1), a formula I posted [EDIT] well before I realized that (2x + 2k - 1) -- where k is the number of edges of a polygon -- yields the following prime values for k = 1 through 7:

3, 7, 13, 23, 41, 73, 127

In order, where K = Kissing Number:

K_1 + 1
K_2 + 1
K_3 + 1
K_4 - 1
K_5 + 1
K_6 + 1
K_7 + 1

One can furthermore only speculate in allo-historical and allo-physical fashion if scientists would now be so dismissive of Bode's Law had he used the Sautoy formula mentioned above for x rather than 2^n in his equation (3x + 4)/10,  given that by doing so he would have been off on his prediction of Neptune by a mere 4/100 ths of an AU (30.06 actual vs. 30.10 modified "prediction"), rather than by more than 8 AU, and also given that the sum total of the first 9 terms (including 2.8 AU) total 69.0 versus the total empirical mean AU distance of the planets from the Sun of 70.389 (Bode's terms sum to 80.1 AU).

Now, are these above observations, relating one well known geometrically precise number progression anchored in Pascal's Triangle (related to Probability, Set theory, the Mathematical Constants, Prime Divisibility, etc.) to the very big, the very small and everything in between (i.e. Quantum, Celestial and Kissing spaces), at all meaningful?

I don't know. But I do know A ) that these relationships are surprising, and B ) that breakthroughs tend to be made in science when observations like these are made and when questions like these are asked in an environment where Federation, not Tyranny is the rule and the law of the land and where civility is viewed by those who lead such Federations as a public virtue worthy of protection.

Best Regards,
Raphie

P.S. You might, I hope, appreciate the irony of my referencing Bode's Law (which I learned about from you), given that I was looking at Bode's Law earlier this past week by way of providing an example of "Bad Numerology" to contrast to the "Good Numerology" of the Balmer Formula.
Raphie Frank
Perhaps related...

Schrödinger equation for convex plane polygons. II. A no‐go theorem for plane waves representation of solutions
http://jmp.aip.org/resource/1/jmapaq/v34/i...isAuthorized=no
V. Amar, M. Pauri, and A. Scotti

Note that the conclusion is negative for the topic of the paper.

- RF
jeremyebert
Great posts Raphie.
jeremyebert
QUOTE (jeremyebert+Nov 8 2011, 10:32 PM)
more eli the ice men

e^ix = cos(x) + i sin(x)

with relation to n

1<=k<=n

time (moments) = t = acos(1-k(2/(n+1)))/ acos((n-1)/(n+1))

frequency = w = acos( (n-1)/(n+1) )

amplitude = a = n^((n+1)/2)


a*e^iwt


http://en.wikipedia.org/wiki/Angular_frequency

http://en.wikipedia.org/wiki/Sine_wave


Related information:


http://en.wikipedia.org/wiki/Angular_frequency
http://en.wikipedia.org/wiki/Angular_velocity
http://en.wikipedia.org/wiki/Angular_momentum
http://en.wikipedia.org/wiki/Land%C3%A9_g-factor

A triangular sequence related to angular momentum: t(n,m)=4*(n*(n-1)-m*(m-1)).
http://oeis.org/A152420

Raphie, notice the multiplication table for odd numbers in this triangle?
jeremyebert
QUOTE (jeremyebert+Nov 8 2011, 10:32 PM)
more eli the ice men

e^ix = cos(x) + i sin(x)

with relation to n

1<=k<=n

time (moments) = t = acos(1-k(2/(n+1)))/ acos((n-1)/(n+1))

frequency = w = acos( (n-1)/(n+1) )

amplitude = a = (n+1)/2


a*e^iwt


http://en.wikipedia.org/wiki/Angular_frequency

http://en.wikipedia.org/wiki/Sine_wave

Interesting wolfram view.

((n+1)/2) * e^iwt where w = acos(1-(2/(n+1)))

Re:

http://www.wolframalpha.com/input/?i=Plot3...-36%2C+36%7D%5D

Im:
http://www.wolframalpha.com/input/?i=Plot3...-36%2C+36%7D%5D

I wish wolfram had a scatter plot function.
jeremyebert
QUOTE (jeremyebert+Nov 11 2011, 12:47 PM)
Interesting wolfram view.

((n+1)/2) * e^iwt where w = acos(1-(2/(n+1)))

Re:

http://www.wolframalpha.com/input/?i=Plot3...-36%2C+36%7D%5D

Im:
http://www.wolframalpha.com/input/?i=Plot3...-36%2C+36%7D%5D

I wish wolfram had a scatter plot function.
jeremyebert
QUOTE (jeremyebert+Nov 14 2011, 08:15 AM)
Nice table of the values:

http://www.wolframalpha.com/input/?i=Table...%7Bk%2C+n%7D%5D

Squaring the imaginary gives us the multiplication table:

http://www.wolframalpha.com/input/?i=Table...%7Bk%2C+n%7D%5D
jeremyebert
QUOTE (jeremyebert+Nov 14 2011, 08:20 AM)
Squaring the imaginary gives us the multiplication table:

http://www.wolframalpha.com/input/?i=Table...%7Bk%2C+n%7D%5D

A little deeper view of how the function works:
http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D
Raphie Frank
QUOTE (jeremyebert+Nov 14 2011, 02:50 PM)
A little deeper view of how the function works:
http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D

2*e^(i*pi/3) --> 1 + sqrt 3 (with real part and imaginary parts summed without reference to direction). Ditto for 2*e^((i*pi)/6). Add them together and you get real part 1 + sqrt 3 and imaginary part i(1 + sqrt 3). Multiply them together and you get 4i. Divide them and you get (sqrt 3)/2 real part and .5i imaginary part.

Spin magnetic moment
http://en.wikipedia.org/wiki/Spin_magnetic_moment

The shortest distance to fully connect the 4 corners of a square is 1 + sqrt 3, not 2 sqrt 2 as one might naturally intuit.
Raphie Frank
((2*e^((i*pi)/3))/(2*e^((i*pi)/6)))^3 + ((2*e^((i*pi)/6))/(2*e^((i*pi)/3)))^3 = 0

If, in a "numerological" frame of mind, one could combine some of these terms thusly...

10/(e^((i*Golden Ratio)^2) + (1 + sqrt 3)/10^5 + (137.03599900683 - 130)/10^8)
= 10/(e^((i*Golden Ratio)^2) + (1 + sqrt 3)/10^5 + (7.03599900683)/10^8)
~ 137.03599900683

Borrowing from the website of Robert Munafo...

2008 Jun2 0.0072973525376(50) 137.035999679(94) CODATA 2006

2008 Jul 0.0072973525692(27) 137.035999084(51) Gabrielse 2008, Hanneke 2008

2010 Dec 0.0072973525717(48) 137.035999037(91) Bouchendira 2010

http://mrob.com/pub/num/n-b137_035.html

Now take these values...
2, 2, 2, 1, 2, 3, 7, 31, 8191

... and input them in to the following formula ((sqrt (4x - 3) - 1)/2)^2. The odd values will all be integers. The even ones will be equal to (1/Golden Ratio)^2

Alternatively, the "Lewis and Carroll Looking Glass" Lattice/Planetary-based derivation for the Fine Structure Constant I told you about works out thusly...

1/10^3*(31*720720)/163 - 7/(10^8)*(2*2*2*1*2*3*5*11*181)
~ 137.035999053

(137.03599900683 + 137.035999053)/2 ~ 137.03599903

- RF

Note: 31 = p_11, 163 = p_38 = p_sigma(p_sigma(11)), 720720 = HCN_37 for HCN denotes "Highly Composite Number." 37 = (p_sigma(11)). sigma(p_(sigma(p_31))) = sigma(p_(128)), by the way, = 720.

2*2*2*1*2*3*5*11*181 are derived from 2, 2, 2, 1, 2, 3, 7, 31, 8191 by the formula: floor [sqrt(4x - 3)]
Raphie Frank
QUOTE (Raphie Frank+Nov 14 2011, 04:03 PM)
1/10^3*(31*720720)/163 - 7/(10^8)*(2*2*2*1*2*3*5*11*181)
~ 137.035999053

(2* 1 * 2 * 4 * 8 * 16 * 31 * 63 * 99 * 130)/48^2

= (31*720720))

Least Common Multiple:
2,1,2,4,8,16,31,63,99,130 = 31 * 720720

3 * 2 - 2 = 4 -- > Mercury*10
3 * 1 + 4 = 7 -- > Venus*10
3 * 2 + 4 = 10 -- > Earth*10
3 * 4 + 4 = 16 -- > Mars*10
3 * 8 + 4 = 28 -- > Asteroid Belt/Ceres*10
3 * 16 + 4 = 52 -- > Jupiter*10
3 * 31 + 4 = 97 -- > Saturn*10
3 * 63 + 4 = 197 -- > Uranus*10
3 * 99 + 4 = 301 -- > Neptune*10
3 * 130 - 2 = 388 -- > Pluto*10

That's empirically-based. But the values all show up in the CUT(n,k) square from rows - 2 through row 9 in absolutely symmetrical manner. You have to skip over a couple rows, though. The ones containing the values 1 and 163, which is where the denominator comes from...

LCM 163 = 163, corresponding with Pluto at Aphelion and Mercury at Perihelion (multiplied by 10/3).

Units: Astronomical Units, which are based upon the "Goldilocks" distance of 1 Earth Distance to the Sun (which potentially has nothing to do with mysticism, much less fairytales or pre-Copernican ignorance, but with habitable zones of Solar Systems in general. Google search: "Planetary Green Zone")

The Correlation Coefficient of this Geometric framing with respect to Empirical Reality is roughly .99999

(And 31*ceiling [e^11/11]/10^5 gives an almost perfect match for the Empirically Determined Planetary Positioning Ration at n = 10 if we set the Asteroid Belt to 2.816 AU).)

3 * 2 - 2 = 4 -- > Mercury*10
-------------------------------------------------
3 * 1 + 0 = 3 --> Mercury*10 @ Perihelion
-------------------------------------------------
3 * 1 + 4 = 7 -- > Venus*10
3 * 2 + 4 = 10 -- > Earth*10
3 * 4 + 4 = 16 -- > Mars*10
3 * 8 + 4 = 28 -- > Asteroid Belt/Ceres*10
--------------------------------------------
3 * 16 + 4 = 52 -- > Jupiter*10
3 * 31 + 4 = 97 -- > Saturn*10
3 * 63 + 4 = 197 -- > Uranus*10
3 * 99 + 4 = 301 -- > Neptune*10
-------------------------------------------------
3 * 163 + 0 = 489 --> Pluto*10 @ Aphelion
-------------------------------------------------
3 * 130 - 2 = 388 -- > Pluto*10

- RF
jeremyebert
QUOTE (Raphie Frank+Nov 14 2011, 12:16 PM)
(2* 1 * 2 * 4 * 8 * 16 * 31 * 63 * 99 * 130)/48^2

= (31*720720))

Least Common Multiple:
2,1,2,4,8,16,31,63,99,130 = 31 * 720720

3 * 2 - 2 = 4 -- > Mercury*10
3 * 1 + 4 = 7 -- > Venus*10
3 * 2 + 4 = 10 -- > Earth*10
3 * 4 + 4 = 16 -- > Mars*10
3 * 8 + 4 = 28 -- > Asteroid Belt/Ceres*10
3 * 16 + 4 = 52 -- > Jupiter*10
3 * 31 + 4 = 97 -- > Saturn*10
3 * 63 + 4 = 197 -- > Uranus*10
3 * 99 + 4 = 301 -- > Neptune*10
3 * 130 - 2 = 388 -- > Pluto*10

That's empirically-based. But the values all show up in the CUT(n,k) square from rows - 2 through row 9 in absolutely symmetrical manner. You have to skip over a couple rows, though. The ones containing the values 1 and 163, which is where the denominator comes from...

LCM 163 = 163, corresponding with Pluto at Aphelion and Mercury at Perihelion (multiplied by 10/3).

Units: Astronomical Units, which are based upon the "Goldilocks" distance of 1 Earth Distance to the Sun (which potentially has nothing to do with mysticism, much less fairytales or pre-Copernican ignorance, but with habitable zones of Solar Systems in general. Google search: "Planetary Green Zone")

The Correlation Coefficient of this Geometric framing with respect to Empirical Reality is roughly .99999

(And 31*ceiling [e^11/11]/10^5 gives an almost perfect match for the Empirically Determined Planetary Positioning Ration at n = 10 if we set the Asteroid Belt to 2.816 AU).)

3 * 2 - 2 = 4 -- > Mercury*10
-------------------------------------------------
3 * 1 + 0 = 3 --> Mercury*10 @ Perihelion
-------------------------------------------------
3 * 1 + 4 = 7 -- > Venus*10
3 * 2 + 4 = 10 -- > Earth*10
3 * 4 + 4 = 16 -- > Mars*10
3 * 8 + 4 = 28 -- > Asteroid Belt/Ceres*10
--------------------------------------------
3 * 16 + 4 = 52 -- > Jupiter*10
3 * 31 + 4 = 97 -- > Saturn*10
3 * 63 + 4 = 197 -- > Uranus*10
3 * 99 + 4 = 301 -- > Neptune*10
-------------------------------------------------
3 * 163 + 0 = 489 --> Pluto*10 @ Aphelion
-------------------------------------------------
3 * 130 - 2 = 388 -- > Pluto*10

- RF
jeremyebert
QUOTE (jeremyebert+Nov 8 2011, 10:32 PM)

1<=k<=n

time (moments) = t = acos(1-k(2/(n+1)))/ acos((n-1)/(n+1))

frequency = w = acos( (n-1)/(n+1) )

amplitude = a = n^((n+1)/2)


a*e^iwt


http://en.wikipedia.org/wiki/Angular_frequency

http://en.wikipedia.org/wiki/Sine_wave

Simplex Simplex Picking seems to be related somehow.
http://mathworld.wolfram.com/SimplexSimplexPicking.html
jeremyebert
Recursion via a variable t:


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

1<=k<=n
1<=n<=t
1<=t<=25

Gives us our fimiliar pattern:

http://dl.dropbox.com/u/13155084/recursion.png
jeremyebert
QUOTE (jeremyebert+Nov 14 2011, 10:11 PM)
Recursion via a variable t:


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

1<=k<=n
1<=n<=t
1<=t<=25

Gives us our fimiliar pattern:

http://dl.dropbox.com/u/13155084/recursion.png
Raphie Frank
QUOTE (jeremyebert+Nov 14 2011, 06:33 PM)
Very interesting Rapie,

Related sequences and info:

http://oeis.org/A003461
http://oeis.org/A131500

I know these sequences, Jeremy, but they all flat out miss on Neptune.

Even assuming the "impossible," that there is a geometric order to the Planetary spacing, 3x + 4 (more or less) may be on the right track, but the key term is x. Miss on that and you have to get all Ptolomaic to fix the error, drawing error correction upon error correction, or you have to engage in "stupid" speculation such as that Neptune, somehow, is an "interloper."

For whatever it's worth, x seems to work best when one relates it to the geometric division of space.

Like, "Wow," that should come as any big surprise?

- RF
jeremyebert
QUOTE (Raphie Frank+Nov 15 2011, 02:23 AM)
I know these sequences, Jeremy, but they all flat out miss on Neptune.

Even assuming the "impossible," that there is a geometric order to the Planetary spacing, 3x + 4 (more or less) may be on the right track, but the key term is x. Miss on that and you have to get all Ptolomaic to fix the error, drawing error correction upon error correction, or you have to engage in "stupid" speculation such as that Neptune, somehow, is an "interloper."

For whatever it's worth, x seems to work best when one relates it to the geometric division of space.

Like, "Wow," that should come as any big surprise?

- RF

smile.gif

Im just saying, maybe OEIS needs another sequence...

A003461 is the classical version.
A131500 is based off the radius of Mercury (includes both Neptune's and Pluto's)
Axxxxxx geometric division of space
jeremyebert
QUOTE (jeremyebert+Nov 14 2011, 10:16 PM)
recursion 36
http://dl.dropbox.com/u/13155084/Recusion36.png

recursion 48
http://dl.dropbox.com/u/13155084/Recusion48.png

To see the recursion in action:
Click "OK"
Press "3" = Recursion Pattern
Press "Space" = Start
Press "v" = Changes view from the standard "square" multiplication table to the "square root" multiplication table via a matrix transform.

**pressing "m" brings up the menu for other options.

http://dl.dropbox.com/u/13155084/PL3D2/P_Lattice_3D_2.html
jeremyebert
where 1<=k<=n
f(n,k) = (n+1)/2 * e^(i cos^(-1)(1-k(2/(n+1))))

Hopefully its obvious now that when f(n,k) is a Gaussian Integer and a Pythagorean Triple, Im[f(n,k)]^2 is a perfect square.

ex:
f(n,k) at n=9 -> {4 + i3} | {3 + i4} | {2 + isqrt(21)} | {1 + i(2*sqrt(6))} | {0 + i5} ...
Im[f(n,k)]^2 at n=9 -> 9,16,21,25...

f(n,k) at n=57 -> ...{21 + i20} | {20 + i21} | {19 + i(4*sqrt(30))} | {18 + isqrt(517)}
IM[f(n,k)]^2 n=57 -> 57,112,165,216,265,312,357,400,441,480,517.....

1, 3, 9, 23, 57, 139, 337, 815, 1969, 4755.....
http://oeis.org/A133654 ( a(n) = 2*A000129(n) - 1)

0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741.....
http://oeis.org/A000129 (Pell Numbers)

1, 5, 29, 169, 985, 5741
http://oeis.org/A001653 (all Pythagorean triples (X,X+1,Z) ordered by increasing Z)

1, 7, 41, 239, 1393, 8119, 47321....
http://oeis.org/A002315 (corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs)

Hopefully now the relation between my f(n) scatter plot and the Pythagorean triple scatter plot is obvious as well.

http://dl.dropbox.com/u/13155084/Recusion36.png

http://upload.wikimedia.org/wikipedia/comm...catterplot2.png

http://demonstrations.wolfram.com/Primitiv...es1ScatterPlot/
jeremyebert
QUOTE (jeremyebert+Nov 15 2011, 12:35 PM)
where 1<=k<=n
f(n,k) = (n+1)/2 * e^(i cos^(-1)(1-k(2/(n+1))))

Hopefully its obvious now that when f(n,k) is a Gaussian Integer and a Pythagorean Triple, Im[f(n,k)]^2 is a perfect square.

ex:
f(n,k) at n=9 -> {4 + i3} | {3 + i4} | {2 + isqrt(21)} | {1 + i(2*sqrt(6))} | {0 + i5} ...
Im[f(n,k)]^2 at n=9 -> 9,16,21,25...

f(n,k) at n=57 -> ...{21 + i20} | {20 + i21} | {19 + i(4*sqrt(30))} | {18 + isqrt(517)}
IM[f(n,k)]^2 n=57 -> 57,112,165,216,265,312,357,400,441,480,517.....

1, 3, 9, 23, 57, 139, 337, 815, 1969, 4755.....
http://oeis.org/A133654 ( a(n) = 2*A000129(n) - 1)

0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741.....
http://oeis.org/A000129 (Pell Numbers)

1, 5, 29, 169, 985, 5741
http://oeis.org/A001653 (all Pythagorean triples (X,X+1,Z) ordered by increasing Z)

1, 7, 41, 239, 1393, 8119, 47321....
http://oeis.org/A002315 (corresponds to the sum of legs of near-isosceles primitive Pythagorean triangles (with consecutive legs)

Hopefully now the relation between my f(n) scatter plot and the Pythagorean triple scatter plot is obvious as well.

http://dl.dropbox.com/u/13155084/Recusion36.png

http://upload.wikimedia.org/wikipedia/comm...catterplot2.png

http://demonstrations.wolfram.com/Primitiv...es1ScatterPlot/

all Pythagorean triples (X,X+1,Z)

seem to be related to the 45 Deg. angle intersection of the dense parabolas in the scatter-plot.

The generator of the dense parabolas off the x axis seem to be related to the triples at

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

where sqrt(n) = {1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363...}

1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363
http://oeis.org/A078057

of course all of this is related to sqrt(2)
jeremyebert
QUOTE (jeremyebert+Nov 15 2011, 02:03 PM)
all Pythagorean triples (X,X+1,Z)

seem to be related to the 45 Deg. angle intersection of the dense parabolas in the scatter-plot.

The generator of the dense parabolas off the x axis seem to be related to the triples at

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

where sqrt(n) = {1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363...}

1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363
http://oeis.org/A078057

of course all of this is related to sqrt(2)
jeremyebert
QUOTE (jeremyebert+Nov 15 2011, 02:03 PM)
all Pythagorean triples (X,X+1,Z)

seem to be related to the 45 Deg. angle intersection of the dense parabolas in the scatter-plot.

The generator of the dense parabolas off the x axis seem to be related to the triples at

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

where sqrt(n) = {1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363...}

1, 3, 7, 17, 41, 99, 239, 577, 1393, 3363
http://oeis.org/A078057

of course all of this is related to sqrt(2)

"Using my Pythagorean Lattice value (1+sqrt(2))*n)^2

http://dl.dropbox.com/u/13155084/45.png"

pi/4 radians is an intersection of two sets of square root parabolas 90 Degrees out of phase. '''Quaternion? http://en.wikipedia.org/wiki/Quaternion

I know I've said this before but, the sqrt(2) and pell numbers are very interesting.
http://en.wikipedia.org/wiki/Pell_number
jeremyebert
QUOTE (jeremyebert+Nov 15 2011, 08:27 PM)
"Using my Pythagorean Lattice value (1+sqrt(2))*n)^2

http://dl.dropbox.com/u/13155084/45.png"

pi/4 radians is an intersection of two sets of square root parabolas 90 Degrees out of phase. '''Quaternion? http://en.wikipedia.org/wiki/Quaternion

I know I've said this before but, the sqrt(2) and pell numbers are very interesting.
http://en.wikipedia.org/wiki/Pell_number

(n-k^2)/2k + i sqrt(n)

the variable k forms the "square root parabolas" in my model.

and the density of perfect squares (Gaussian integers) on the parabola is the
highest on integers related to sqrt(2)

{1, 1, 9, 49, 289, 1681, 9801, 57121, 332929...} http://oeis.org/A090390
{x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057

so the kth "square root parabola" parabola in my model that has the most Gaussian integers along its curve is when:

k = {x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057

follow that?
jeremyebert
QUOTE (jeremyebert+Nov 15 2011, 09:58 PM)
(n-k^2)/2k + i sqrt(n)

the variable k forms the "square root parabolas" in my model.

and the density of perfect squares (Gaussian integers) on the parabola is the
highest on integers related to sqrt(2)

{1, 1, 9, 49, 289, 1681, 9801, 57121, 332929...} http://oeis.org/A090390
{x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057

so the kth "square root parabola" in my model that has the most Gaussian integers along its curve is when:

k = {x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057

follow that?

scatter-plot (n-k^2)/2k + i sqrt(n) where 1<=k<=n

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


Table[{(n-k^2)/2k + i sqrt(n)},{n,25},{k,n}]

http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D
Raphie Frank
QUOTE (jeremyebert+Nov 15 2011, 12:47 PM)
smile.gif

Im just saying, maybe OEIS needs another sequence...

A003461 is the classical version.
A131500 is based off the radius of Mercury (includes both Neptune's and Pluto's)
Axxxxxx geometric division of space

So go ahead and submit it, Jeremy. If you provide a link, it's not like you'd be taking any credit not your due.

Seems you've been busy making a number of connections...
jeremyebert
QUOTE (jeremyebert+Nov 15 2011, 08:27 PM)
"Using my Pythagorean Lattice value (1+sqrt(2))*n)^2

http://dl.dropbox.com/u/13155084/45.png"

pi/4 radians is an intersection of two sets of square root parabolas 90 Degrees out of phase. '''Quaternion? http://en.wikipedia.org/wiki/Quaternion

I know I've said this before but, the sqrt(2) and pell numbers are very interesting.
http://en.wikipedia.org/wiki/Pell_number

EDIT
"so the kth "square root parabola" in my model that has the most Gaussian integers along its curve is when:

k = {x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057"

should read sqrt(k) = {x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057

k = {1, 1, 9, 49, 289, 1681, 9801, 57121, 332929, 1940449...} http://oeis.org/A090390

Referencing the Quaternion relation...

http://oeis.org/A090390
"Sequences Related to Floretions"
http://www.mrob.com/pub/math/seq-floretion.html

Which is used in algorithms to draw curves in space.
http://sourceforge.net/projects/cranborg/
jeremyebert
Hopefully this equivalence obvious now

((n+1)/2)-k + i sqrt(k(n+1-k)) = ((n+1)/2) * e^(i arccos(1-k(2/(n+1))))

Table ((n+1)/2)-k + i sqrt(k(n+1-k))
http://www.wolframalpha.com/input/?i=table...C%7Bk%2Cn%7D%5D

Table ((n+1)/2) * e^(i arccos(1-k(2/(n+1))))
http://www.wolframalpha.com/input/?i=Table...%7Bk%2C+n%7D%5D

Table
((n+1)/2)-k + i sqrt(k(n+1-k)) - ((n+1)/2) * e^(i arccos(1-k(2/(n+1))))
http://www.wolframalpha.com/input/?i=Table...%7Bk%2C+n%7D%5D
jeremyebert
QUOTE (jeremyebert+Nov 16 2011, 10:02 AM)
EDIT
"so the kth "square root parabola" in my model that has the most Gaussian integers along its curve is when:

k = {x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057"

should read sqrt(k) = {x, 1, 3, 07, 017, 0041, 0099, 00239, 000577...} http://oeis.org/A078057

k =  {1, 1, 9, 49, 289, 1681, 9801, 57121, 332929, 1940449...} http://oeis.org/A090390

Referencing the Quaternion relation...

http://oeis.org/A090390
"Sequences Related to Floretions"
http://www.mrob.com/pub/math/seq-floretion.html

Which is used in algorithms to draw curves in space.
http://sourceforge.net/projects/cranborg/

I beleive these square root parabolas are of the form
i * sqrt(4(k/2)( n + (k/2)))


4(k/2)( n + (k/2)) is a pretty interesting function itself"
http://www.wolframalpha.com/input/?i=4%28k...+%28k%2F2%29%29

and its triangle... what is this?????

Table[{4(k/2)( n + (k/2))} ,{n,25},{k,n}]

http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D
jeremyebert
QUOTE (jeremyebert+Nov 16 2011, 04:00 PM)
I believe these square root parabolas are of the form
i * sqrt(4(k/2)( n + (k/2)))


4(k/2)( n + (k/2)) is a pretty interesting function itself"
http://www.wolframalpha.com/input/?i=4%28k...+%28k%2F2%29%29

and its triangle... what is this?????

Table[{4(k/2)( n + (k/2))} ,{n,25},{k,n}]

http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D

Very interesting triangle,,,,

Table[{4(k/2)( n + (k/2))} ,{n,25},{k,n}]
http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D


rows = n(n+2k) row sums = 2*n^3 + n^2 = n^2*(2*n+1).

http://oeis.org/A099721 "n^2*(2*n+1)"
http://www.wolframalpha.com/input/?i=n%5E2+%2B+2+n%5E3




1st left diag http://oeis.org/A033428 "3*n^2 0, 3, 12, 27, 48, 75, 108, 147, 192..."

2nd left diag http://oeis.org/A045944 "3*n^2 + 2n 0, 5, 16, 33, 56, 85, 120, 161, 208..Rhombic matchstick numbers"

3rd left diag http://oeis.org/A140676 "3*n^2 + 4n 0, 7, 20, 39, 64, 95, 132, 175, 224...."

4th left diag http://oeis.org/A067725 "3*n^2 + 6n 0, 9, 24, 45, 72, 105, 144, 189, 240..."



3*n^2 + (2(n-1))n = 3, 16, 39, 72, 115, 168, 231, 304, 387, 480....

http://www.wolframalpha.com/input/?i=3*n%5...82%28n-1%29%29n




0, 3, 16, 39, 72, 115, 168, 231, 304.... = (5*(n-1)-7)*(n-2)
http://oeis.org/A147874 "Appears to be related to various other sequences"....
jeremyebert
QUOTE (jeremyebert+Nov 16 2011, 07:08 PM)
Very interesting triangle,,,,

Table[{4(k/2)( n + (k/2))} ,{n,25},{k,n}]
http://www.wolframalpha.com/input/?i=Table...C%7Bk%2Cn%7D%5D


rows = n(n+2k)  row sums = 2*n^3 + n^2 = n^2*(2*n+1).

http://oeis.org/A099721 "n^2*(2*n+1)"
http://www.wolframalpha.com/input/?i=n%5E2+%2B+2+n%5E3




1st left diag http://oeis.org/A033428 "3*n^2    0, 3, 12, 27, 48, 75, 108, 147, 192..."

2nd left diag http://oeis.org/A045944 "3*n^2 + 2n  0, 5, 16, 33, 56, 85, 120, 161, 208..Rhombic matchstick numbers"

3rd left diag http://oeis.org/A140676 "3*n^2 + 4n 0, 7, 20, 39, 64, 95, 132, 175, 224...."

4th left diag http://oeis.org/A067725 "3*n^2 + 6n 0, 9, 24, 45, 72, 105, 144, 189, 240..."



3*n^2 + (2(n-1))n = 3, 16, 39, 72, 115, 168, 231, 304, 387, 480....

http://www.wolframalpha.com/input/?i=3*n%5...82%28n-1%29%29n




0, 3, 16, 39, 72, 115, 168, 231, 304.... =   (5*(n-1)-7)*(n-2)
http://oeis.org/A147874 "Appears to be related to various other sequences"....

hopefully the equivalency is apparent:

divisons
(n - k^2)/(2 k) + i Sqrt[n]
ListPlot[Flatten[Table[{(n - k^2)/(2 k), Sqrt[n]}, {n, 25}, {k, n}], 1], AspectRatio -> .2]
http://dl.dropbox.com/u/13155084/parabolas25.png


square root parabolas
(2 k n + k^2)^(1/2) = (4 (k/2) (n + (k/2)))^(1/2)
Plot[Evaluate[Table[(2 k*n + k^2)^(1/2), {k, -25, 25}]], {n, -12.5, 12.5}, AspectRatio -> 1.5, Mesh -> 49, MeshStyle -> PointSize[Medium]]
http://dl.dropbox.com/u/13155084/parabolas12.png

Mathematica is awesome
Raphie Frank
QUOTE (jeremyebert+Nov 18 2011, 01:51 AM)
Mathematica is awesome

It's a great social equalizer, Jeremy. Many of the statements I make upon this forum now are no different than those I made 3 or 4 years ago. But now, I can "prove" those statements. For instance, it would take you somewhere around 55 steps of simple but involved algebra, if I recall correctly, to "prove" what someone like me or you intuits immediately without need of those 55 steps [each step involving, let's just say, 20 or 30 operations (20*55 = 1100)].

e.g. All elements of Binomial C(n,4) are (Generalized) Pentagonal Numbers.

Anyone want to challenge me on that or label me "delusional" or "dense" or "stupid" or a "crank" or a "numerologist" as I have been labeled upon this forum in the past? Fine. Go ahead. Whether you are Dr. R*****t P****r or Dr. G****e Weath****l. You'll have to argue, not with me, but with Stephen Wolfram and his product.

And, yes, Jeremy, I DO at this point have a bit of a chip, a really pretty big chip actually, on my shoulder (If I were to post my past feedback upon this forum, which I have backed up, I would open myself up to charges of "information terrorism" and/or "public indecency"). But... for good reason in my own view, as I believe you might "get."

My respect for professional mathematicians and physicists, based upon my online experiences upon this forum and one other, frankly, is just this side of that I have for those who beat up old ladies and steal their money.

WITH ONE PROVISO.

I am thankful to RPenner for letting me and you simply have the ability to talk to one another without interference.

Best,
Raphie Frank
917-202-2610
100 Metropolitan Ave. #6
Brooklyn, NY 11211

P.S. And have you yet had even one professional confirm that you rediscovered the Dirichlet Divisor Sum?

I rest my case.

They, whoever that "they" may be, are either too stupid to recognize it, too vain to admit it, or too selfish and/or self-involved to help a young talent such as yourself. So... the chip on my shoulder is more than a little bit for you too, Jeremy. It irritates the heck out of me that you are stating the obvious and yet no professional will confirm it. It's not as if they could not.
jeremyebert
QUOTE (jeremyebert+Nov 17 2011, 08:51 PM)
hopefully the equivalency is apparent:

divisions
(n - k^2)/(2 k) + i Sqrt[n]
ListPlot[Flatten[Table[{(n - k^2)/(2 k), Sqrt[n]}, {n, 25}, {k, n}], 1], AspectRatio -> .2]
http://dl.dropbox.com/u/13155084/parabolas25.png


square root parabolas
(2 k n + k^2)^(1/2) = (4 (k/2) (n + (k/2)))^(1/2)
Plot[Evaluate[Table[(2 k*n + k^2)^(1/2), {k, -25, 25}]], {n, -12.5, 12.5}, AspectRatio -> 1.5, Mesh -> 49, MeshStyle -> PointSize[Medium]]
http://dl.dropbox.com/u/13155084/parabolas12.png

Mathematica is awesome


So basically my equation is a conformal map of the standard number grid.


http://dl.dropbox.com/u/13155084/parabolas12.png
http://en.wikipedia.org/wiki/Conformal_map

Correct?
jeremyebert
QUOTE (Raphie Frank+Nov 18 2011, 04:35 AM)
It's a great social equalizer, Jeremy. Many of the statements I make upon this forum now are no different than those I made 3 or 4 years ago. But now, I can "prove" those statements. For instance, it would take you somewhere around 55 steps of simple but involved algebra, if I recall correctly, to "prove" what someone like me or you intuits immediately without need of those 55 steps [each step involving, let's just say, 20 or 30 operations (20*55 = 1100)].

e.g. All elements of Binomial C(n,4) are (Generalized) Pentagonal Numbers.

Anyone want to challenge me on that or label me "delusional" or "dense" or "stupid" or a "crank" or a "numerologist" as I have been labeled upon this forum in the past? Fine. Go ahead. Whether you are Dr. R*****t P****r or Dr. G****e Weath****l. You'll have to argue, not with me, but with Stephen Wolfram and his product.

And, yes, Jeremy, I DO at this point have a bit of a chip, a really pretty big chip actually, on my shoulder (If I were to post my past feedback upon this forum, which I have backed up, I would open myself up to charges of "information terrorism" and/or "public indecency"). But... for good reason in my own view, as I believe you might "get."

My respect for professional mathematicians and physicists, based upon my online experiences upon this forum and one other, frankly, is just this side of that I have for those who beat up old ladies and steal their money.

WITH ONE PROVISO.

I am thankful to RPenner for letting me and you simply have the ability to talk to one another without interference.

Best,
Raphie Frank
917-202-2610
100 Metropolitan Ave. #6
Brooklyn, NY 11211

P.S. And have you yet had even one professional confirm that you rediscovered the Dirichlet Divisor Sum?

I rest my case.

They, whoever that "they" may be, are either too stupid to recognize it, too vain to admit it, or too selfish and/or self-involved to help a young talent such as yourself. So... the chip on my shoulder is more than a little bit for you too, Jeremy. It irritates the heck out of me that you are stating the obvious and yet no professional will confirm it. It's not as if they could not.

I'm with ya Raphie. Dealing with people like that is VERY frustrating, and I haven't been doing it near as long as you have. At least we have a place to discus our ideas though. In my experience, conversations like ours can leave you lonely at times. I'm just grateful to have a dialog going with you about this stuff.
jeremyebert
QUOTE (jeremyebert+Nov 18 2011, 04:03 PM)
jeremyebert
QUOTE (jeremyebert+Nov 20 2011, 04:31 PM)
PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.
To quit out of "lo-fi" mode and return to the regular forums, please click here.