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jeremyebert
QUOTE (Raphie Frank+Jul 29 2011, 03:10 PM)
Jeremy, I believe it would be helpful if you compiled a list of some of the relevant number progressions in regards to what you're doing, put them all in one place with the first several terms of each, and showed how they relate to one another. The number of progressions you are working with are starting to pile up and it's getting difficult to keep track...

And if it's getting difficult for me to keep track, how difficult do you think it will be for others?

- RF

Sorry, gettin a little carried away Wiil do.
jeremyebert
QUOTE (Raphie Frank+Jul 29 2011, 03:00 PM)
In relation to recent conversations, Jeremy, I have found some very interesting relationships to Maximal Lattice Sphere Packings that I want to show you. But first, let's define a finite set of Integers that lies at the heart of those relationships; one that we can refer back to later...

Let {p'_n} denote the set {0, 1 U the prime numbers}
--> n in N | d(n) < 3

(indexed from -1)

Let {c} = -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6
(indexed from -5)

These are all valid Definitions for {c}...

for Z is the set of all positive and negative integers, plus 0
Definition #1: n in Z | totient(|n|) < 3
Definition #2: n in Z | p'_|n| divides 196560
Definition #3: n in Z | |n|!/|n-1|! are proper divisors of 12
Definition #4: n in Z | d(p'_|n| - 1) = |n|
Definition #5: n in Z | 2*cos (((2*pi)*|n-1|!)/|n|!) is in N

#1: Totient |c| = {2, 2, 2, 1, 1, 0, 1, 1, 2, 2, 2}
#2: p'_|c| = {13, 7, 5, 3, 2, 1, 2, 3, 5, 7, 13} --> Unique Divisors of Unique Prime Divisors of the Leech Lattice
#3: Elements of {c} all divide 12 when placed into the form |n|!/|n-1|!
#4: d(p'_|n| - 1) = d ({12, 6, 4, 2, 1, 0, 1, 2, 4, 6, 12}) = {6, 4, 3, 2, 1, 0, 1, 2, 3, 4, 6}
#5: 2*cos (((2*pi)*|n-1|!)/|n|!) = {1, 0, -1, -2, 2, 2, 2, -2, -1, 0, 1} --> {0 U Allowable (periodic) n-fold rotational symmetries by the Crystallographic restriction theorem}

Crystallographic Restriction Theorem: Short Trigonometry Proof
http://en.wikipedia.org/wiki/Crystallograp...gonometry_proof

- RF

I'm gonna have to think about these functions for a bit but you can continue if you like.
jeremyebert
QUOTE (Raphie Frank+Jul 29 2011, 03:00 PM)
In relation to recent conversations, Jeremy, I have found some very interesting relationships to Maximal Lattice Sphere Packings that I want to show you. But first, let's define a finite set of Integers that lies at the heart of those relationships; one that we can refer back to later...

Let {p'_n} denote the set {0, 1 U the prime numbers}
--> n in N | d(n) < 3

(indexed from -1)

Let {c} = -6, -4, -3, -2, -1, 0, 1, 2, 3, 4, 6
(indexed from -5)

These are all valid Definitions for {c}...

for Z is the set of all positive and negative integers, plus 0
Definition #1: n in Z | totient(|n|) < 3
Definition #2: n in Z | p'_|n| divides 196560
Definition #3: n in Z | |n|!/|n-1|! are proper divisors of 12
Definition #4: n in Z | d(p'_|n| - 1) = |n|
Definition #5: n in Z | 2*cos (((2*pi)*|n-1|!)/|n|!) is in N

#1: Totient |c| = {2, 2, 2, 1, 1, 0, 1, 1, 2, 2, 2}
#2: p'_|c| = {13, 7, 5, 3, 2, 1, 2, 3, 5, 7, 13} --> Unique Divisors of Unique Prime Divisors of the Leech Lattice
#3: Elements of {c} all divide 12 when placed into the form |n|!/|n-1|!
#4: d(p'_|n| - 1) = d ({12, 6, 4, 2, 1, 0, 1, 2, 4, 6, 12}) = {6, 4, 3, 2, 1, 0, 1, 2, 3, 4, 6}
#5: 2*cos (((2*pi)*|n-1|!)/|n|!) = {1, 0, -1, -2, 2, 2, 2, -2, -1, 0, 1} --> {0 U Allowable (periodic) n-fold rotational symmetries by the Crystallographic restriction theorem}

Crystallographic Restriction Theorem: Short Trigonometry Proof
http://en.wikipedia.org/wiki/Crystallograp...gonometry_proof

- RF
Raphie Frank
Here's a new sequence on OEIS, as of March of 2011, Jeremy

a(n) = (-1 + prime(n)^2)/24.
1, 2, 5, 7, 12, 15, 22, 35, 40, 57, 70, 77, 92, 117, 145, 155, 187, 210, 222, 260, 287, 330, 392, 425, 442, 477, 495, 532, 672, 715, 782, 805, 925, 950, 1027, 1107, 1162, 1247, 1335, 1365, 1520, 1552, 1617, 1650, 1855, 2072, 2147, 2185, 2262, 2380, 2420, 2625, 2752, 2882, 3015
http://oeis.org/A024702

You can do a search for "Ken Ono and Factoring" and it'll bring you to a discussion from February of 2011 on another forum. Go there and you'll also see how I found out about Euler's Divisor summing technique.

Observation: 1,2,5,7,15,22 & 77 are partition numbers of index # [0,1],2,4,5,7,8,12

I don't know if there are any others as I have not checked. But perhaps you'll notice that 0,1,2,5,7,12, the first 6 Generalized Pentagonal Numbers, show up in that series. I'll show you in another post how to map this sub-set of integers to Kissing Numbers in Dimension 0,0,1,4,8,24

- RF

P.S. I'm so good at "cribbing" materials from others, Jeremy, that sometimes I can even do it in advance!

e.g. Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number. (see comments dated June 25, 2011)
http://oeis.org/A124174
jeremyebert
QUOTE (Raphie Frank+Jul 30 2011, 10:52 AM)
Here's a new sequence on OEIS, as of March of 2011, Jeremy

a(n) = (-1 + prime(n)^2)/24.
1, 2, 5, 7, 12, 15, 22, 35, 40, 57, 70, 77, 92, 117, 145, 155, 187, 210, 222, 260, 287, 330, 392, 425, 442, 477, 495, 532, 672, 715, 782, 805, 925, 950, 1027, 1107, 1162, 1247, 1335, 1365, 1520, 1552, 1617, 1650, 1855, 2072, 2147, 2185, 2262, 2380, 2420, 2625, 2752, 2882, 3015
http://oeis.org/A024702

You can do a search for "Ken Ono and Factoring" and it'll bring you to a discussion from February of 2011 on another forum. Go there and you'll also see how I found out about Euler's Divisor summing technique.

Observation: 1,2,5,7,15,22 & 77 are partition numbers of index # [0,1],2,4,5,7,8,12

I don't know if there are any others as I have not checked. But perhaps you'll notice that 0,1,2,5,7,12, the first 6 Generalized Pentagonal Numbers, show up in that series. I'll show you in another post how to map this sub-set of integers to Kissing Numbers in Dimension 0,0,1,4,8,24

- RF

P.S. I'm so good at "cribbing" materials from others, Jeremy, that sometimes I can even do it in advance!

e.g. Sophie Germain triangular numbers tr: 2*tr+1 is also a triangular number. (see comments dated June 25, 2011)
http://oeis.org/A124174

The Dedekind eta function is truly a beautiful thing.

I wonder what it looks like whith the complex numbers sets:

Set{l}

Real part = 1-(k(2/(n+1))
Imaginary part = sqrt(4d(n+1-k))/(n+1)

Set{m}

Real part = ((n+1)/2) - k
Imaginary part = sqrt(k(n+1-k))

Set{n}

Real part = (n-k^2)/(2k)
Imaginary part = sqrt(n)

Where

n = 1--->LIM
k = 1--->n
jeremyebert
QUOTE (jeremyebert+Jul 30 2011, 12:50 PM)

The Dedekind eta function is truly a beautiful thing.

I wonder what it looks like whith the complex numbers sets:

Set{l}

Real part = 1-(k(2/(n+1))
Imaginary part = sqrt(4d(n+1-k))/(n+1)

Set{m}

Real part = ((n+1)/2) - k
Imaginary part = sqrt(k(n+1-k))

Set{n}

Real part = (n-k^2)/(2k)
Imaginary part = sqrt(n)

Where

n = 1--->LIM
k = 1--->n

oops!!
Edit:
Set{l}

Real part = 1-(k(2/(n+1))
Imaginary part = sqrt(4k(n+1-k))/(n+1)
Raphie Frank
Some relevant information for you Jeremy...

Cyclic Group
via Wikipedia
The fundamental theorem of cyclic groups states that if G is a cyclic group of order n then every subgroup of G is cyclic. Moreover, the order of any subgroup of G is a divisor of n and for each positive divisor k of n the group G has exactly one subgroup of order k. This property characterizes finite cyclic groups: a group of order n is cyclic if and only if for every divisor d of n the group has at most one subgroup of order d. Sometimes the refined statement is used: a group of order n is cyclic if and only if for every divisor d of n the group has exactly one subgroup of order d.
http://en.wikipedia.org/wiki/Cyclic_group

And a bit more in relation to the Crystallographic Restriction Theorem...

Cyclic Group
via Wikipedia
In 2D and 3D the symmetry group for n-fold rotational symmetry is C_n, of abstract group type Z_n. In 3D there are also other symmetry groups which are algebraically the same, see Symmetry groups in 3D that are cyclic as abstract group...

Point Groups In Three Dimensions
Symmetry groups in 3D that are cyclic as abstract group
http://en.wikipedia.org/wiki/Point_groups_..._abstract_group

Symmetry groups in 3D that are dihedral as abstract group
http://en.wikipedia.org/wiki/Point_groups_..._abstract_group

Also...
Crystallographic Point Group
http://en.wikipedia.org/wiki/Crystallographic_point_group
(At bottom of page there is a link to a nice graphic of the 32 crystallographic point groups...)

- RF

P.S. Impostant to know: Cyclic Groups are intimately related to the Euler Totient Function.

[Moderator: Suspended 40 days.]
jeremyebert
QUOTE (Raphie Frank+Jul 29 2011, 03:10 PM)
Jeremy, I believe it would be helpful if you compiled a list of some of the relevant number progressions in regards to what you're doing, put them all in one place with the first several terms of each, and showed how they relate to one another. The number of progressions you are working with are starting to pile up and it's getting difficult to keep track...

And if it's getting difficult for me to keep track, how difficult do you think it will be for others?

- RF

Here is the number progressions I've been working with as of late:

sum_{k divides n} abs((n-k^2)/(2k)) = (1/2)*Sum_{d divides n} abs(n/d-d)

A079667 (1/2)*Sum_{d divides n} abs(n/d-d).
A006874 Mu-atoms of period n on continent of Mandelbrot set.
A009188 Short leg of more than one Pythagorean triangle.

A009188 00, 00, 00, 00, 00, 00, 00, 00, 09, 00, 00, 12, 00, 00, 15, 16, 00, 18, 00, 20,
A079667 00, 01, 02, 03, 04, 06, 06, 09, 08, 12, 10, 16, 12, 18, 16, 21, 16, 27, 18, 28,
A006874 01, 01, 02, 03, 04, 06, 06, 09, 10, 12, 10, 22, 12, 18, 24, 27, 16, 38, 18, 44,

The index of A009188 only misses on 01 and 41 for the values available on OEIS
jeremyebert
QUOTE (jeremyebert+Jul 30 2011, 12:50 PM)

The Dedekind eta function is truly a beautiful thing.

I wonder what it looks like whith the complex numbers sets:

Set{l}

Real part = 1-(k(2/(n+1))
Imaginary part = sqrt(4d(n+1-k))/(n+1)

Set{m}

Real part = ((n+1)/2) - k
Imaginary part = sqrt(k(n+1-k))

Set{n}

Real part = (n-k^2)/(2k)
Imaginary part = sqrt(n)

Where

n = 1--->LIM
k = 1--->n

RPenner,

Would this be considered a circle group?

Set{l}

Real part = 1-(k(2/(n+1))
Imaginary part = sqrt(4(n+1-k))/(n+1)

where

n = 1--->LIM
k = 1--->n

With the harmonic number relation you've shown it seems to tie in greatly with Pontryagin duality and Lie groups as well. Is this a correct view?

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

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

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

jeremyebert
Here is a link to the original 2D animation. I find it useful in determining some of the areas and ratios involved with this equation.
http://dl.dropbox.com/u/13155084/2D/Fourier.html
jeremyebert
Does anyone at least see the difference between squares being an odd number directly relates to a square wave?

http://en.wikipedia.org/wiki/Square_wave
jeremyebert
QUOTE (jeremyebert+Aug 2 2011, 09:10 PM)
Does anyone at least see the difference between squares being an odd number directly relates to a square wave?

http://en.wikipedia.org/wiki/Square_wave
jeremyebert
QUOTE (jeremyebert+Jul 20 2011, 04:15 PM)
I've mentioned this before but it seems that ((n-1)/(n+1))^(-n/2) converges pretty quickly to e.

RPenner,

Thanks again for all or you help. I have another question.

Since (n-1)/(n+1) ^(n/2) converges to e

which one can transform

(n-1)/(n+1)

into

SUM{k-->n}(n-k^2)/2k

directly relating to the harmonic sequence as you have shown,

shouldn't there be an exponential ^(n/2) link into the Euler–Mascheroni constant in addition to the

"( ln n + 2 (Euler-Mascheroni Constant) - 1)(n/4) - 1/24"

**"It is defined as the limiting difference between the harmonic series and the natural logarithm"?
rpenner
The relative factor of 2 arises from the change of variables:
u² = n
in
∑ (k=1 .. ⌊√n⌋) (n − k²)/(2 k)                                 {Expression A}
= ∑ (k=1 .. u) (u² − k²)/(2 k)
= (u²/2) ∑ (k=1 .. u) 1/k − (1/2) ∑ (k=1 .. u) k
= (u²/2) ∑ (k=1 .. u) 1/k − (u/4)(u+1)
= (u²/2) [ γ + ln u + 1/(2 u) − 1/(12 u²) + 1/(120 u⁴) − 1/(252 u⁶) + 1/(240 u⁸) − .... ] − (u/4) − u²/4
= u² γ / 2 − u²/4 + u^2 (ln u) / 2 + u/4 − u/4 − 1/24 + 1/(240 u²) − 1/(504 u⁴) + 1/(480 u⁶) − ...
= n γ / 2 − n/4 + n (ln n ) / 4 − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³) − ...
= (n/4)[ (2γ − 1) + ln n ] − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³) − ...
≈ (n/4)[ 2γ − 1 + ln n ] − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³)                                 {Expression B}
≈ (n/4)[ 2γ − 1 + ln n ] − 1/24                                 {Expression C}

Where γ is the Euler-Mascheroni constant and the term in the square brackets dates back to Euler.

http://www.physforum.com/index.php?showtop...80&#entry489635
As claimed, the following table illustrates that Expression C is very close to Expression A when n is a perfect square and large. Expression B is quite a lot better when 1 < n < 100 and square, but pretty complicated.
CODE

n        Expression A          Expression B        Expression C
1        0                  0.0012070+         -0.0030588+
4        1.5                1.5000092+          1.4990590+
9        5.25               5.2500004+          5.2495591+
16       11.6666667-        11.6666667+         11.6664135+
25       21.0416667-        21.0416667-         21.0415031-
100      118.9484127-       118.9484127-        118.9483712+
10000    23411.8875882-     23411.8875882-      23411.8875878-
1000000  3492485.4302752-   3492485.4302752-    3492485.4302752-
jeremyebert
QUOTE (rpenner+Aug 8 2011, 06:48 PM)
The relative factor of 2 arises from the change of variables:
u² = n
in
∑ (k=1 .. ⌊√n⌋) (n − k²)/(2 k)                                 {Expression A}
= ∑ (k=1 .. u) (u² − k²)/(2 k)
= (u²/2) ∑ (k=1 .. u) 1/k − (1/2) ∑ (k=1 .. u) k
= (u²/2) ∑ (k=1 .. u) 1/k − (u/4)(u+1)
= (u²/2) [ γ + ln u + 1/(2 u) − 1/(12 u²) + 1/(120 u⁴) − 1/(252 u⁶) + 1/(240 u⁸) − .... ] − (u/4) − u²/4
= u² γ / 2 − u²/4 + u^2 (ln u) / 2 + u/4 − u/4 − 1/24 + 1/(240 u²) − 1/(504 u⁴) + 1/(480 u⁶) − ...
= n γ / 2 − n/4 + n (ln n ) / 4 − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³) − ...
= (n/4)[ (2γ − 1) + ln n ] − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³) − ...
≈ (n/4)[ 2γ − 1 + ln n ] − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³)                                 {Expression B}
≈ (n/4)[ 2γ − 1 + ln n ] − 1/24                                 {Expression C}

Where γ is the Euler-Mascheroni constant and the term in the square brackets dates back to Euler.

http://www.physforum.com/index.php?showtop...80&#entry489635
As claimed, the following table illustrates that Expression C is very close to Expression A when n is a perfect square and large. Expression B is quite a lot better when 1 < n < 100 and square, but pretty complicated.
CODE

n        Expression A          Expression B        Expression C
1        0                  0.0012070+         -0.0030588+
4        1.5                1.5000092+          1.4990590+
9        5.25               5.2500004+          5.2495591+
16       11.6666667-        11.6666667+         11.6664135+
25       21.0416667-        21.0416667-         21.0415031-
100      118.9484127-       118.9484127-        118.9483712+
10000    23411.8875882-     23411.8875882-      23411.8875878-
1000000  3492485.4302752-   3492485.4302752-    3492485.4302752-

You never never cease to amaze me with your insight RPenner. So the similarities I'm seeing between the series:

k-->n
ABS( ( (n-k^2)/(2k) ) - ( (n-1)/2 ) )

here:

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

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

and the difference between terms:

http://dl.dropbox.com/u/13155084/constant_numb2.png

http://dl.dropbox.com/u/13155084/constant2.png

and the Euler-Mascheroni constant:

http://en.wikipedia.org/wiki/File:Gamma-area.svg

are explained by this?

jeremyebert
QUOTE (Raphie Frank+Jul 26 2011, 05:07 PM)
In a trivial way, the Pentagonal Numbers, insofar as they can be related to squares, are already embedded in your model in at least 4 obvious ways...

A ) Penta_(z - 2) = Binomial (n,4) = 0, 1, 5, 15, 35, 70...

70 - 15 = 55 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2
35 - 5 = 30 = 1^2 + 2^2 + 3^2 + 4^2

Therefore... (70 - 15) - (35 - 5) = 25 = 5^2 --> 4 Terms

B ) Penta_n = 1, 2, 5, 7, 12, 15, 22, 26, 35, 40

22 - 6 = 16 = 4^2
15 - 6 = 9 = 3^2
12 - 3 = 9 = 3^2
7 - 3 = 4 = 2^2
5 - 1 = 4 = 2^2
2 - 1 = 1 = 1^2
1 - 1 = 0 = 0^2

for 1, 3, 6, 10... are Triangular Numbers (Binomial (n, 2)), also expressible in terms of Pentagonal Numbers...

57 - 12 = 45 = T_9
51 - 15 = 36 = T_8
40 - 12 = 28 = T_7
26 - 5 = 21 = T_6
22 - 7 = 15 = T_5
15 - 5 = 10 = T_4
7 - 1 = 6 = T_3
5 - 2 = 3 = T_2
2 - 1 = 1 = T_1

Therefore... 35 - (15 - 5) = 25 = 5^2 --> 3 Terms

C ) Also, of course, as you already know, you can get squares and cubes from the + and - Pentagonal Pyramid Numbers...

40 + 24 = 64 = 4^3
40 - 24 = 16 = 4^2

75 + 50 = 125 = 5^3
75 - 50 = 25 = 5^2

D) All n in N are expressible as the difference between two Pentagonal Numbers. Squares are obviously a subset of that... --> 2 Terms

Whereas the Dirichlet Divisor Sum introduces a new variable every time you get to a square, Euler's Summing technique introduces a new variable every time you get to a Generalized Pentagonal Number.

- RF

Raphie/RPenner,
I just found a forumal that shows a direct dependency on the relationship between triangular numbers and pentagonal numbers in relation to the difference between terms of of my:

k-->n
ABS( ( (n-k^2)/(2k) ) - ( (n-1)/2 ) )

with this

http://dl.dropbox.com/u/13155084/constant2.png

http://dl.dropbox.com/u/13155084/constant_numb2.png

http://dl.dropbox.com/u/13155084/constant%20triangle.png

http://dl.dropbox.com/u/13155084/constant%...gle%20x%204.png

and 4 times a pentagonal pyramidal number

http://dl.dropbox.com/u/13155084/divide%20...gle%20x%204.png

jeremyebert
QUOTE (jeremyebert+Aug 9 2011, 09:12 PM)
Raphie/RPenner,
I just found a forumal that shows a direct dependency on the relationship between triangular numbers and pentagonal numbers in relation to the difference between terms of of my:

k-->n
ABS( ( (n-k^2)/(2k) ) - ( (n-1)/2 ) )

with this

http://dl.dropbox.com/u/13155084/constant2.png

http://dl.dropbox.com/u/13155084/constant_numb2.png

http://dl.dropbox.com/u/13155084/constant%20triangle.png

http://dl.dropbox.com/u/13155084/constant%...gle%20x%204.png

and 4 times a pentagonal pyramidal number

http://dl.dropbox.com/u/13155084/divide%20...gle%20x%204.png

This is pretty simple but to me it shows some interesting relationships.

basically

k=1...n

ABS( ( (n-k^2)/(2k) ) - ( (n-1)/2 ) ) {Sequence A}

= (k-1) + ( (n-k) * ( (k-1) / (2(k-1) + 2) ) ) {Sequence B}

= (k-1) + ( (n-k) * ( ((k-1)/2)^2 / T_(k-1) ) ) {Sequence C}

= (k-1) + ( (n-k) * ( (k-1)^2 / 4T_(k-1) ) ) {Sequence D}

where T_(k-1) is the kth -1 Triangular number

ex. n=6

{Sequence D}
00 + (05 * 00/00) = 0.00
01 + (04 * 01/04) = 2.00
02 + (03 * 04/12) = 3.00
03 + (02 * 09/24) = 3.75
04 + (01 * 16/40) = 4.40
05 + (00 * 25/60) = 5.00

{Sequence C}
00 + (05 * 0.00/00) = 0.00
01 + (04 * 0.25/01) = 2.00
02 + (03 * 1.00/03) = 3.00
03 + (02 * 2.25/06) = 3.75
04 + (01 * 4.00/10) = 4.40
05 + (00 * 6.25/15) = 5.00

{Sequence B}
00 + (05 * 0.00/02) = 0.00
01 + (04 * 1.00/04) = 2.00
02 + (03 * 2.00/06) = 3.00
03 + (02 * 3.00/08) = 3.75
04 + (01 * 4.00/10) = 4.40
05 + (00 * 5.00/12) = 5.00
jeremyebert
QUOTE (jeremyebert+Aug 10 2011, 01:30 PM)
This is pretty simple but to me it shows some interesting relationships.

basically

k=1...n

ABS( ( (n-k^2)/(2k) ) - ( (n-1)/2 ) ) {Sequence A}

= (k-1) + ( (n-k) * ( (k-1) / (2(k-1) + 2) ) ) {Sequence B}

= (k-1) + ( (n-k) * ( ((k-1)/2)^2 / T_(k-1) ) ) {Sequence C}

= (k-1) + ( (n-k) * ( (k-1)^2 / 4T_(k-1) ) ) {Sequence D}

where T_(k-1) is the kth -1 Triangular number

ex. n=6

{Sequence D}
00 + (05 * 00/00) = 0.00
01 + (04 * 01/04) = 2.00
02 + (03 * 04/12) = 3.00
03 + (02 * 09/24) = 3.75
04 + (01 * 16/40) = 4.40
05 + (00 * 25/60) = 5.00

{Sequence C}
00 + (05 * 0.00/00) = 0.00
01 + (04 * 0.25/01) = 2.00
02 + (03 * 1.00/03) = 3.00
03 + (02 * 2.25/06) = 3.75
04 + (01 * 4.00/10) = 4.40
05 + (00 * 6.25/15) = 5.00

{Sequence B}
00 + (05 * 0.00/02) = 0.00
01 + (04 * 1.00/04) = 2.00
02 + (03 * 2.00/06) = 3.00
03 + (02 * 3.00/08) = 3.75
04 + (01 * 4.00/10) = 4.40
05 + (00 * 5.00/12) = 5.00

So basically we have a ratio of a number squared divided by 4 times its Triangular number:

n^2 / 4T_n

So from this we could say every time

(n-k) * (k-1)^2 / (4 * T_(k-1)) = 0 mod 0.5

k is a divisor of n

all half integer results are also divisible by 2

or cleaned up a bit

(n-k) * (k-1)^2 / (2k(k-1)) = 0 mod 0.5

jeremyebert
QUOTE (jeremyebert+Aug 10 2011, 01:30 PM)
This is pretty simple but to me it shows some interesting relationships.

basically

k=1...n

ABS( ( (n-k^2)/(2k) ) - ( (n-1)/2 ) )                         {Sequence A}

= (k-1) + (  (n-k) * ( (k-1) / (2(k-1) + 2) )  )           {Sequence B}

= (k-1) + (  (n-k) * ( ((k-1)/2)^2  / T_(k-1) )  )       {Sequence C}

= (k-1) + (  (n-k) * ( (k-1)^2  / 4T_(k-1) )  )           {Sequence D}

where T_(k-1) is the kth -1 Triangular number

ex. n=6

{Sequence D}
00 + (05 * 00/00) = 0.00
01 + (04 * 01/04) = 2.00
02 + (03 * 04/12) = 3.00
03 + (02 * 09/24) = 3.75
04 + (01 * 16/40) = 4.40
05 + (00 * 25/60) = 5.00

{Sequence C}
00 + (05 * 0.00/00) = 0.00
01 + (04 * 0.25/01) = 2.00
02 + (03 * 1.00/03) = 3.00
03 + (02 * 2.25/06) = 3.75
04 + (01 * 4.00/10) = 4.40
05 + (00 * 6.25/15) = 5.00

{Sequence B}
00 + (05 * 0.00/02) = 0.00
01 + (04 * 1.00/04) = 2.00
02 + (03 * 2.00/06) = 3.00
03 + (02 * 3.00/08) = 3.75
04 + (01 * 4.00/10) = 4.40
05 + (00 * 5.00/12) = 5.00

another trivial observation:

k=1...n

when k*(k-1) = n-k

then

k=sqrt(n)

simple but it shows when

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

then

k=sqrt(n)

which indirectly connects "some how" to the exponential relation n^(1/2) = sqrt(n).

RPenner, do you see the relation I'm trying to draw?
jeremyebert
RPenner/Raphie,
I've been looking at a way to describe the relation between a square lattice and my parabolic lattice as a type of "Stress–energy tensor”.
If you could just reply with something constructive even if it is just a "No, I don't see the connection, that is quite a stretch." I just need to know if how I'm learning (teaching myself, with your help) is close to right.

related info:

http://en.wikipedia.org/wiki/Mode_of_vibra...#Elastic_solids

and

http://en.wikipedia.org/wiki/Lattice_vibration
jeremyebert
QUOTE (jeremyebert+Aug 14 2011, 06:38 PM)
RPenner/Raphie,
I've been looking at a way to describe the relation between a square lattice and my parabolic lattice as a type of "Stress–energy tensor”.
If you could just reply with something constructive even if it is just a "No, I don't see the connection, that is quite a stretch." I just need to know if how I'm learning (teaching myself, with your help) is close to right.

related info:

http://en.wikipedia.org/wiki/Mode_of_vibra...#Elastic_solids

and

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

The term (1/2)hυ represents the "zero-point energy",

can't the zero point also be at the nodes of a standing wave?

jeremyebert
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
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

RPenner,

Do you not see a direct relation to the "simplex simplex picking" equation?

http://dl.dropbox.com/u/13155084/Pythagorean%20lattice.pdf

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

http://mathworld.wolfram.com/SimplexSimplexPicking.html

I need to know.
jeremyebert
QUOTE (jeremyebert+Aug 2 2011, 12:44 PM)
RPenner,

The Dedekind eta function is truly a beautiful thing.

I wonder what it looks like whith the complex numbers sets:

Set{l}

Real part =  1-(k(2/(n+1))
Imaginary part = sqrt(4d(n+1-k))/(n+1)

Set{m}

Real part =  ((n+1)/2) - k
Imaginary part = sqrt(k(n+1-k))

Set{n}

Real part =  (n-k^2)/(2k)
Imaginary part = sqrt(n)

Where

n = 1--->LIM
k = 1--->n

Feels like cheatin' but it answers a lot of questions...
http://dl.dropbox.com/u/13155084/dedekindeta.png

http://www.wolframalpha.com/

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

Looking at it exponentially bares interesting results.

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

Looking at it exponentially bares very interesting results.

n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )
jeremyebert
QUOTE (jeremyebert+Aug 30 2011, 06:53 PM)
Looking at it exponentially bares very interesting results.

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

RPenner,

why is it that the largest value of n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) ) seems to always be when k=36?
jeremyebert
QUOTE (jeremyebert+Aug 30 2011, 07:36 PM)
RPenner,

why is it that the largest value of n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1)  ) seems to always be when k=36?

Well, I feel real dumb... Nevermind this post, I had an issue with the application I was using to compute the results. Sorry to all who spent any time on this.
jeremyebert
QUOTE (jeremyebert+Aug 30 2011, 07:36 PM)
RPenner,

why is it that the largest value of n^( (((n-k)*(k-1)/(2k)) + (k-1))/(n-1) ) seems to always be when k=36?

Interesting function none the less.

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

n^(t)

when k=1 then t=0 and n^(t)=k

when k=n^(1/2) then t=0.5 and n^(t)=k

when k=n then t=1 and n^(t)=k
jeremyebert
QUOTE (jeremyebert+Aug 31 2011, 03:23 PM)
Interesting function none the less.

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

n^(t)

when k=1 then t=0 and n^(t)=k

when k=n^(1/2) then t=0.5 and n^(t)=k

when k=n then t=1 and n^(t)=k

WolframAlpha shows some interesting results at well. The series expansion shows terms related to the first 6 double factorial numbers.

http://www.wolframalpha.com/input/?i=n%5E%...%2F%28n-1%29%29
jeremyebert
QUOTE (jeremyebert+Sep 1 2011, 07:39 AM)
WolframAlpha shows some interesting results at well. The series expansion shows terms related to the first 6 double factorial numbers.

http://www.wolframalpha.com/input/?i=n%5E%...%2F%28n-1%29%29

Slight tweak shows a series expansion built from double factorial numbers:

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

http://www.wolframalpha.com/input/?i=n%5E%...8k-1%29%29+%29+
jeremyebert
QUOTE (jeremyebert+Sep 1 2011, 05:38 PM)
Looking at its deviation from k is very interesting:

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

http://dl.dropbox.com/u/13155084/nt-k-4.png
http://dl.dropbox.com/u/13155084/nt-k-9.png
http://dl.dropbox.com/u/13155084/nt-k-16.png
http://dl.dropbox.com/u/13155084/nt-k-25.png
http://dl.dropbox.com/u/13155084/nt-k-36.png
http://dl.dropbox.com/u/13155084/nt-k-49.png

so basically the roots of the function:

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

are

k=1
k=n^(1/2)
k=n

Is this a correct statement Rpenner?
jeremyebert
More WolframAlpha:

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

http://www.wolframalpha.com/input/?i=cos%2...%2F%28n-1%29%29
Maxila
Hi jermyebert:

Maxila
jeremyebert
QUOTE (Maxila+Sep 6 2011, 12:04 PM)
Hi jermyebert:

Maxila

Maxila,
Yes, it is quite discouraging. I did have quite a nice dialog going with RPenner and Raphie but I believe some egos seeped into the conversations resulting in RPenner suspending Raphie from the forum. If you have any suggestions on where to go for productive dialog I would be very grateful. As I have mentioned before, all I am trying to do teach myself. It's very hard to do that without 2-way communication.

Thanks again for your response Maxila.

Jeremy
jeremyebert
QUOTE (jeremyebert+Sep 6 2011, 02:28 PM)
Maxila,
Yes, it is quite discouraging. I did have quite a nice dialog going with RPenner and Raphie but I believe some egos seeped into the conversations resulting in RPenner suspending Raphie from the forum. If you have any suggestions on where to go for productive dialog I would be very grateful. As I have mentioned before, all I am trying to do teach myself. It's very hard to do that without 2-way communication.

Thanks again for your response Maxila.

Jeremy
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

The divisor symmetry still shows up nicely. For example 36:

(ln(x)/ln(36)) - (((36-x)*(x-1)/(2*x)) + (x-1))/(36-1))

http://dl.dropbox.com/u/13155084/36.png
jeremyebert
QUOTE (jeremyebert+Sep 4 2011, 04:16 PM)
so basically the roots of the function:

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

are

k=1
k=n^(1/2)
k=n

Is this a correct statement Rpenner?

for the function

f(n,k) = ( (ln(k)/ln(n)) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

the

local minimum = (((n-1)/2)-sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))
local maximum = (((n-1)/2)+sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))

ex: n=49

49 minimum = (24-sqrt(576-49 log^2(7)))/(log(7))
49 maximum = (24+sqrt(576-49 log^2(7)))/(log(7))

and min * max = n
Raphie Frank
QUOTE (Raphie Frank+Aug 2 2011, 04:00 AM)
P.S. Impostant to know: Cyclic Groups are intimately related to the Euler Totient Function.

[Moderator: Suspended 40 days.]

I'm really, really sorry for the typo RPenner. "Impostant" should have read "Important."

Best,
Raphie Frank
jeremyebert
the contour plot shows the divisor function very nicely:

(n-k^2)/2k mod .5

http://www.wolframalpha.com/input/?i=Conto...C+-4%2C+4%7D%5D
jeremyebert
QUOTE (jeremyebert+Sep 14 2011, 05:03 PM)
the contour plot shows the divisor function very nicely:

(n-k^2)/2k mod .5

http://www.wolframalpha.com/input/?i=Conto...C+-4%2C+4%7D%5D

for those who have problems with wolframalpha.com

http://dl.dropbox.com/u/13155084/mod%20point%205.png
jeremyebert
QUOTE (Raphie Frank+Aug 1 2011, 11:00 PM)
Some relevant information for you Jeremy...

Cyclic Group
via Wikipedia
The fundamental theorem of cyclic groups states that if G is a cyclic group of order n then every subgroup of G is cyclic. Moreover, the order of any subgroup of G is a divisor of n and for each positive divisor k of n the group G has exactly one subgroup of order k. This property characterizes finite cyclic groups: a group of order n is cyclic if and only if for every divisor d of n the group has at most one subgroup of order d. Sometimes the refined statement is used: a group of order n is cyclic if and only if for every divisor d of n the group has exactly one subgroup of order d.
http://en.wikipedia.org/wiki/Cyclic_group

And a bit more in relation to the Crystallographic Restriction Theorem...

Cyclic Group
via Wikipedia
In 2D and 3D the symmetry group for n-fold rotational symmetry is C_n, of abstract group type Z_n. In 3D there are also other symmetry groups which are algebraically the same, see Symmetry groups in 3D that are cyclic as abstract group...

Point Groups In Three Dimensions
Symmetry groups in 3D that are cyclic as abstract group
http://en.wikipedia.org/wiki/Point_groups_..._abstract_group

Symmetry groups in 3D that are dihedral as abstract group
http://en.wikipedia.org/wiki/Point_groups_..._abstract_group

Also...
Crystallographic Point Group
http://en.wikipedia.org/wiki/Crystallographic_point_group
(At bottom of page there is a link to a nice graphic of the 32 crystallographic point groups...)

- RF

P.S. Impostant to know: Cyclic Groups are intimately related to the Euler Totient Function.

[Moderator: Suspended 40 days.]

As well as the DFT which I have mentioned in previous threads.

http://www.katjaas.nl/rootsofunity/rootsofunity.html
jeremyebert
Really nice paper on the parabolas created by Pythagorean triples. It coincides nicely with the relations I've been trying to point out.

http://conservancy.umn.edu/bitstream/4878/1/438.pdf

http://dl.dropbox.com/u/13155084/2D/Fourier.html

http://dl.dropbox.com/u/13155084/Pythagorean%20lattice.pdf
Raphie Frank
The very irrational sword of Damocles prohibits (or, rather, "strongly suggestively" suggests against...) any reasonable or rational response to your exploratory formulations, Jeremy. FYI.

For the benefit of "Damocles," (aka "The Moderator") and for the PUBLIC record, would you mind commenting on how it is you initially came to be interested in the Dirichlet Divisor function in relation to your model?

Furthermore, would you care to provide a list of all the professional mathematicians who have come forward to verify the correctness of your insights?

Best,
Raphie Frank
917-202-2610
100 Metropolitan Ave. # 6
Kings County, New York.
jeremyebert
QUOTE (Raphie Frank+Sep 22 2011, 01:20 AM)
The very irrational sword of Damocles prohibits (or, rather, "strongly suggestively" suggests against...) any reasonable or rational response to your exploratory formulations, Jeremy. FYI.

For the benefit of "Damocles," (aka "The Moderator") and for the PUBLIC record, would you mind commenting on how it is you initially came to be interested in the Dirichlet Divisor function in relation to your model?

Furthermore, would you care to provide a list of all the professional mathematicians who have come forward to verify the correctness of your insights?

Best,
Raphie Frank
917-202-2610
100 Metropolitan Ave. # 6
Kings County, New York.

While searching for answers on another forum, you pointed me right to it Raphie.

4-21-2011

"Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.

Best,
Raphie"

As for verification from professional mathematicians, that list is lacking.
jeremyebert
QUOTE (jeremyebert+Sep 22 2011, 07:04 PM)
While searching for answers on another forum, you pointed me right to it Raphie.

4-21-2011

"Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.

Best,
Raphie"

As for verification from professional mathematicians, that list is lacking.

My model shows:

Dirichlet Divisor function

sum ((2*floor[(n - k^2)/k]) + 1) where k=1 to floor[sqrt(n)]

= A006218 http://oeis.org/A006218

sum ( ((n-k^2)/(2k)) - (((n-k^2)/(2k)) mod .5) ) * -2 where k=1 to n

= A161664 http://oeis.org/A161664

I would think this would be of great interest to professional mathematicians, maybe not...

jeremyebert
QUOTE (jeremyebert+Sep 23 2011, 10:26 AM)
My model shows:

Dirichlet Divisor function

sum ((2*floor[(n - k^2)/k]) + 1) where k=1 to floor[sqrt(n)]

= A006218 http://oeis.org/A006218

sum ( ((n-k^2)/(2k)) - (((n-k^2)/(2k)) mod .5) ) * -2 where k=1 to n

= A161664 http://oeis.org/A161664

I would think this would be of great interest to professional mathematicians, maybe not...

Borrowing from Rpenner ‘s insight on Harmonic Numbers…

(-1/2 (2 n H(n)-n^2-n) ) + (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

Dirichlet Divisor function =
n H(n) - (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

So I guess I’m open to suggestions on what to call the function:

f(n) = sum ( (((n-k^2)/(2k)) mod .5) ) * 2 where k=1 to n
jeremyebert
QUOTE (jeremyebert+Sep 24 2011, 09:08 AM)
Borrowing from Rpenner ‘s insight on Harmonic Numbers…

(-1/2 (2 n H(n)-n^2-n) ) + (sum (  (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

Dirichlet Divisor function =
n H(n)  - (sum (  (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

So I guess I’m open to suggestions on what to call the function:

f(n) = sum (  (((n-k^2)/(2k)) mod .5) ) * 2 where k=1 to n

This might be easier to understand for some:

tau(n) = sigma(0,n)

c(n) = sum((n-k^2)/(2k)) * -2 where k = 1 to n

H(n) = nth Harmonic Number

T(n) = nth Triangular Numner

j(n) = sum((n-k^2)/(2k) mod .5) * 2 where k = 1 to n

non-divisor base = c(n)

divisor base = n*H(n)

c(n) + j(n) = Cicada function

n*H(n) - j(n) = sum(tau(n))

n*H(n) + c(n) = T(n)
jeremyebert
I need some help guys.... I'm looking for a way to get a single formula for the sigma function. So far I have got it down to 2.

sigma(0,n) = ceiling [ SUM(2*(((((n-u)-k^2)/(2k)) mod .5) - (((n-k^2)/(2k)) mod .5 ))) ] where k=1 to n, u = 0.000001/n

sigma(x,n) = ceiling [ SUM(((2*k*((((n-u)-k^2)/(2k)) mod .5)) - (n mod k))^x) ] where k=1 to n, u = 0.0000001/n, x>0
jeremyebert
QUOTE (Raphie Frank+Sep 22 2011, 01:20 AM)
The very irrational sword of Damocles prohibits (or, rather, "strongly suggestively" suggests against...) any reasonable or rational response to your exploratory formulations, Jeremy. FYI.

For the benefit of "Damocles," (aka "The Moderator") and for the PUBLIC record, would you mind commenting on how it is you initially came to be interested in the Dirichlet Divisor function in relation to your model?

Furthermore, would you care to provide a list of all the professional mathematicians who have come forward to verify the correctness of your insights?

Best,
Raphie Frank
917-202-2610
100 Metropolitan Ave. # 6
Kings County, New York.

I'm trying to formalize my ideas here guys...

I think my parabolic version of the D(n) on a square lattice equals the Pythagorean Theorem basically...
( c - b ) < a
( c + b ) > a
( c - b ) * ( c + b ) = a^2 , where a^2 = n an element of Z. This forms the lattice points Hyperbola description for the Dirichlet Divisor function.

image:
http://dl.dropbox.com/u/13155084/wolframal...28191510717.gif

I love wolfram by the way.
Is this correct statement?
jeremyebert
QUOTE (jeremyebert+Sep 29 2011, 03:36 PM)
I'm trying to formalize my ideas here guys...

I think my parabolic version of the D(n) on a square lattice equals the Pythagorean Theorem basically...
( c - b ) < a
( c + b ) > a
, where a^2 = n an element of Z. This forms the lattice points Hyperbola description for the Dirichlet Divisor function.

image:
http://dl.dropbox.com/u/13155084/wolframal...28191510717.gif

I love wolfram by the way.
Is this correct statement?

( c - b ) * ( c + b ) = a^2 (3D Cone)

and

( c - b ) * ( c + b ) = z (hyperbolic paraboloid)

jeremyebert
QUOTE (jeremyebert+Sep 29 2011, 04:17 PM)

( c - b ) * ( c  + b )  = a^2 (3D Cone)

and

( c - b ) * ( c  + b )  = z (hyperbolic paraboloid)

jeremyebert
QUOTE (jeremyebert+Sep 29 2011, 04:17 PM)

( c - b ) * ( c + b ) = a^2 (3D Cone)

and

( c - b ) * ( c + b ) = z (hyperbolic paraboloid)

c and b are a set of integers and half integers

c = Z and Z+1/2
b = Z and Z+1/2
a = R
n = Z
Raphie Frank
QUOTE (jeremyebert+Sep 29 2011, 09:45 PM)
the multiplication table:

http://en.wikipedia.org/wiki/Paraboloid#Multiplication_table

sweet:

For z = x^2/a^2 - y^2/b^2...

Note the special case for a = b = sqrt 2 (The Pythagorean Constant):

(x^2 - y^2)/2...

... which, as the Wikipedia Page notes is congruent to the surface z = xy.

Observations:
A ) The convergents to the square root of two are Pell sequence related (related also to "The Silver Ratio" = 1 + sqrt 2).
B ) The covering radius of the Leech Lattice is sqrt 2.

- RF

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

P.S. Create p^n, q^(k-n) triangles based on powers of...
(0,0)
(0,1)
(1,1)
(1,2)
(2,3)

And then take the triangle sums of their associated divisors. In order you will get...

Binomial (n+0,0), Binomial (n+1,1), Binomial (n+2,2), Binomial (n+3,3), Binomial (n+4,4)

All are multiplication tables.

e.g.

p^n, q^(k-n); p = 2, q = 3

01
02 03
04 06 09
08 12 18 27
16 24 36 54 81

Divisors
---------
01
02 02
03 04 03
04 06 06 04
05 08 09 08 05

Triangle Sum = Binomial(n+4, 4)

Not so for (2,2), the p^n, p^(k-n) triangle, the shallow diagonals of which sum to Generalized Pentagonal Numbers.

Specific values for p and q only become relevant when you calculate sigma(n) or totient(n), but general relationships remain scale-free. For example, for any p in N, then the sum of:

p^2 + totient (p^2) + sigma(p^2) + d(p^2) = 3p^2 + 4

e.g. 11^2 + totient (11^2) + sigma(11^2) + d(11^2) = 121 + 110 + 133 + 3 = 367 = 3*11^2 + 4

For lack of better terminology, I call this sum the "Gross Arithmetic Sum" [GAS(x)], as opposed, for instance, to the "Gross Polygonal Sum" [GPS(x)] which I use to refer to the maximal number of combined points, intersections, connections and areas of a fully connected polygon.

e.g.
Let y = GAS(x); x = p^2
Then y = 3x + 4
Raphie Frank
QUOTE (jeremyebert+Sep 23 2011, 12:04 AM)
While searching for answers on another forum, you pointed me right to it Raphie.

4-21-2011

"Jeremy, given the manner of observations you are reporting, I very much believe you would behoove yourself, contextually speaking, to at least begin to familiarize yourself with arithmetic functions. For instance, check out the Dirichlet Divisor function. A little research will make it manifest the relationship between this, the Riemann Hypothesis/Zeta Function and Lattice Points under a hyperbola.

Best,
Raphie"

As for verification from professional mathematicians, that list is lacking.

Thank you for this, Jeremy.

Best,
Raphie
jeremyebert
QUOTE (Raphie Frank+Oct 1 2011, 12:55 PM)
For z = x^2/a^2 - y^2/b^2...

Note the special case for a = b = sqrt 2 (The Pythagorean Constant):

(x^2 - y^2)/2...

... which, as the Wikipedia Page notes is congruent to the surface z = xy.

Observations:
A ) The convergents to the square root of two are Pell sequence related (related also to "The  Silver Ratio" = 1 + sqrt 2).
B ) The covering radius of the Leech Lattice is sqrt 2.

- RF

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

P.S. Create p^n, q^(k-n) triangles based on powers of...
(0,0)
(0,1)
(1,1)
(1,2)
(2,3)

And then take the triangle sums of their associated divisors. In order you will get...

Binomial (n+0,0), Binomial (n+1,1), Binomial (n+2,2), Binomial (n+3,3), Binomial (n+4,4)

All are multiplication tables.

e.g.

p^n, q^(k-n); p = 2, q = 3

01
02 03
04 06 09
08 12 18 27
16 24 36 54 81

Divisors
---------
01
02 02
03 04 03
04 06 06 04
05 08 09 08 05

Triangle Sum = Binomial(n+4, 4)

Not so for (2,2), the p^n, p^(k-n) triangle, the shallow diagonals of which sum to Generalized Pentagonal Numbers.

Specific values for p and q only become relevant when you calculate sigma(n) or totient(n), but general relationships remain scale-free. For example, for any p in N, then the sum of:

p^2 + totient (p^2) + sigma(p^2) + d(p^2) = 3p^2 + 4

e.g. 11^2 + totient (11^2) + sigma(11^2) + d(11^2) = 121 + 110 + 133 + 3 = 367 = 3*11^2 + 4

For lack of better terminology, I call this sum the "Gross Arithmetic Sum" [GAS(x)], as opposed, for instance, to the "Gross Polygonal Sum" [GPS(x)] which I use to refer to the maximal number of combined points, intersections, connections and areas of a fully connected polygon.

e.g.
Let y = GAS(x); x = p^2
Then y = 3x + 4

Raphie,
I’ve mentioned this on another forum but the 45 degree intersection of the parabolas in my model fall at 1+sqrt(2))*n.
As for the p.i.c.a. and GAS equation, I find it very interesting.

jeremyebert
QUOTE (jeremyebert+Oct 3 2011, 07:32 PM)
Raphie,
I’ve mentioned this on another forum but the 45 degree intersection of the parabolas in my model fall at 1+sqrt(2))*n.
As for the p.i.c.a. and GAS equation, I find it very interesting.

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

http://dl.dropbox.com/u/13155084/45.png
Raphie Frank
QUOTE (jeremyebert+Oct 4 2011, 03:03 PM)
Using my Pythagorean Lattice value (1+sqrt(2))*n)^2

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

The Silver Ratio can be thought of as a special case of the following:

(sqrt (n^2 + 4) + n))/2

For n = 1, the Golden Ratio (sqrt (1^2 + 4) + 1)/2
For n = 2, the Silver Ratio (sqrt (2^2 + 4) + 2)/2
For n = 3, the Bronze Ratio (sqrt (3^2 + 4) + 3)/2

Coinciding, respectively, with the recursive construction rules

1A + 1B = C
1A + 2B = C
1A + 3B = C

The form n^2 + 4 on OEIS
http://oeis.org/A087475

All elements of such number progressions can be partitioned quite sensibly into multiples of Binomial Coefficients.

For example...

I.
Triangle with Construction Rule 1A + 1B = C
-- Shallow diagonals sum to the Fibonacci Series
-- Rows sum to Powers of 2

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

II.
Triangle with Construction Rule 1A + 2B = C
-- Shallow diagonals sum to the Pell Numbers
-- Rows Sum to Powers of 3

01
02 01
04 04 01
08 12 06 1
16 32 24 8 1

III.
Triangle with Construction Rule 1A + 3B = C
-- Shallow diagonals sum to the ? Numbers
-- Rows Sum to Powers of 4

001
003 001
009 006 001
027 027 009 001
081 108 054 012 001

As you may well have noticed, the leftmost value in each triangle is equal to the row sum of the triangle with construction rule 1A + (n-1)B, beginning with a triangle with the construction rule 1A + 0B...

-- Shallow diagonals sum to Grandi's Series
-- Rows Sum to Powers of 1

1
0 1
0 0 1
0 0 0 1
0 0 0 0 1

As such, all these triangles from Pascal's Triangle on up can be generated recursively from the Binomial Coefficients.

e.g.
Terms from Row 4 of Triangle II. above...
1 * (1 + 4 + 6 + 4 + 1) = 16
4 * (1 + 3 + 3 + 1) = 32
6 * (1 + 2 + 1) = 24
4 * ( 1 + 1) = 8
1 * (1) = 1

1 st Term of Row 4 of Triangle III above...
16 + 32 + 24 + 8 + 1 = 81 (= 3^4)

All powers of all n in N can thusly be constructed although it would get quite cumbersome quite fast. Nevertheless, it becomes quite apparent that the individual terms of the Riemann Zeta Function can be constructed entirely from Binomial Coefficients utilizing only basic binary operations.

Riemann Zeta Function
http://en.wikipedia.org/wiki/Riemann_zeta_function

- RF
Raphie Frank
QUOTE (Raphie Frank+Oct 5 2011, 09:38 PM)
For n = 1, the Golden Ratio (sqrt (1^2 + 4) + 1)/2
For n = 2, the Silver Ratio  (sqrt (2^2 + 4) + 2)/2
For n = 3, the Bronze Ratio (sqrt (3^2 + 4) + 3)/2

Coinciding, respectively, with the recursive construction rules

1A + 1B = C
1A + 2B = C
1A + 3B = C

Of course, a far simpler way to recursively generate powers of n is to apply the recursive construction rule:

0A + nB = C

A_0 = 0
B_0 = 1

e.g.

0A + 2B
[0 1] 2 4 8 16 32...

0A + 3B
[0 1] 3 9 27 81 243

etc.

But then you miss the relationship with the Binomial Coefficients and number progressions such as the Fibonacci and Pell Series...

- RF

P.S. Just for fun, instead of 0A + nB = C, try the construction rule nA + 0B = C.

P.P.S. Also see in relation to this post and the prior one: Fibonacci Polynomials http://mathworld.wolfram.com/FibonacciPolynomial.html
jeremyebert
QUOTE (jeremyebert+Oct 4 2011, 10:03 AM)
Using my Pythagorean Lattice value (1+sqrt(2))*n)^2

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

Here is a scatter plot of the recursion algorithm.

http://dl.dropbox.com/u/13155084/divisor%20invert.bmp
jeremyebert
QUOTE (Raphie Frank+Oct 5 2011, 04:38 PM)

All powers of all n in N can thusly be constructed although it would get quite cumbersome quite fast. Nevertheless, it becomes quite apparent that the individual terms of the Riemann Zeta Function can be constructed entirely from Binomial Coefficients utilizing only basic binary operations.

The Generalized Harmonic Numbers converge to the Zeta function as well.
H(n,m) = Zeta(m)

http://en.wikipedia.org/wiki/Harmonic_numb...armonic_numbers

Does the Binomial Coefficient method work for Complex Exponentials?
Raphie Frank
QUOTE (jeremyebert+Oct 7 2011, 01:57 PM)
Does the Binomial Coefficient method work for Complex Exponentials?

Unfortunately, I don't have an answer to offer you at the moment, assuming the question you are asking is whether or not Complex Exponentials can be constructed from Binomial Coefficients. It's not something I've given any thought to.

But here's a paper (mentioned on the page you linked to) that you might want to take a look at...

Arthur T. Benjamin, Gregory O. Preston, Jennifer J. Quinn, A Stirling Encounter with Harmonic Numbers, (2002) Mathematics Magazine, 75 (2) pp 95–103.
http://www.math.hmc.edu/~benjamin/papers/harmonic.pdf

Also, see the section on the page you linked to:
"Generalization to the complex plane"

- RF
Raphie Frank
QUOTE (Raphie Frank+Oct 5 2011, 10:40 PM)
Of course, a far simpler way to recursively generate powers of n is to apply the recursive construction rule:

0A + nB = C

A_0 = 0
B_0 = 1

e.g.

0A + 2B
[0 1] 2 4 8 16 32...

0A + 3B
[0 1] 3 9 27 81 243

And with a little bit of modification, you can also derive factorial numbers recursively using a similar form...

POWERS OF TWO
A_0 = 1
B_0 = 1

0A + (2 + 0n)B = C
[1 1] 2 4 8 16 32...
SUM (0A + (2 + 0n)B )
[1 2] 4 8 16 32 64...

SUMMATION OF |n - 1|!
A_0 = 1
B_0 = 1

0A + (1 + 1n)B = C
[1 1] 1 2 6 24 120...
SUM (0A + (1 + 1n)B )
[1 2] 3 5 11 35 155...

A kind of cool little property of the summation above (for the range shown above only) is this:

pi (1, 2, 3, 5, 11, 35, 155) = 0, 1, 2, 3, 5, 11, 36

pi (36) = 11
pi (11) = 5
pi (5) = 3
pi (3) = 2
pi (2) = 1
pi (1) = 0

Also, FWIW, which may not be much, {0, 1, 2, 3, 5, 11, 36} + 1 = {1, 2, 3, 4, 6, 12, 37} are, in order, the first 7 divisors of 444 (= p_12 *12 = 37*pi(37))

[n | 444] = 1, 2, 3, 4, 6, 12, 37, 74, 111, 148, 222, 444

- RF

Note: 444/1 = 444, 444/2 = 222, 444/3 = 148, 444/4 = 111, 444/6 = 74, 444/12 = 37, 444/37 = 12
Raphie Frank
Hi Jeremy. Extending the logic of the previous two posts, the following mathematical statement appears to hold...

Let...

C = xA + yB

Note: C = xA + yB is the form of the linear Diophantine Equation

A and B are recursive "anchors" (...and are elements of Progression C from which subsequent terms are constructed. Think of them as a(n-1) and a(n))
x and y are "anchor coefficients"

Let...

A = T((r -1),(p - 1))
B = T((r -1),(p - 0))
C = T(r,p)
x = (r^b )
y = n in N

Where...

b = 0 or 1 (aka a "binary switch" --> "yes" = 1 or "no" = 0)
r = row number of any Pascal-type triangle (START: r = 0)
p = position within row (START: p = 0)
T(r, p) = Triangle of r and p

Then..

for...
A_0 = 0 (or the Null Set)
B_0 = 1 = T(r,p) for r and p = 0

... and b = 0, then...
I. ROWSUM (C = xA + yB) = (y + 1)^n
(CONSTANT COEFFICIENTS)

... and for b = 1, then
II. ROWSUM (C = xA + yB) = r!*y-hedral_(r+1)
("SLIDING" x COEFFICIENT)

e.g.
2-hedral_(r+1) = 1, 3, 6, 10, 15, 21... = T_(n+1) (Triangular Numbers)
3-hedral_(r+1) = 1, 4, 10, 20, 35, 56... = Tetra_(n+1) (Tetrahedral Numbers)

For the special case of y = 1, then...

for b = 0, (C = xA + yB) --> PASCAL'S TRIANGLE
for b = 1, (C = xA + yB) --> (REVERSED) STIRLING I TRIANGLE
(Unsigned)

A130534 Stirling I Triangle see: http://oeis.org/A130534/table

Take the Product of I & II and you get a nice little, compact, all-in-one, recursively based mathematical description for sets of integers divisible by Powers of n, Factorials and Binomial Coefficients.

- RF

============================================================
RELATED PAPER
Finite, closed-form expressions for the partition function and for Euler, Bernoulli, and Stirling numbers
Jerome Malenfant
(Submitted on 8 Mar 2011 (v1), last revised 24 May 2011 (this version, v6))

We find general solutions to the generating-function equation sum c_q^{(X)} z^q = F(z)^X, where X is a complex number and F(z) is a convergent power series with |F(0)| >0. We then use these results to derive finite expressions containing only integers or simple fractions for partition functions and for Euler, Bernoulli, and Stirling numbers.
============================================================
Raphie Frank
Just for fun, take a look at the Stirling I Triangle and see if you can spot how the following sequence (related to exponential integrals and, thus, the complex plane) is almost certainly embedded within it (I only checked the first several values...) just as powers of 11 are embedded within Pascal's Triangle...

A051431 (n+10)!/10!
1, 11, 132, 1716, 24024, 360360, 5765760, 98017920, 1764322560, 33522128640, 670442572800, 14079294028800, 309744468633600, 7124122778572800, 170978946685747200, 4274473667143680000, 11113631534573568000
http://oeis.org/A051431

Stirling Triangle (Reversed)
http://oeis.org/A094638/table

Also see: Exponential Integral
http://mathworld.wolfram.com/ExponentialIntegral.html
Raphie Frank
QUOTE (Raphie Frank+Oct 10 2011, 03:41 AM)
For the special case of y = 1, then...

for b = 0, (C = xA + yB) --> PASCAL'S TRIANGLE
for b = 1, (C = xA + yB) --> (REVERSED) STIRLING I TRIANGLE
(Unsigned)

A130534 Stirling I Triangle see: http://oeis.org/A130534/table

Set x to p^b instead of r^b, and for the special case of y = 1 and b = 1, you'll get the Stirling Numbers of the Second Kind Triangle, the row sums of which are equal to the Bell Numbers.

In expanded form...

ROWSUM (C = xA + yB) = ROWSUM (T(r,p)) = ROWSUM (p^1*T((r -1),(p - 1)) + 1*T((r -1),(p - 0)))
= Bell_n

Stirling Numbers of the Second Kind
http://en.wikipedia.org/wiki/Stirling_numb...the_second_kind

- RF
jeremyebert
Raphie,
I'd like to work these triangular multiplication tables into layered overlays on my existing geometries; I think a scatter plot with a different color for each table will show some very interesting results.

http://dl.dropbox.com/u/13155084/divisor%20invert.bmp
jeremyebert
QUOTE (Raphie Frank+Oct 7 2011, 01:37 PM)
Unfortunately, I don't have an answer to offer you at the moment, assuming the question you are asking is whether or not Complex Exponentials can be constructed from Binomial Coefficients. It's not something I've given any thought to.

But here's a paper (mentioned on the page you linked to) that you might want to take a look at...

Arthur T. Benjamin, Gregory O. Preston, Jennifer J. Quinn, A Stirling Encounter with Harmonic Numbers, (2002) Mathematics Magazine, 75 (2) pp 95–103.
http://www.math.hmc.edu/~benjamin/papers/harmonic.pdf

Also, see the section on the page you linked to:
"Generalization to the complex plane"

- RF

Just starting on this but I love the playing card visual, it’s used in Prime Obsession as well.

http://en.wikipedia.org/wiki/Prime_Obsession
Raphie Frank
QUOTE (jeremyebert+Oct 10 2011, 03:10 PM)
Raphie,
I'd like to work these triangular multiplication tables into layered overlays on my existing geometries; I think a scatter plot with a different color for each table will show some very interesting results.

http://dl.dropbox.com/u/13155084/divisor%20invert.bmp

Jeremy, this will require further modification, I already know, because the formulation below does not accommodate Polygonal Numbers [P_(n,k)] which follow the recursive construction rule:

SUM (-1A + 2B); A_0 = 1, B_0 = (n-1)

Nevertheless, utilizing a "sliding" y-coefficient...

If...
-------------------
A = T((r-1),(p-1)
B = T((r-1),(p-0)
C = T(r,p)

Then...
-------------------
C = xA + yB

For...
-------------------
A_0 = 0
B_0 = 1
x = n in N
y = (p+1)^b''(r+0)^b'

where...
r = row #
p = Position within row
b'' = T_(n-1)(mod 3)
b' = T_(n-0)(mod 3)
T(r,p) = Triangle of r,p

Where x = 1, there are 3 possible cases...

I. p = 0, r = 0 --> Pascal's Triangle
II. p = 0, r = 1 --> Stirling I Triangle
II. p = 1, r = 0 --> Stirling II Triangle

Best,
Raphie
Raphie Frank
QUOTE (Raphie Frank+Oct 11 2011, 08:21 AM)

Jeremy, this will require further modification, I already know, because the formulation below does not accommodate Polygonal Numbers [P_(n,k)] which follow the recursive construction rule:

SUM (-1A + 2B); A_0 = 1, B_0 = (n-1)

Nevertheless, utilizing a "sliding" y-coefficient...

If...
-------------------
A = T((r-1),(p-1)
B = T((r-1),(p-0)
C = T(r,p)

Then...
-------------------
C = xA + yB

For...
-------------------
A_0 = 0
B_0 = 1
x = n in N
y = (p+1)^b''(r+0)^b'

where...
r = row #
p = Position within row
b'' = T_(n-1)(mod 3)
b' = T_(n-0)(mod 3)
T(r,p) = Triangle of r,p

Where x = 1, there are 3 possible cases...

I. p = 0, r = 0 --> Pascal's Triangle
II. p = 0, r = 1 --> Stirling I Triangle
II. p = 1, r = 0 --> Stirling II Triangle

Best,
Raphie

I was referring here (in I. II. and III.) , of course, to the exponents associated with p and r, not p and r themselves...

- RF
jeremyebert
QUOTE (jeremyebert+Sep 24 2011, 09:08 AM)
Borrowing from Rpenner ‘s insight on Harmonic Numbers…

(-1/2 (2 n H(n)-n^2-n) ) + (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

Dirichlet Divisor function =
n H(n) - (sum ( (((n-k^2)/(2k)) mod .5) ) * 2) where k=1 to n

So I guess I’m open to suggestions on what to call the function:

f(n) = sum ( (((n-k^2)/(2k)) mod .5) ) * 2 where k=1 to n
jeremyebert
QUOTE (jeremyebert+Oct 11 2011, 08:26 PM)
A nice little chart:
http://dl.dropbox.com/u/13155084/chart.pdf

n log(n) - n H(n)

series expantion

-gamma n-1/2 + 1/(12 n) - 1/(120 n^3) + 1/(252 n^5) - 1/(240 n^7) + 1/(132 n^9) - 691/(32760 n^11) + 1/(12 n^13) - 3617/(8160 n^15) + 43867/(14364 n^17)...

the coefficients showing a direct relation to Bernoulli numbers:

Bernoulli(2n)/(2n)

**Euler–Mascheroni constant = gamma = y (chart)
jeremyebert
QUOTE (rpenner+Aug 8 2011, 06:48 PM)

The relative factor of 2 arises from the change of variables:
u² = n
in
∑ (k=1 .. ⌊√n⌋) (n − k²)/(2 k)                                 {Expression A}
= ∑ (k=1 .. u) (u² − k²)/(2 k)
= (u²/2) ∑ (k=1 .. u) 1/k − (1/2) ∑ (k=1 .. u) k
= (u²/2) ∑ (k=1 .. u) 1/k − (u/4)(u+1)
= (u²/2) [ γ + ln u + 1/(2 u) − 1/(12 u²) + 1/(120 u⁴) − 1/(252 u⁶) + 1/(240 u⁸) − .... ] − (u/4) − u²/4
= u² γ / 2 − u²/4 + u^2 (ln u) / 2 + u/4 − u/4 − 1/24 + 1/(240 u²) − 1/(504 u⁴) + 1/(480 u⁶) − ...
= n γ / 2 − n/4 + n (ln n ) / 4 − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³) − ...
= (n/4)[ (2γ − 1) + ln n ] − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³) − ...
≈ (n/4)[ 2γ − 1 + ln n ] − 1/24 + 1/(240 n) − 1/(504 n²) + 1/(480 n³)                                 {Expression B}
≈ (n/4)[ 2γ − 1 + ln n ] − 1/24                                 {Expression C}

Where γ is the Euler-Mascheroni constant and the term in the square brackets dates back to Euler.

http://www.physforum.com/index.php?showtop...80&#entry489635
As claimed, the following table illustrates that Expression C is very close to Expression A when n is a perfect square and large. Expression B is quite a lot better when 1 < n < 100 and square, but pretty complicated.
CODE

n        Expression A          Expression B        Expression C
1        0                  0.0012070+         -0.0030588+
4        1.5                1.5000092+          1.4990590+
9        5.25               5.2500004+          5.2495591+
16       11.6666667-        11.6666667+         11.6664135+
25       21.0416667-        21.0416667-         21.0415031-
100      118.9484127-       118.9484127-        118.9483712+
10000    23411.8875882-     23411.8875882-      23411.8875878-
1000000  3492485.4302752-   3492485.4302752-    3492485.4302752-

RPenner,
I see now that you also noticed the Bernoulli(2n)/(2n) coefficients.

If n is a complex number then the complex roots seem to be related to the roots of unity,

(n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5)

example contour plots:

-1/(12 n)+1/(120 n^3)-1/(252 n^5)+1/(240 n^7)
Re
http://dl.dropbox.com/u/13155084/Bernoulli%20Re%20n7.png
Im
http://dl.dropbox.com/u/13155084/Bernoulli%20Im%20n7.png

-1/(12 n)+1/(120 n^3)-1/(252 n^5)+1/(240 n^7)-1/(132 n^9)
Re
http://dl.dropbox.com/u/13155084/Bernoulli%20Re%20n9.png
Im
http://dl.dropbox.com/u/13155084/Bernoulli%20Im%20n9.png

My question RPenner:
Is there a know connection with Bernoulli numbers and roots of unity or cyclic groups?
jeremyebert
QUOTE (jeremyebert+Oct 11 2011, 08:26 PM)
A nice little chart:
http://dl.dropbox.com/u/13155084/chart.pdf

I fixed the chart for a(n).

Basic question I think,

since;

(n log(n)) * (n/log(n)) = n^2

and

(n H(n)) * (n/H(n)) = n^2

and

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

shouldn’t a(n) directly be related to the prime counting functions "growth rate" (n/log(n)) by squares?

http://en.wikipedia.org/wiki/Prime-counting_function

http://en.wikipedia.org/wiki/Prime_number_theorem
jeremyebert
QUOTE (jeremyebert+Oct 12 2011, 09:51 AM)
RPenner,
I see now that you also noticed the Bernoulli(2n)/(2n) coefficients.

If n is a complex number then the complex roots seem to be related to the roots of unity,

(n*H(n) ) - (n*log(n))  - (EulerGamma*n + .5)

example contour plots:

-1/(12 n)+1/(120 n^3)-1/(252 n^5)+1/(240 n^7)
Re
http://dl.dropbox.com/u/13155084/Bernoulli%20Re%20n7.png
Im
http://dl.dropbox.com/u/13155084/Bernoulli%20Im%20n7.png

-1/(12 n)+1/(120 n^3)-1/(252 n^5)+1/(240 n^7)-1/(132 n^9)
Re
http://dl.dropbox.com/u/13155084/Bernoulli%20Re%20n9.png
Im
http://dl.dropbox.com/u/13155084/Bernoulli%20Im%20n9.png

My question RPenner:
Is there a know connection with Bernoulli numbers and roots of unity or cyclic groups?

Here is the basic equivalency

(n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5)

= sum( -Bernoulli(2m) / (2m n^(2m-1)) )
jeremyebert
QUOTE (jeremyebert+Oct 13 2011, 12:14 PM)
Here is the basic equivalency

(n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5)

= sum( -Bernoulli(2m) / (2m n^(2m-1))  )

Something else I find very interesting:

1/(-n*(((n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5) ) )) converges to 12 very quickly
Raphie Frank
QUOTE (jeremyebert+Oct 13 2011, 06:21 PM)
Something else I find very interesting:

1/(-n*(((n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5) ) )) converges to 12 very quickly

Jeremy,

I'd love to see some expansions on your data and/or formulas that could be plugged in to Wolfram Alpha. For instance, Progression H_n -- Does that denote Harmonic Number? (I assume, yes) -- A little bit of data can make such questions far easier to answer, not just for me, but also for others. And, for instance, even if you know what a harmonic number is, and know that this is what you are referencing, it still takes time to track down those H_n values.

Lacking that, I'm left, as per your charts, with something like:

Red - Blue = Green

Purple is greater than Green, although Green comes really, really close to Purple sometimes.

;-)

- RF
Raphie Frank
Thought this might interest you, Jeremy...

If...
A = T((r - 1), (k - 1))
B = T((r - 1), (k - 0))
C = T(r,k)

r = row number
k = position within row
T(r,k) = Triangle at row "r" and position "k"
(The General Form of any Pascal-type triangle...)

And...
C = xA + yB
T(0,0) = 1

Then...

=====================================================
Eulerian Triangle
"T(n,k)=number of permutations of [n] with k runs."

(x,y) Recursion Scheme
x = r^1 + 1 - k^1
y = k^1 + r^0
x + y - r^1 = 2
=========
(1,1)
(2,1) (1,2)
(3,1) (2,2) (1,3)
(4,1) (3,2) (2,3) (1,4)
(5,1) (4,2) (3,3) (2,4) (1,5)
(6,1) (5,2) (4,3) (3,4) (2,5) (1,6)
(7,1) (6,2) (5,3) (4,4) (3,5) (2,6) (1,7)

VALUES GENERATED
1 xxx xxxx xxxx xxxx xxx x -- ROWSUM = 0001 = 1 * 0!
1 001 xxxx xxxx xxxx xxx x -- ROWSUM = 0002 = 2 * 1!
1 004 0001 xxxx xxxx xxx x -- ROWSUM = 0006 = 3 * 2!
1 011 0011 0001 xxxx xxx x -- ROWSUM = 0024 = 4 * 3!
1 026 0066 0026 0001 xxx x -- ROWSUM = 0120 = 5 * 4!
1 057 0302 0302 0057 001 x -- ROWSUM = 0720 = 6 * 5!
1 120 1191 2416 1191 120 1 -- ROWSUM = 5040 = 7 * 6!

e.g. 4*26 + 3*66 = 104 + 198 = 302

ROWSUMS --> "Order of symmetric group S_n, number of permutations of n letters." http://oeis.org/A001710

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

2-Restricted Stirling Triangle (Unsigned)
"This sequence... represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}."

(x,y) Recursion Scheme
x = r^0 + 1 - k^0
y = r^1 + k^0
x + y - r^1 = 2
=========
(1,1)
(1,2) (1,2)
(1,3) (1,3) (1,3)
(1,4) (1,4) (1,4) (1,4)
(1,5) (1,5) (1,5) (1,5) (1,5)
(1,6) (1,6) (1,6) (1,6) (1,6) (1,6)
(1,7) (1,7) (1,7) (1,7) (1,7) (1,7) (1,7)

VALUES GENERATED
0001 xxxx xxxx xxxx xxx xx x -- ROWSUM = 00001 = 01 * 0!
0002 0001 xxxx xxxx xxx xx x -- ROWSUM = 00003 = 03 * 1!
0006 0005 0001 xxxx xxx xx x -- ROWSUM = 00012 = 06 * 2!
0024 0026 0009 0001 xxx xx x -- ROWSUM = 00060 = 10 * 3!
0120 0154 0071 0014 001 xx x -- ROWSUM = 00360 = 15 * 4!
0720 1044 0580 0155 020 01 x -- ROWSUM = 02520 = 21 * 5!
5040 8028 5104 1665 295 27 1 -- ROWSUM = 20160 = 28 * 6!

e.g. 1*154 + 6*71 = 154 + 426 = 580

ROWSUMS --> "Order of alternating group A_n, or number of even permutations of n letters." http://oeis.org/A000142
=====================================================

These triangles, as opposed, for instance, to the Pell Triangle where x + y - r^0 = 2 and rows sum to powers of 3

I've already generalized some of what I showed you previously, btw. More soon. I'll also have to introduce you to what I call the GDS aka "The Gross Diophantine Sum." You're already getting a taste of how I use it to "callibrate" number progressions and/or triangles.

- RF

P.S. If you had a hankering to, those x and y values up above could all be interpreted in terms of some of those divisor tables we have looked at. For instance, the y values of the 2-Restricted Stirling Triangle corresponds 1-1 with the divisors of the p^n*p^(k-n) Triangle. Add those y values along the diagonal (left to right and up) and you get the following progression I believe you will be more than a bit familiar with: 1, 2, 5, 7, 12, 15, 22, 26... etc.
jeremyebert
QUOTE (Raphie Frank+Oct 13 2011, 11:46 PM)
Thought this might interest you, Jeremy...

If...
A = T((r - 1), (k - 1))
B = T((r - 1), (k - 0))
C = T(r,k)

r = row number
k = position within row
T(r,k) = Triangle at row "r" and position "k"
(The General Form of any Pascal-type triangle...)

And...
C = xA + yB
T(0,0) = 1

Then...

=====================================================
Eulerian Triangle
"T(n,k)=number of permutations of [n] with k runs."

(x,y) Recursion Scheme
x = r^1 + 1 - k^1
y = k^1 + r^0
x + y - r^1 = 2
=========
(1,1)
(2,1) (1,2)
(3,1) (2,2) (1,3)
(4,1) (3,2) (2,3) (1,4)
(5,1) (4,2) (3,3) (2,4) (1,5)
(6,1) (5,2) (4,3) (3,4) (2,5) (1,6)
(7,1) (6,2) (5,3) (4,4) (3,5) (2,6) (1,7)

VALUES GENERATED
1 xxx xxxx xxxx xxxx xxx x -- ROWSUM = 0001 = 1 * 0!
1 001 xxxx xxxx xxxx xxx x -- ROWSUM = 0002 = 2 * 1!
1 004 0001 xxxx xxxx xxx x -- ROWSUM = 0006 = 3 * 2!
1 011 0011 0001 xxxx xxx x -- ROWSUM = 0024 = 4 * 3!
1 026 0066 0026 0001 xxx x -- ROWSUM = 0120 = 5 * 4!
1 057 0302 0302 0057 001 x -- ROWSUM = 0720 = 6 * 5!
1 120 1191 2416 1191 120 1 -- ROWSUM = 5040 = 7 * 6!

e.g. 4*26 + 3*66 = 104 + 198 = 302

ROWSUMS --> "Order of symmetric group S_n, number of permutations of n letters." http://oeis.org/A001710

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

2-Restricted Stirling Triangle (Unsigned)
"This sequence... represents the number of permutations in the alternating group An of length k, where the length is taken with respect to the generators set {(12)(ij)}."

(x,y) Recursion Scheme
x = r^0 + 1 - k^0
y = r^1 + k^0
x + y - r^1 = 2
=========
(1,1)
(1,2) (1,2)
(1,3) (1,3) (1,3)
(1,4) (1,4) (1,4) (1,4)
(1,5) (1,5) (1,5) (1,5) (1,5)
(1,6) (1,6) (1,6) (1,6) (1,6) (1,6)
(1,7) (1,7) (1,7) (1,7) (1,7) (1,7) (1,7)

VALUES GENERATED
0001 xxxx xxxx xxxx xxx xx x -- ROWSUM = 00001 = 01 * 0!
0002 0001 xxxx xxxx xxx xx x -- ROWSUM = 00003 = 03 * 1!
0006 0005 0001 xxxx xxx xx x -- ROWSUM = 00012 = 06 * 2!
0024 0026 0009 0001 xxx xx x -- ROWSUM = 00060 = 10 * 3!
0120 0154 0071 0014 001 xx x -- ROWSUM = 00360 = 15 * 4!
0720 1044 0580 0155 020 01 x -- ROWSUM = 02520 = 21 * 5!
5040 8028 5104 1665 295 27 1 -- ROWSUM = 20160 = 28 * 6!

e.g. 1*154 + 6*71 = 154 + 426 = 580

ROWSUMS --> "Order of alternating group A_n, or number of even permutations of n letters." http://oeis.org/A000142
=====================================================

These triangles, as opposed, for instance, to the Pell Triangle where x + y - r^0 = 2 and rows sum to powers of 3

I've already generalized some of what I showed you previously, btw. More soon. I'll also have to introduce you to what I call the GDS aka "The Gross Diophantine Sum." You're already getting a taste of how I use it to "callibrate" number progressions and/or triangles.

- RF

P.S. If you had a hankering to, those x and y values up above could all be interpreted in terms of some of those divisor tables we have looked at. For instance, the y values of the 2-Restricted Stirling Triangle corresponds 1-1 with the divisors of the p^n*p^(k-n) Triangle. Add those y values along the diagonal (left to right and up) and you get the following progression I believe you will be more than a bit familiar with: 1, 2, 5, 7, 12, 15, 22, 26... etc.

Very nice Raphie!

The Recursion Scheme
=========
(1,1)
(2,1) (1,2)
(3,1) (2,2) (1,3)
(4,1) (3,2) (2,3) (1,4)
(5,1) (4,2) (3,3) (2,4) (1,5)
(6,1) (5,2) (4,3) (3,4) (2,5) (1,6)
(7,1) (6,2) (5,3) (4,4) (3,5) (2,6) (1,7)

my model is based off this as well. (it forms the circles)

Notice the numbers under the center circle...
http://dl.dropbox.com/u/13155084/2D/Fourier.html

=========(horizontal lines at the sqrt(n) ) = (n-k^2)/(2k)
(1,1)
(1,2) (1,2)
(1,3) (1,3) (1,3)
(1,4) (1,4) (1,4) (1,4)
(1,5) (1,5) (1,5) (1,5) (1,5)
(1,6) (1,6) (1,6) (1,6) (1,6) (1,6)
(1,7) (1,7) (1,7) (1,7) (1,7) (1,7) (1,7)

This will help me greatly in the overlays I wish to do.

Also,
The series expansion of -Bernoulli(2m)/(2m n^(2m-1)) involves Sterling Numbers of the second kind which I believe are involved with the Roots of Unity relations I'm seeing. My next post will have more explanation.

http://www.wolframalpha.com/input/?i=-Bern....*Meters.dflt--
jeremyebert
QUOTE (Raphie Frank+Oct 13 2011, 09:57 PM)
Jeremy,

I'd love to see some expansions on your data and/or formulas that could be plugged in to Wolfram Alpha. For instance, Progression H_n -- Does that denote Harmonic Number? (I assume, yes) -- A little bit of data can make such questions far easier to answer, not just for me, but also for others. And, for instance, even if you know what a harmonic number is, and know that this is what you are referencing, it still takes time to track down those H_n values.

Lacking that, I'm left, as per your charts, with something like:

Red - Blue = Green

Purple is greater than Green, although Green comes really, really close to Purple sometimes.

;-)

- RF

I'll simplify a bit...

Roots of Unity relation.

Example of 2(x-1) Roots in the complex plane:

0 = SUM(-Bernoulli(2k)/(2k n^(2k-1))) where k=1 to x, x = 5

http://www.wolframalpha.com/input/?i=0+%3D...here+k%3D1+to+5
jeremyebert
QUOTE (jeremyebert+Oct 13 2011, 01:21 PM)
Something else I find very interesting:

1/(-n*(((n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5) ) )) converges to 12 very quickly

SUM(-Bernoulli(2k)/(2k n^(2k-1))) = (n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5)

or roughly

(n*log(n)) + (EulerGamma*n + .5)

overshoots

n H(n)

by 1/12n

1/(-n*(((n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5) ) )) where n = 100

http://www.wolframalpha.com/input/?i=+1%2F...og-_*Log.Log10-
jeremyebert
QUOTE (jeremyebert+Oct 14 2011, 08:56 AM)

SUM(-Bernoulli(2k)/(2k n^(2k-1))) = (n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5)

or roughly

(n*log(n)) + (EulerGamma*n + .5)

overshoots

n H(n)

by 1/12n

1/(-n*(((n*H(n) ) - (n*log(n)) - (EulerGamma*n + .5) ) )) where n = 100

http://www.wolframalpha.com/input/?i=+1%2F...og-_*Log.Log10-

Using the Bernoulli form:

1/SUM(-Bernoulli(2k)/(2k n^(2k-1))) * n where k=1 to 10

or

1/SUM(-Bernoulli(2k)/(2k n^(2k-2))) where k=1 to 10

http://www.wolframalpha.com/input/?i=+1%2F...ere+k%3D1+to+10
jeremyebert
QUOTE (Raphie Frank+Oct 13 2011, 09:57 PM)
Jeremy,

I'd love to see some expansions on your data and/or formulas that could be plugged in to Wolfram Alpha. For instance, Progression H_n -- Does that denote Harmonic Number? (I assume, yes) -- A little bit of data can make such questions far easier to answer, not just for me, but also for others. And, for instance, even if you know what a harmonic number is, and know that this is what you are referencing, it still takes time to track down those H_n values.

Lacking that, I'm left, as per your charts, with something like:

Red - Blue = Green

Purple is greater than Green, although Green comes really, really close to Purple sometimes.

;-)

- RF

n = a + bi (complex number)

SUM(-Bernoulli(2k)/(2k n^(2k-1))) where k=1 to 6

http://www.wolframalpha.com/input/?i=SUM%2...here+k%3D1+to+6
Raphie Frank
QUOTE (jeremyebert+Oct 14 2011, 12:41 PM)
Very nice Raphie!

The Recursion Scheme
=========
(1,1)
(2,1) (1,2)
(3,1) (2,2) (1,3)
(4,1) (3,2) (2,3) (1,4)
(5,1) (4,2) (3,3) (2,4) (1,5)
(6,1) (5,2) (4,3) (3,4) (2,5) (1,6)
(7,1) (6,2) (5,3) (4,4) (3,5) (2,6) (1,7)

my model is based off this as well. (it forms the circles)

Notice the numbers under the center circle...
http://dl.dropbox.com/u/13155084/2D/Fourier.html

Well, then, Jeremy, it would seem that, however unknowingly, you have been, in some manner we've yet to fully nail down, mapping the "rises and falls" of permutation runs associated with the Eulerian Triangle and Symmetric Group S(n).

Could you give a bit better explanation of those numbers under the center circle in relation to your model?
Raphie Frank
jeremyebert
QUOTE (Raphie Frank+Oct 14 2011, 02:42 PM)

Hmmm..
http://en.wikipedia.org/wiki/Coupon_collec...the_expectation

http://en.wikipedia.org/wiki/Coupon_collec...ng_the_variance

wolfram calculated it to be

SUM(-Bernoulli(2k)/(2k n^(2k-1)))

right?

which = 1/(12n) at n = LIM

correct?
jeremyebert
More chart explanations...

a(n) = 2 * SUM((n-2k)/(2k)) mod 0.5 where k = 1 to n = "sum of the distance from the divisors. divisors are the set of integers and half integers"

a(n) = "indivisibility value of n, max at primes"

a(n) MIN = n H(n) - ("""Dirichlet's leading behavior = n log(n) + n (2 EulerGamma - 1) + sqrt(n)""") *** min at highly composite numbers

a(n) MAX = n H(n) - "Dirichlet divisor problem"

http://dl.dropbox.com/u/13155084/chart.pdf
Raphie Frank
QUOTE (Raphie Frank+Oct 14 2011, 04:46 AM)
Thought this might interest you, Jeremy...

If...
A = T((r - 1), (k - 1))
B = T((r - 1), (k - 0))
C = T(r,k)

r = row number
k = position within row
T(r,k) = Triangle at row "r" and position "k"
(The General Form of any Pascal-type triangle...)

And...
C = xA + yB
T(0,0) = 1

Then...

....

I've already generalized some of what I showed you previously, btw. More soon. I'll also have to introduce you to what I call the GDS aka "The Gross Diophantine Sum." You're already getting a taste of how I use it to "callibrate" number progressions and/or triangles.

Before I forget, Jeremy, here is one general rule, one which is not yet fully generalized...

If x + y = r^b + n

b = 0 or 1 (a "binary switch" 0 = "no" 1 = "yes")

Then, for b = 0...

|ROW SUM (T(xA + yB) = C)| = (n + 1)^r = (x + y)^r

READ: "The Absolute Value of the row sum of the triangle constructed by the recursion/construction rule (xA + yB) = C [where A, B & C are defined in the above quoted] is equal to the sum of of the A & B coefficients (x+y) [for any x and y in Z] to the 'roweth' power"...

And for b = 1...

|ROW SUM (T(xA + yB) = C)| = C(n,r)r! = C((x + y - r),r)r!

Note: C(n,r)r! reads also as "Binomial(n,r)*r!"

If review should suggest the above to be false, it is my typing and mathematically induced "dyslexia" that is at fault, not the general premise.

I will follow near-term with a couple "hands-on" examples of the above to make the already obvious and trivial a bit more clear.

- RF
jeremyebert
QUOTE (jeremyebert+Oct 14 2011, 03:59 PM)
Hmmm..
http://en.wikipedia.org/wiki/Coupon_collec...the_expectation

http://en.wikipedia.org/wiki/Coupon_collec...ng_the_variance

wolfram calculated it to be

SUM(-Bernoulli(2k)/(2k n^(2k-1)))

right?

which = 1/(12n) at n = LIM

correct?
jeremyebert
n = 36

e(pi i ( ((n-k^2)/(2k)) / ((n-1)/2) ) )

http://dl.dropbox.com/u/13155084/unit%20circle.gif

sum (((n-k^2)/(2k))/((n-1)/2))) where k = 1 to n = -(n (-2 H(n)+n+1))/(2 (n-1))

e^(pi i -(n (-2 H(n)+n+1))/(2 (n-1)))

interesting function example :

e^(pi i -(n (-2 H(n)+n+1))/(2 (n-1))) where n = 6
http://www.wolframalpha.com/input/?i=e%5E%...HeavisideTheta-

shouldn'ld the max dev of 2*SUM((n-k^2)/(2k)) mod 0.5 where k = 1 to n

be related to -(n (-2 H(n)+n+1))/(2 (n-1)) in some way?
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

for the function

f(n,k) = ( (ln(k)/ln(n)) - ((((n-k)*(k-1)/(2k)) + (k-1))/(n-1) )

the

local minimum = (((n-1)/2)-sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))
local maximum = (((n-1)/2)+sqrt(((n-1)/2)^2-(n*ln^2(sqrt(n)))))/log(sqrt(n))

ex: n=49

49 minimum = (24-sqrt(576-49 log^2(7)))/(log(7))
49 maximum = (24+sqrt(576-49 log^2(7)))/(log(7))

and min * max = n

make any sence?
jeremyebert
The divisor symmetry shows up nicely. For example 36:

(ln(x)/ln(36)) - (((36-x)*(x-1)/(2*x)) + (x-1))/(36-1))

http://dl.dropbox.com/u/13155084/36.png
jeremyebert
Last one for now...

http://www.wolframalpha.com/input/?i=Plot3...-5%2C+100%7D%5D
jeremyebert
QUOTE (Raphie Frank+Oct 14 2011, 02:18 PM)
Well, then, Jeremy, it would seem that, however unknowingly, you have been, in some manner we've yet to fully nail down, mapping the "rises and falls" of permutation runs associated with the Eulerian Triangle and Symmetric Group S(n).

Could you give a bit better explanation of those numbers under the center circle in relation to your model?

Here is an old GIF I made before the flash stuff. It shows some of the beginnings of the equation.

http://dl.dropbox.com/u/13155084/sqrt.html
jeremyebert
QUOTE (jeremyebert+Oct 18 2011, 05:42 PM)
Here is an old GIF I made before the flash stuff. It shows some of the beginnings of the equation.

http://dl.dropbox.com/u/13155084/sqrt.html

More detail...

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

how many unique right triangles can be formed with the height equal to the sqrt(n) and the base and hypotenuse are both either integer or half-integer solutions?

sigma(0,n) / 2 where n is not a perfect square
(sigma(0,n)-1) / 2 where n is a perfect square

the parabolas are formed by tracing the right triangles with a hypotenuse-base difference of k and a height of sqrt(n).

when the base and hypotenuse are both either integers or half-integers then k is a divisor of n.
jeremyebert
QUOTE (jeremyebert+Oct 19 2011, 07:43 PM)
More detail...

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

how many unique right triangles can be formed with the height equal to the sqrt(n) and the base and hypotenuse are both either integer or half-integer solutions?

sigma(0,n) / 2 where n is not a perfect square
(sigma(0,n)-1) / 2 where n is a perfect square

the parabolas are formed by tracing the right triangles with a hypotenuse-base difference of k and a height of sqrt(n).

when the base and hypotenuse are both either integers or half-integers then k is a divisor of n.

how many unique right triangles with side lengths less than n,,can be formed with the height equal to the sqrt(n) and the base and hypotenuse are both either integer or half-integer solutions?
Raphie Frank
QUOTE (jeremyebert+Oct 21 2011, 01:53 PM)

I am following the thread Jeremy. Maybe others are, maybe not...

Regardless, the work you are doing [PREDICTION]: will be meaningful.

Not now. Not just yet.

But later.

If lucky, any contributions you might make to mathematics, specifically with regards to the uncovering of the hidden fractal (and eminently geometric) distribution of the primes, will be recognized in this, not the next (IMHO non-existent) lifetime. And hopefully, you, not the guy who steals your ideas, will be the one to get credit.

[Just a little moral support for a guy who managed to rediscover (with a little help) the Dirichlet Divisor Sum without even knowing the basic laws of mathematics and yet exists in such a world where not one single "expert" is yet willing to acknowledge, or capable of acknowledgong, it.]

Best,
Raphie Frank
jeremyebert
QUOTE (Raphie Frank+Oct 22 2011, 04:15 AM)
I am following the thread Jeremy. Maybe others are, maybe not...

Regardless, the work you are doing [PREDICTION]: will be meaningful.

Not now. Not just yet.

But later.

If lucky, any contributions you might make to mathematics, specifically with regards to the uncovering of the hidden fractal (and eminently geometric) distribution of the primes, will be recognized in this, not the next (IMHO non-existent) lifetime. And hopefully, you, not the guy who steals your ideas, will be the one to get credit.

[Just a little moral support for a guy who managed to rediscover (with a little help) the Dirichlet Divisor Sum without even knowing the basic laws of mathematics and yet exists in such a world where not one single "expert" is yet willing to acknowledge, or capable of acknowledgong, it.]

Best,
Raphie Frank

Thanks Raphie, you've been immensely helpful in my personal solidification of ideas. I understand so much more today than I did when I started down this rabbit hole, especially about all the areas I don't understand yet. That’s why I love to think about numbers, their aspects are without limit, there is always something to learn.

Speaking of which, do you think that a type of (stress) Tensor could be used to describe the circle and square root vectors in my equation? If you look at the dots on the circle on the x dimension (x y coordinate system) they are evenly distributed. The dots on the square root line are un-evenly distributed but ordered by a seemingly basic logarithmic function that is symmetrical with log(k)/log(n) and equal at 1, sqrt(n) and n.

http://dl.dropbox.com/u/13155084/divisor%20semmetry.png
http://en.wikipedia.org/wiki/Tensor
http://en.wikipedia.org/wiki/Euclidean_vec...nal_derivatives
http://en.wikipedia.org/wiki/Stress_(mecha...3_stress_tensor
jeremyebert
a pendulums velocity is at maximum at its equilibrium position.

http://en.wikipedia.org/wiki/Pendulum
http://en.wikipedia.org/wiki/Mechanical_equilibrium

the stress would be max at that vector as well.

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