Parabolic Coordinates And Prime Roots

Raphie Frank

14th July 2011 - 07:38 PM

Additional values for A028246 Triangular array of numbers a(n,k) = (1/k)*Sum_{i=0..k} (-1)^(k-i)*C(k,i)*i^n; n >= 1, 1<=k<=n. http://oeis.org/A028246

aka The Worpitzky Triangle

xx -- 0 0001 000002 00000003 000000004 0000000005 0000000006 0000000007 00000000008 00000000009 0000000010 0000000011 000000012

00 -- 1
01 -- 1 0001
02 -- 1 0003 000002
03 -- 1 0007 000012 00000006
04 -- 1 0015 000050 00000060 000000024
05 -- 1 0031 000180 00000390 000000360 0000000120
06 -- 1 0063 000602 00002100 000003360 0000002520 0000000720
07 -- 1 0127 001932 00010206 000025200 0000031920 0000020160 0000005040
08 -- 1 0255 006050 00046620 000166824 0000317520 0000332640 0000181440 00000040320
09 -- 1 0511 018660 00204630 001020600 0002739240 0004233600 0003780000 00001814400 00000362880
10 -- 1 1023 057002 00874500 005921520 0021538440 0046070640 0059875200 00046569600 00019958400 0003628800
11 -- 1 2047 173052 03669006 033105600 0158838240 0451725120 0801496080 00898128000 00618710400 0239500800 0039916800
12 -- 1 4095 523250 15195180 180204024 1118557440 4115105280 9574044480 14495120640 14270256000 8821612800 3113510400 479001600

RECURSIVE RULE
If we call the first column, Column 0 (C_0) and the first row, Row 0 (R_0) and use W (R_n, C_k) to designate position within the triangle, then:

((n+1)*W (R_n, C_k)) + ((n+2)*W (R_n, C_k+1)) = W (R_n+1, C_k+1))

e.g for n = 11 and k = 7, then
((7+1)*W (R_11, C_7)) + ((7+2)*W (R_11, C_7+1)) = W (R_12, C_8))
= 8*801496080 + 9*898128000 = 14495120640

TYPO & RECURSIVE RULE CHECK
Do we get the Bernoulli Numbers?

B_8 = 1/1 - 255/2 + 6050/3 - 46620/4 + 166824/5 - 317520/6 + 332640/7 - 181440/8 + 40320/9 = -1/30
B_09 = 1/1 - 511/2 + 18660/3 - 204630/4 + 1020600/5 - 2739240/6 + 4233600/7 - 3780000/8 + 1814400/9 - 362880/10 = 0
B_10 = 1/1 - 1023/2 + 57002/3 - 874500/4 + 5921520/5 - 21538440/6 + 46070640/7 - 59875200/8 + 46569600/9 - 19958400/10 + 3628800/11 = 5/66
B_11 = 1/1 - 2047/2 + 173052/3 - 3669006/4 + 33105600/5 - 158838240/6 + 451725120/7 - 801496080/8 + 898128000/9 - 618710400/10 + 239500800/11 - 39916800/12 = 0
B_12 = 1/1 - 4095/2 + 523250/3 - 15195180/4 + 180204024/5 - 1118557440/6 + 4115105280/7 - 9574044480/8 + 14495120640/9 - 14270256000/10 + 8821612800/11 - 3113510400/12 + 479001600/13 = -691/2730
Answer: YES
See: Bernoulli Numbers http://en.wikipedia.org/wiki/Bernoulli_number

HYPOTHESIS EXTENSION....
At Row p-1 are all values not first or last congruent to 0(mod p)?

(1, 1023, 57002, 874500, 5921520, 21538440, 46070640, 59875200, 46569600, 19958400, 3628800) mod 11
= {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10}

(1, 4095, 523250, 15195180, 180204024, 1118557440, 4115105280, 9574044480, 14495120640, 14270256000, 8821612800, 3113510400,479001600) mod 13
= {1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 12}
Answer: YES

QUOTE (Raphie Frank+Jul 12 2011, 08:14 PM)

HYPOTHESIZED GENERAL RULE (for entries of the Worpitzky triangle at row p-1):
* 1st term of each row at (p-1) is congruent to 1 (mod p)
* Last term of each row at (p-1) is congruent to -1 (mod p)
* ELSE: All other terms are congruent to 0(mod p)

Information required to hypothesize the above: The first 7 rows of the Worpitzky Triangle.

- RF

RELATED PAPER
Quantitative Analysis of Neural Networks as Universal Function Approximators
S. Trenn 2006 - 10 - 04
research.stephantrenn.de/tl_files/Misc/DiplInf061002.pdf

Raphie Frank

14th July 2011 - 09:33 PM

QUOTE (Raphie Frank+Jul 14 2011, 07:38 PM)

RECURSIVE RULE
If we call the first column, Column 0 (C_0) and the first row, Row 0 (R_0) and use W (R_n, C_k) to designate position within the triangle, then:

((n+1)*W (R_n, C_k)) + ((n+2)*W (R_n, C_k+1)) = W (R_n+1, C_k+1))

e.g for n = 11 and k = 7, then
((7+1)*W (R_11, C_7)) + ((7+2)*W (R_11, C_7+1)) = W (R_12, C_8))
= 8*801496080 + 9*898128000 = 14495120640

Firstly, ((n+1)*W (R_n, C_k)) + ((n+2)*W (R_n, C_k+1)) = W (R_n+1, C_k+1)) should have read:
((k+1)*W (R_n, C_k)) + ((k+2)*W (R_n, C_k+1))

Secondly, to derive Stirling Numbers of the Second Kind, just divide each value in the Worpitzky Triangle by (k)!

e.g
14495120640/8! = 359502
9574044480/7! = 1899612

A011562 Stirling numbers of second kind S2(13,n).
1, 4095, 261625, 2532530, 7508501, 9321312, 5715424, 1899612, 359502, 39325, 2431, 78, 1
http://oeis.org/A011562

Easy as pie when you get right down to it with the added bonus that we can now plainly see how to derive Bernoulli Numbers from Stirling Numbers of the Second Kind:

e.g.
B_12 = (1*0!)/1 - (4095*1!)/2 + (261625*2!)/3 - (2532530*3!)/4 + (7508501*4!)/5 - (9321312*5!)/6 + (5715424*6!)/7 - (1899612*7!)/8 + (359502*8!)/9 - (39325*9!)/10 + (2431*10!)/11 - (78*11!)/12 + (1*12!)/13 = -691/2730

- RF

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

jeremyebert

15th July 2011 - 02:30 AM

QUOTE (Raphie Frank+Jul 14 2011, 04:33 PM)

Raphie, I’m trying to keep up but I have a question. What do you think about a type of Ruppert's algorithm to define a min and max "triangulation area" of n given the "Pythagorean lattice"?. I know its a vague question, I'm trying to formulate it better...

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

http://en.wikipedia.org/wiki/Ruppert%27s_algorithm

jeremyebert

15th July 2011 - 03:03 PM

I have come up with a pretty basic algorithm for the Divisor summatory function. JavaScript source code is available by the "viewing source" browser function.

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

Raphie Frank

15th July 2011 - 04:02 PM

QUOTE (jeremyebert+Jul 15 2011, 03:03 PM)

Very, very cool, Jeremy. And pretty darn impressive too, since I only suggested a couple months or so ago that you take a look at the Divisor function in relation to your project.

Not sure if this will be helpful or tie in to what you're doing in any manner, but the following is related to what is known as the "Explicit Formula"

Mangoldt Function
http://mathworld.wolfram.com/MangoldtFunction.html

Explicit Formula
http://mathworld.wolfram.com/ExplicitFormula.html

No opinion to offer on Ruppert's Algorithm.

- RF

jeremyebert

15th July 2011 - 07:48 PM

QUOTE (Raphie Frank+Jul 15 2011, 11:02 AM)

Thanks Raphie, like I've said before, you have helped me immensely. I really do appreciate it.
Can this algorithm be put into a closed form expression? If, so would this resolve the Dirichlet's divisor problem? With a little optimization (for example the first term is equal to 2n-1) it seems very fast, especially because you only need to figure the square root down to an integer instead of a decimal.

jeremyebert

15th July 2011 - 09:34 PM

Every row that starts succession of 3’s is a pronic number due to the fact that its counting lattice points from the center out in a V shape. Each new V is an integers multiplication table centered at the square of that integer and the first pronic above the square defining the top of the V and first angle of that V.
Each V is related to its column in my lattice counting, Dsum algorithm. Make any sence?

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

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

Raphie Frank

16th July 2011 - 08:39 AM

QUOTE (jeremyebert+Jul 15 2011, 07:48 PM)

Can this algorithm be put into a closed form expression?

Yes. Or rather, "maybe" with bias towards the positive.

- RF

P.S. But what you've shown so far is no more a "proof" than what Euler observed way back when. Nevertheless, it is very, very suggestive.

jeremyebert

16th July 2011 - 04:37 PM

QUOTE (Raphie Frank+Jul 16 2011, 03:39 AM)

Sorry, only have a few minutes this morning but I think this is on the right track:
u = 1,2,3..floor(sqrt(n))
D(n) = ( floor( (n-u^2) / u ) * 2 ) + 1

Raphie Frank

16th July 2011 - 06:49 PM

QUOTE (jeremyebert+Jul 16 2011, 04:37 PM)

Sorry, only have a few minutes this morning but I think this is on the right track:
u = 1,2,3..floor(sqrt(n))
D(n) = ( floor( (n-u^2) / u ) * 2 ) + 1

Yes, the question here is how to compress your algorithm. The number of terms necessary to calculate the divisor sum (sigma_1) is equal to floor [sqrt n]. And currently, each term, it seems, must be individually calculated. Although the individual terms are trivial to calculate on a case by case basis, as you get to larger numbers there will be a lot of them.

Before going too far in trying to come up with a general (closed) formula, which I see as problematic using only basic algebra, I'd like to see you test this up to at least 1,000 since data is readily available for you to cross-check against.

It would also give me more data to help you spot recurring patterns that could serve as "jump-off" points to at least set upper and/or lower bounds for specific ranges.

- RF

jeremyebert

16th July 2011 - 07:24 PM

QUOTE (Raphie Frank+Jul 16 2011, 01:49 PM)

Here is up to 10,000
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Raphie Frank

16th July 2011 - 07:39 PM

QUOTE (jeremyebert+Jul 16 2011, 07:24 PM)

Here is up to 10,000
http://dl.dropbox.com/u/13155084/DSUMv2.htm

Do have a data source to cross check against? So you know, if you've got this right up to 10,000 I won't be at all surprised.

What you are doing is not at all trivial, Jeremy. In fact, it's quite significant for reasons [EDIT]not only mathematical, but, even if perchance were it not to be original, also social.[/EDIT] And here's your competition:

The Polymath project (see links) sketches an algorithm for computing a(n) in essentially cube root time. [From Charles R Greathouse IV, Oct 10 2010]
http://oeis.org/A006218

- RF

P.S. Do a Google search for the definition of "Polymath."

jeremyebert

16th July 2011 - 07:56 PM

QUOTE (Raphie Frank+Jul 16 2011, 02:39 PM)

Do have a data source to cross check against? So you know, if you've got this right up to 10,000 I won't be at all surprised.

What you are doing is not at all trivial, Jeremy. In fact it's quite significant for reasons not only mathematical, but also social. And here's your competition:

The Polymath project (see links) sketches an algorithm for computing a(n) in essentially cube root time. [From Charles R Greathouse IV, Oct 10 2010]
http://oeis.org/A006218

- RF

P.S. Do a Google search for the definition of "Polymath."

I haven't found up to 10,000 yet but here is 1000 from oies which I'm sure you've seen.
http://oeis.org/A006218/b006218.txt

Raphie Frank

16th July 2011 - 08:11 PM

Your algorithm predicts, for instance, that 9194 should have 85319 - 85315 = 4 divisors, 6709 should have 60151 - 60149 = 2 Divisors, and 6688 should have 59949 - 59925 = 24 divisors.

2*4597 = 9194
Divisors = 4

6709 is prime.
Divisors = 2

2^5 * 11 * 19 = 6688
Divisors = 24

While allowing for the possibility that this is some random fluke of enormous proportion, let me just simply state that I think you've nailed it.

- RF

Raphie Frank

16th July 2011 - 08:44 PM

QUOTE (jeremyebert+Jul 16 2011, 04:37 PM)

Sorry, only have a few minutes this morning but I think this is on the right track:
u = 1,2,3..floor(sqrt(n))
D(n) = ( floor( (n-u^2) / u ) * 2 ) + 1

Could you provide a few expansions of this formula?

Raphie Frank

16th July 2011 - 10:12 PM

With respect to the first term of your algorithm, it suggests you might also get interesting results if you alternately add and then subtract your terms by row...

Leibniz formula for pi/4
http://en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80

In fact, every single odd term in your formula should be able to be related to pi in some manner, while the even terms will cancel out and sum to 0 ala' the Bernoulli Numbers I showed you. You could also add n terms, then subtract n terms, in which case, for example, for the special case of n = 4, then column 4 would converge to pi.

- RF

RELATED READING
Metamath!
by Gregory Chaitin
Chapter 5 -- The Labyrinth of the Continuum
Chapter 3 -- Digital Information: DNA/Software/Leibniz
Chapter 2 -- Three Strange Loves: Primes/Godel/LISP

jeremyebert

17th July 2011 - 05:09 AM

QUOTE (Raphie Frank+Jul 16 2011, 03:44 PM)

Could you provide a few expansions of this formula?

Does this help?
######01##02##03# #4#
D( 19 )= 37 + 15 + 07 + 01 = 60

01(37)=01,01,02,01,03,01,04,01,05,01,06,01,07,01,08,01,09,01,10,01,11,01,12,01,13,01,14,01,15,01,16,01,17,01,18,01,19=(19-1)/1, (18*2)+1

02(15)=04,02,06,02,08,02,10,02,12,02,14,02,16,02,18=(19-4)/2, (7*2)+1

03(07)=09,03,12,03,15,03,18=(19-9)/3, (3*2)+1

04(01)=16=(19-16)/4, (0*2)+1=1

u=(1,2,3..floor(sqrt(n)))
D(n)=(floor((n-u^2)/u)*2)+1

basically its counting lattice points from the square of the integer u.

Raphie Frank

17th July 2011 - 07:22 PM

Here is a simple way to state your algorithm as a partial sum, Jeremy...

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

e.g. for n = 17, then...
2*floor[(17 - 1)/1]+1 + 2*floor[(17 - 4)/2]+1 + 2*floor[(17 - 9)/3]+1 + 2*floor[(17 - 16)/4]+1 = 52

Some research needs to be done to ensure this is not a case of reinventing the wheel and, of course, much higher values need to be tested.

As for finding a closed sum, this is what I meant when I suggested that I believe finding a closed form will be "problematic..."

Divisor summatory function
http://en.wikipedia.org/wiki/Divisor_summatory_function
Finding a closed form for this summed expression seems to be beyond the techniques available, but it is possible to give approximations.

We need to figure out how to relate what you've found so far to Euler's work regarding Pentagonal Numbers. If we can do that, then the closed form expression can probably be determined at some point utilizing not yet devised methods that will inevitably follow from the work of Ono, Malenfant, et al.

- RF

P.S. Start thinking about the following because I've a hunch what you're doing may be applicable to this problem as well: "In mathematics, the Gauss circle problem is the problem of determining how many lattice points there are in a circle centred at the origin and with radius r. The first progress on a solution was made by Carl Friedrich Gauss, hence its name." http://en.wikipedia.org/wiki/Gauss%27s_circle_problem

RELATED CONCEPT
Annulus
http://mathworld.wolfram.com/Annulus.html
http://en.wikipedia.org/wiki/Annulus_(mathematics)

Raphie Frank

17th July 2011 - 08:19 PM

QUOTE (Raphie Frank+Jul 17 2011, 07:22 PM)

Some research needs to be done to ensure this is not a case of reinventing the wheel and, of course, much higher values need to be tested.

Jeremy, I hate to break it to you, but now that I've put your algorithm into partial sum form, it looks suspiciously like the formula you'll find at the bottom of this section:

Divisor Summatory Function - Definition
http://en.wikipedia.org/wiki/Divisor_summa...tion#Definition

e.g. for n = 17, then...

2*(floor[17/1] + floor[17/2] + floor[17/3] + floor[17/4]) - 4^2 = 52

- RF

jeremyebert

17th July 2011 - 08:37 PM

QUOTE (Raphie Frank+Jul 17 2011, 02:22 PM)

Yea, in the definition it kinda states what my equation looks like:

"This quantity can be visualized as the count of the number of lattice points fenced off by a hyperbolic surface in k dimensions"
http://dl.dropbox.com/u/13155084/prime.png
http://en.wikipedia.org/wiki/Divisor_summa...tion#Definition

Raphie Frank

17th July 2011 - 08:47 PM

QUOTE (jeremyebert+Jul 17 2011, 08:37 PM)

I know it's disappointing, but the good news, at least, is that this shows your model is in accord with "mathematical reality."

Moreover, however unintentionally, you've broken down an old formula and recombined it in a new way. Don't forget what I said about how pi relates to that.

- RF

jeremyebert

17th July 2011 - 09:26 PM

QUOTE (jeremyebert+Jul 11 2011, 02:37 PM)

When I said it was kinda close, this is what I did.

r = (n-1)/2

v = 2*ASIN( (n-1)/(n+1) )

T_a = ( (r^2 * v) - (SQRT(n) * (n-1)) ) / 2

A = (r^2 * v) / (T_a + (T_a/18))

T = n * ( (n+1)/2 )

D(n) ~ T - (T*A)

My area idea seems to be basically stated here:
http://en.wikipedia.org/wiki/Gauss%27s_cir..._and_conjecture

"since circles are more efficient at enclosing space than squares"

Triangles are more efficient than squares as well. The smallest triangle area being at a square and pronic and the largest at the 2*(n-1) edge. Given those bounds there should be a way to calculate the maximum number of triangles in the area above the square root. Directly related to the Triangular number ratio to the Dsum algorithm but much faster .
What do you think?

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

jeremyebert

17th July 2011 - 09:48 PM

QUOTE (Raphie Frank+Jul 17 2011, 03:47 PM)

yea, my equation does the Dirichlet's divisor problem with a parabola instead of a hyperbola.

jeremyebert

17th July 2011 - 09:53 PM

QUOTE (jeremyebert+Jul 17 2011, 04:26 PM)

Some interesting stuff here:
http://en.wikipedia.org/wiki/Isoperimetric_inequality

Raphie Frank

17th July 2011 - 10:16 PM

QUOTE (jeremyebert+Jul 17 2011, 09:48 PM)

yea, my equation does the Dirichlet's divisor problem with a parabola instead of a hyperbola.

And insofar as this is the case, it is still, quite possibly, an original discovery. You've created an algorithm that produces sums identical to D(n) (the Dirichlet Divisor sum) by mapping to a related, but entirely distinct, conic section. Which is still pretty cool.

Anyway, gotta shoot off to a picnic now.

- RF

jeremyebert

17th July 2011 - 10:22 PM

QUOTE (Raphie Frank+Jul 17 2011, 05:16 PM)

Thanks for the kind words and all your help thus far Raphie.

jeremyebert

18th July 2011 - 01:38 PM

nevermind...

jeremyebert

18th July 2011 - 02:36 PM

Raphie,
Some more insight into this equation. The lattice points mentioned in the Divisor summatory function definition are square.
The lattice points in our version are based on parabolic coordinates.
Also the terms in our version give the count of the number of ways that integers <=n can be written as a product of k

OURS:

n=10

u=floor(sqrt(n))=3

k=1,2,u

SUM ((2*floor[(n - k^2)/k]) + 1)

_______________________+1____1___2____3____4____5____6_____7_____8_____9__
2*[(10-1)/1]=(2*9)+1 = 19 = 01^2 (01,02)(01,03)(01,04)(01,05)(01,06)(01,07)(01,08)(01,09)(01,10)
2*[(10-4)/2]=(2*3)+1 = 7 = 02^2 (02,06)(02,08)(02,10)
2*[(10-9)/3]=(2*0)+1 = 1 = 03^2
_________________________
27

19 =number of combinations of 1 <=10
07 =number of combinations of 2 <=10
01 =number of combinations of 3 <=10

WIKI'S:

n=10

u=floor(sqrt(n))=3

k=1,2,u

SUM 2*(SUM(floor(n/k))-u^2

fl(10/1)=10
fl(10/2)=5
fl(10/3)=3
2*(10+5+3)=36
-3^2
_______________
27

jeremyebert

18th July 2011 - 03:28 PM

Also I've been thinking...
Gauss's Circle Problem on a parabolic coordinate lattice
radius r=(n+1)/2
N ( r ) = upper limit = n(n+1), lower limit = n(n+1)-2n

jeremyebert

19th July 2011 - 12:36 AM

Just another question... I've been mentioning this for a while. Does anyone else see the possible connection to the shell theorem?

http://upload.wikimedia.org/wikipedia/comm...diag-1-anim.gif

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

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

Raphie Frank

19th July 2011 - 12:39 PM

QUOTE (jeremyebert+Jul 18 2011, 02:36 PM)

I'd love to see a visual on this in order to see the differences in how the lattice points are being counted.

- RF

Raphie Frank

19th July 2011 - 02:55 PM

QUOTE (Raphie Frank+Jul 19 2011, 12:39 PM)

I'd love to see a visual on this in order to see the differences in how the lattice points are being counted.

- RF

Questions to think about regarding these differences:

A ) Is there any clear relationship to figurate numbers (especially Pentagonal Numbers, Triangular Numbers, Squares & Pronics)?
B ) Are there any signposts suggesting...
I. when/where the next prime will occur?
II. when/where the next non-trivial zero will occur?

I really don't know much at all in a technical way -- and certainly not as much as I'd like to -- about the Non-Trivial Zeroes of the Riemann Hypothesis, but still thought to put it out there. Meanwhile, a very related page you should take a look at...

Double counting (proof technique)
In combinatorics, double counting, also called counting in two ways, is a combinatorial proof technique for showing that two expressions are equal by demonstrating that they are two ways of counting the size of one set. In this technique, which van Lint & Wilson (2001) call “one of the most important tools in combinatorics,” one describes a finite set X from two perspectives leading to two distinct expressions for the size of the set. Since both expressions equal the size of the same set, they equal each other.
http://en.wikipedia.org/wiki/Double_counti...roof_technique)

The technique in action...

Proofs of Fermat's Little Theorem
http://en.wikipedia.org/wiki/Proofs_of_Fer..._little_theorem

The "Double Counting" Page gives other examples, and all of them, more or less, are at least tangentially related to the explorations upon this thread.

- RF

jeremyebert

19th July 2011 - 03:12 PM

QUOTE (Raphie Frank+Jul 19 2011, 07:39 AM)

I'd love to see a visual on this in order to see the differences in how the lattice points are being counted.

- RF

Does this help?
http://dl.dropbox.com/u/13155084/DSUM.png

Also the hyperbola version:

http://demonstrations.wolfram.com/LatticeP...nderAHyperbola/

Raphie Frank

19th July 2011 - 05:47 PM

QUOTE (jeremyebert+Jul 19 2011, 03:12 PM)

Does this help?
http://dl.dropbox.com/u/13155084/DSUM.png

Also the hyperbola version:

http://demonstrations.wolfram.com/LatticeP...nderAHyperbola/

My OS seems to be too outdated to install the necessary software to view the Wolfram demonstration. Regardless, it would be great if you could translate that demo into a graphic in similar manner to your Pythagorean lattice so that we could compare apples to apples.

The real question here, for me, is whether or not you are partitioning the count in either A ) a more precise manner and/or B ) a manner which actually counts differently or is simply a result of bending straight lines. I suspect A ) over B ), which would become then an issue of whether or not you have found a way to "slice" the count into more pieces than is typically done.

For instance, what happens if you straighten out the curves of your parabolas so that they become effectively the x and y axes? Do the straight lines upon which you are currently counting become "half-hyperbolas" (or approximations thereof)? Conversely, what happens if you take concentric circles (emanating from the origin) and overlay them on top of the hyperbola?

Not sure if this is clear...

- RF

jeremyebert

19th July 2011 - 07:42 PM

QUOTE (Raphie Frank+Jul 19 2011, 12:47 PM)

If you bend my square root line into a hyperbola it becomes the wiki definition visualization.

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

jeremyebert

19th July 2011 - 09:39 PM

QUOTE (jeremyebert+Jul 19 2011, 02:42 PM)

If you bend my square root line into a hyperbola it becomes the wiki definition visualization.

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

So I guess the basic summation of my equation is that it a transformation of the D(n) function from a hyperbola over a square lattice to Gauss’s Circle Problem over a parabolic lattice. I don’t think this has been done before, maybe I’m wrong. I do know parabolic coordinate systems are used in quantum physics, the inverse square law, orbital equations and Central Force as well, which as you know Raphie, I’ve mentioned aspects of these before in relation to this equation.
http://en.wikipedia.org/wiki/Parabolic_cyl...asic_definition
I also think it makes an obvious connection to the exponential function e
http://en.wikipedia.org/wiki/Exponential_function

Sound right?

jeremyebert

19th July 2011 - 09:44 PM

QUOTE (jeremyebert+Jun 29 2011, 04:29 PM)

After you click ok;

press "1"

press "space bar"
wait 5 sec...

press "2"
wait 5 sec....

press "space bar"

now play around with the views, 3D and zoom.

Can you see how I'm trying to relate it to the "shell theorem" and the "Inverse-square law"?

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

http://en.wikipedia.org/wiki/Inverse-square_law

Raphie,
I really wish you could see this thing in 3D....
http://dl.dropbox.com/u/13155084/PL3D2SPHE..._3D_Sphere.html

Raphie Frank

19th July 2011 - 09:58 PM

QUOTE (jeremyebert+Jul 19 2011, 09:44 PM)

Raphie,
I really wish you could see this thing in 3D....
http://dl.dropbox.com/u/13155084/PL3D2SPHE..._3D_Sphere.html

It seems that the computer Gods heard you Jeremy. I don't know if you did any tweaks or if my computer is simply being nice to me today, but I was just able to view it it in 3-D. It's pretty amazing and, yes, I do think you are looking at an old problem from a new angle that could be revealing in many ways as yet unforeseen.

In terms of physics concepts, I think you'll understand if I refrain from voicing my views for now on how this might be applicable to physical and/or social systems, in spite of the fact that it is these very views that have (indirectly) allowed me to see the sense in what you are doing.

But in the general abstract, what you are working on has a lot to do with how space gets filled and/or left unfilled.

And it also, mark my words, has a lot to do with fractal geometry and both periodic and aperiodic rotational symmetry.

- RF

jeremyebert

19th July 2011 - 10:32 PM

QUOTE (Raphie Frank+Jul 19 2011, 04:58 PM)

Yes!!! Thank you Raphie!

Some very intersting stuff here:
http://en.wikipedia.org/wiki/Laplace%E2%80...0%93Lenz_vector

and the recusion pattern of the 2D version:
http://dl.dropbox.com/u/13155084/PL3D2/P_Lattice_3D_2.html

shows up here

http://en.wikipedia.org/wiki/File:Kepler_h...raph_family.png

Raphie Frank

19th July 2011 - 10:48 PM

In 3 dimensions, Jeremy, the only allowable n-fold rotational symmetries under the Crystallographic Restriction Theorem are 1, 2, 3, 4, 6 [the solutions to 2*cos ((2*pi)/n) is in N], corresponding with rotations by 360, 180, 120, 90 and 60 degrees around a circle. (1,2,3,4,& 6 being the proper divisors of 12).

Because symmetry is viewed as a "process," this basically takes the 360 degree rotation out of the picture in terms of qualifying as "symmetry." (But personally I view the 360 degree rotation as the point of origin and associate each single rotation by 360 with a temporal component...)

In any case, the point is this: In your 2-D model, 2,3,4 & 6-fold rotational symmetries [as well as their combinatorial derivatives (such as 2*6 = 12)] might well make themselves manifest in comprehensible, repeating manner (yet to be determined) while allowable rotations in higher dimensions (that are not also allowable in lower ones) might flit in and out of the picture as if, metaphorically speaking, particles flashing into and out of existence. The higher the dimension, again metaphorically, the briefer the "flash" and the shorter the "trail" through 2-D space. (In statistics, such brief flashes with short trails might get cast aside as "outliers," but in mathematics you don't get such a luxury. If an anomaly is there, it must be accounted for...).

This general, and as yet rather "undifferentiated" hypothesis could, of course, be wildly wrong, or right for a time and then wrong thereafter, but I believe it's something to at least think about.

- RF

jeremyebert

20th July 2011 - 12:54 AM

QUOTE (Raphie Frank+Jul 19 2011, 05:48 PM)

(But personally I view the 360 degree rotation as the point of origin and associate each single rotation by 360 with a temporal component...)

I believe Euler did too. Isn’t that considered a period of a sign wave?

http://en.wikipedia.org/wiki/File:Unfasor.gif

niels

20th July 2011 - 07:57 AM

QUOTE (Raphie Frank+Jul 19 2011, 10:48 PM)

Is this the same as saying that accuracy cannot be absolute in calculus - that calculus represents the "frozen" states of something dynamically changing in shorter and shorter flashes - the smaller /higher the dimension is, and that 2D popularly speaking is the flash and 3D is the frozen state - everything in a respective scale of course.

Raphie Frank

20th July 2011 - 02:31 PM

QUOTE (niels+Jul 20 2011, 07:57 AM)

Is this the same as saying that accuracy cannot be absolute in calculus - that calculus represents the "frozen" states of something dynamically changing in shorter and shorter flashes

As far as I'm concerned Calculus works and achieves absolutely accurate results. We could get into an entire philosophical discussion about this, but it would take this thread far off-topic which I would much prefer not to do.

Dispense with the idea of absolute accuracy in Calculus and, for instance, you couldn't show quite simply how to derive pi and phi from the Bernoulli Triangle [aka "The Pascal Knight Square" (Personal Terminology: "The Division of Space Square" aka "CUT(n,k)")]

- RF

jeremyebert

20th July 2011 - 06:12 PM

Some simple D(n) triangles:

01 =01
02 01 =03
03 01 01 =05
04 02 01 01 =08
05 02 01 01 01 =10
06 03 02 01 01 01 =14
07 03 02 01 01 01 01 =16
08 04 02 02 01 01 01 01 =20
09 04 03 02 01 01 01 01 01 =23
10 05 03 02 02 01 01 01 01 01 =27
11 05 03 02 02 01 01 01 01 01 01 =29
12 06 04 03 02 02 01 01 01 01 01 01 =35

01 =01
03 00 =03
05 00 00 =05
07 01 00 00 =08
09 01 00 00 00 =10
11 03 00 00 00 00 =14
13 03 00 00 00 00 00 =16
15 05 00 00 00 00 00 00 =20
17 05 01 00 00 00 00 00 00 =23
19 07 01 00 00 00 00 00 00 00 =27
21 07 01 00 00 00 00 00 00 00 00 =29
23 09 03 00 00 00 00 00 00 00 00 00 =35

Raphie Frank

20th July 2011 - 06:33 PM

QUOTE (jeremyebert+Jul 19 2011, 09:39 PM)

I also think it makes an obvious connection to the exponential function e
http://en.wikipedia.org/wiki/Exponential_function

Sound right?

The Binomial Coefficients n!/((n-k)!k!) are derived via the expansion of...

(x^1 + 1^1)^n

Substitute n for x and instead add the reciprocals of the two terms above and you can see the relationship between Euler's Number (e) and the Binomial Coefficients quite plainly...

(n^-1 + 1^-1)^n

So, for instance...

#1
1/3^3 + 3/3^2 + 3/3^1 + 1/3^0 = 64/27 = 2.370...
1/4^4 + 4/4^3 + 6/4^2 + 4/4^1 + 1/4^0 = 625/256 = 2.441...
1/5^5 + 5/5^4 + 10/5^3 + 10/5^2 + 5/5^1 + 1/5^0 = 7776/3125 = 2.488...
1/6^6 + 6/6^5 + 15/6^4 + 20/6^3 + 15/6^2 + 6/6^1 + 1/6^0 = 117649/46656 = 2.522...
1/7^7 + 7/7^6 + 21/7^5 + 35/7^4 + 35/7^3 + 21/7^2 + 7/7^1 + 1/7^0 = 2097152/823543 = 2.546...

Related Sequences:
Number of labeled rooted trees with n nodes: n^(n-1). START: n = 1
http://oeis.org/A000169
x, 1, 2, 9, 64, 625, 7776, 117649, 2097152, 43046721, 1000000000, 25937424601, 743008370688, 23298085122481, 793714773254144, 29192926025390625...
A000312 Number of labeled mappings from n points to themselves (endofunctions): n^n. START: n = 0
http://oeis.org/A000312
1, 1, 4, 27, 256, 3125, 46656, 823543, 16777216, 387420489, 10000000000, 285311670611, 8916100448256, 302875106592253, 11112006825558016...

Thus...
lim n --> infinity (n+1)^(n)/(n^n) = e

This is a pretty slowly converging series. Maybe why A000169 doesn't even mention the relationship to Euler's Number and A00312 makes only a passing mention of it as of about a month ago...
lim_{n->infinity} A000169(n+1)/a(n) = exp(1). Convergence is slow, e.g., it takes n > 74 to get one decimal place correct and n > 163 to get two of them. - From Alonso del Arte, Jun 20 2011

Alternatively, a far faster way to get from the binomial coefficients to an approximation of e...

#2
(1*0! + 4*1! + 6*2! + 4*3! + 1*4!)/4! = 65/24 = 2.708333...
(1*0! + 5*1! + 10*2! + 10*3! + 5*4! + 1*5!)/5! = 163/60 = 2.71666...
(1*0! + 6*1! + 15*2! + 20*3! + 15*4! + 6*5! + 1*6!)/6! = 1957/720 = 2.7180555...
etc.

Simple stuff, but not a lot of people know it. And even fewer know of the relationship between such as this and fractals consequence of the work of Ono, Malenfant et al.

- RF

P.S. And if you'd like to see how Gamma (Euler's Constant) fits into the above as a consequence of "approximating two ways" and then taking a look at the ratio between the two approximations, then plug the following into the Wolfram Computational Engine http://www.wolframalpha.com.. (see "alternative representations").

((1*0! + 6*1! + 15*2! + 20*3! + 15*4! + 6*5! + 1*6!)/6!) /(1/6^6 + 6/6^5 + 15/6^4 + 20/6^3 + 15/6^2 + 6/6^1 + 1/6^0)

RELATED LINK
Incomplete Gamma Function
http://mathworld.wolfram.com/IncompleteGammaFunction.html

Raphie Frank

20th July 2011 - 08:42 PM

QUOTE (Raphie Frank+Jul 20 2011, 06:33 PM)

#1
1/3^3 + 3/3^2 + 3/3^1 + 1/3^0 = 64/27 = 2.370...
1/4^4 + 4/4^3 + 6/4^2 + 4/4^1 + 1/4^0 = 625/256 = 2.441...
1/5^5 + 5/5^4 + 10/5^3 + 10/5^2 + 5/5^1 + 1/5^0 = 7776/3125 = 2.488...
1/6^6 + 6/6^5 + 15/6^4 + 20/6^3 + 15/6^2 + 6/6^1 + 1/6^0 = 117649/46656 = 2.522...
1/7^7 + 7/7^6 + 21/7^5 + 35/7^4 + 35/7^3 + 21/7^2 + 7/7^1 + 1/7^0 = 2097152/823543 = 2.546...

And if one wishes to associate the denominators of #1 above with the binomial theorem, this too can be done in elementary manner...

e.g.
=================================================
1/7^7 + 7/7^6 + 21/7^5 + 35/7^4 + 35/7^3 + 21/7^2 + 7/7^1 + 1/7^0 = 2097152/823543 = 2.546...
=================================================

1/(1*6^7 + 7*6^6 + 21*6^5 + 35*6^4 + 35*6^3 + 21*6^2 + 7*6^1 + 1*6^0)
+
7/(1*6^6 + 6*6^5 + 15*6^4 + 20*6^3 + 15*6^2 + 6*6^1 + 1*6^0)
+
21/(1*6^5 + 5*6^4 + 10*6^3 + 10*6^2 + 5*6^1 + 1*6^0)
+
35/(1*6^4 + 4*6^3 + 6*6^2 + 4*6^1 + 1*6^0)
+
35/(1*6^3 + 3*6^2 + 3*6^1 + 1*6^0)
+
21/(1*6^2 + 2*6^1 + 1*6^0)
+
7/(1*6^1 + 1*6^0)
+
1/(1*6^0)
=
2097152/823543
=
2.546...

It's a simple matter to further break this formula down into it's component parts by substituting all the powers of 6 with Binomial expressions based on powers of 5 and then to substitute all the power of 5 terms with Binomial expressions based on powers of 4, etc...

Visually speaking you'd end up with a whole bunch of triangles within triangles if you completed the process all the way down to powers of 0. And not a single exponent or factorial expression to be found in the lot. Every one of those elements contained in the various triangles would be a binomial coefficient.

And now if you really wanted to get ambitious, then you could, via substitution, replace each and all of those binomial coefficients with the respective formulas based on Stirling Numbers that generate them...

Do that and then with a proper transform (Stirling Numbers of the first and second kinds are intimately related) you could now express Euler's Number (or any partial sum approximation of Euler's Number based on #1 above) in fractal terms thanks to Ono, Bruinier, Malenfant et al. One of these days I'm going to have to try to figure out the specifics of the maths involved in their work. Which, unfortunately, is not elementary.

- RF

jeremyebert

20th July 2011 - 09:15 PM

QUOTE (Raphie Frank+Jul 20 2011, 01:33 PM)

I've mentioned this before but it seems that ((n-1)/(n+1))^(-n/2) converges pretty quickly to e.

Raphie Frank

21st July 2011 - 12:06 AM

QUOTE (jeremyebert+Jul 20 2011, 09:15 PM)

I've mentioned this before but it seems that ((n-1)/(n+1))^(-n/2) converges pretty quickly to e.

"Quickly" is a relative term...

((56-1)/(56+1))^(-56/2) = 2.718570832...

This converges more quickly than #1 above, but not more quickly than #2 although #2 requires more terms than the formula you present...

(1*0! + 6*1! + 15*2! + 20*3! + 15*4! + 6*5! + 1*6!)/6! = 2.7180555...

e = 2.718281828459...

But as we know (in human terms) from the story of the tortoise and the hare, faster is not always "better."

- RF

Raphie Frank

21st July 2011 - 12:25 AM

QUOTE (jeremyebert+Jul 20 2011, 09:15 PM)

I've mentioned this before but it seems that ((n-1)/(n+1))^(-n/2) converges pretty quickly to e.

But what I like about your framing is this: (n-1)*(n+1) = n^2 - 1

For n is a prime greater than 3, then p^2 -1 is a Pentagonal Number *24. And, in general, n^2 - 1 (if I recall correctly) is twice the harmonic average of two consecutive Triangular numbers.

In the context of this thread that makes it potentially interesting.

- RF

Raphie Frank

21st July 2011 - 04:18 PM

A paper I suggest you take a look at, Jeremy...

Euler and the pentagonal number theorem
by Jordan Bell
http://front.math.ucdavis.edu/math.HO/0510054
"In this paper we give the history of Leonhard Euler's work on the pentagonal number theorem, and his applications of the pentagonal number theorem to the divisor function, partition function and divergent series..."

You will see many familiar topics discussed, including the relationship of Pentagonal Numbers to sigma_1 (the sum of divisors of a single integer. p.11). What's particularly nice is that they are all discussed in one place in pretty concise manner.

NOTE: Imagine your 2-D lattice with height determined by the sum of divisors of the individual integers.

- RF

jeremyebert

21st July 2011 - 07:24 PM

QUOTE (Raphie Frank+Jul 21 2011, 11:18 AM)

Thanks Raphie, really great stuff in there. I really like the geometric progression on page 13. Just so you know, there is a direct connection to the area under the circular segment with this parabolic lattice and pentagonal numbers. When the square root line y value is 3n+1 the area under it is 9*(n^2*(n+1)/2) or 9 times a Pentagonal pyramidal number.

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

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

jeremyebert

21st July 2011 - 07:48 PM

QUOTE (jeremyebert+Jul 21 2011, 02:24 PM)

Also in the visualization, the lattice points are the same. My lattice points are around a circle not a pentagon but they are the same none the less.
http://upload.wikimedia.org/wikipedia/comm...onal_number.gif

jeremyebert

22nd July 2011 - 05:01 PM

QUOTE (jeremyebert+Jul 21 2011, 02:48 PM)

When I said the lattice points are the same, I was referring to how the multiplication table overlays the lattice. I suspect this is part of the reason for the divisor function relation to pentagonal numbers.
http://dl.dropbox.com/u/13155084/Pentagonal.png

Raphie Frank

22nd July 2011 - 09:54 PM

No formula, provisional or otherwise, to offer for this yet, but it seems possible that not only Binomial (n,4), the summation of the Tetrahedral Numbers, but also Binomial (n,5), the summation of the summation of the Tetrahedral Numbers, is embedded in the Generalized Pentagonal Numbers...

Binomial (n,5)
http://oeis.org/A000389
[... 0, 0, 0] 0, 1, 6, 21, 56, 126, 252, 462, 792, 1287, 2002, 3003, 4368, , 6188, 8568, 11628, 15504, 20349, 26334

Generalized pentagonal numbers: n(3n-1)/2
http://oeis.org/A001318
0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335...

0 = 1 - 1 = (Penta_1 - Penta_1); 0^3 + 1
0 = 2 - 2 = (Penta_2 - Penta_2); 1^3 + 1
21 = 26 - 5 = (Penta_8 - Penta_3); 8 = 2^3 + 0
252 = 287 - 35 = (Penta_27 - Penta_9); 27 = 3^3 + 0
1287 = 1617 - 330 = (Penta_65 - Penta_29); 65 = 4^3 + 1
4368 = 5985 - 1617 = (Penta_126 - Penta_65); 126 = 5^3 + 1
11628 = 17550 - 5922 = (Penta_216 - Penta_125); 216 = 6^3 + 0
The relationship to cubes, offhand, seems to begin to break down after this point...
e.g. 26634 = 43776 - 17442 = (Penta_341 - Penta_215); 341 = 7^3 - 2

0 = 1 - 1 = (Penta_1 - Penta_1)
1 = 2 - 1 = (Penta_2 - Penta_1)
56 = 126 - 70 = (Penta_18 - Penta_13)
462 = 532 - 70 = (Penta_37 - Penta_13)
2002 = ?
6188 = ?

0 = 0 + 0 = (Penta_0 + Penta_0)
6 = 5 + 1 = (Penta_3 + Penta_1)
126 = 100 + 26 = (Penta_16 + Penta_8)
792 = 532 + 260 = (Penta_37 + Penta_26)
3003 = ?
8568 = ?

Will have to check further. If true, which it might well not be, it's probably quite elementary although I doubt you will find mention of it anywhere except upon this thread.

Binomial (n,5) can be viewed as the first difference between Division of a 4-Space and Division of a 5-Space. One of these days, Jeremy, I will show you why that particularly interests me, but I'll have to delve a little bit into what others view as "numerology" in order to do that.

In the meantime, another book suggestion: "The Grand Design" by Steven Hawking and Leonard Mlodinow (2010). In this book, Hawking and Mlodinow present an argument in favor of the strong anthropic principle based on the evolutionary principle of Natural Selection.

- RF

P.S. It's very, very easy to come up with formulas relating Triangular Numbers [Binomial (n,2)] and Counting Numbers [Binomial (n,1)] to the Pentagonal Numbers as first differences between two Pentagonal Numbers.

Raphie Frank

22nd July 2011 - 11:23 PM

QUOTE (Raphie Frank+Jul 22 2011, 09:54 PM)

0 = 1 - 1 = (Penta_1 - Penta_1); 0^3 + 1
0 = 2 - 2 = (Penta_2 - Penta_2); 1^3 + 1
21 = 26 - 5 = (Penta_8 - Penta_3); 8 = 2^3 + 0
252 = 287 - 35 = (Penta_27 - Penta_9); 27 = 3^3 + 0
1287 = 1617 - 330 = (Penta_65 - Penta_29); 65 = 4^3 + 1
4368 = 5985 - 1617 = (Penta_126 - Penta_65); 126 = 5^3 + 1
11628 = 17550 - 5922 = (Penta_216 - Penta_125); 216 = 6^3 + 0
The relationship to cubes, offhand, seems to begin to break down after this point...
e.g. 26634 = 43776 - 17442 = (Penta_341 - Penta_215); 341 = 7^3 - 2

And 53130 = 95382 - 42252 = (Penta_504 - Penta_335) = Binomial (25,5)

504 = 8^3 - 8
335 = 7^3 - 8

Binomial (n,5) = (1*n^5 - 10*n^4 + 35*n^3 - 50*n^2 + 24*n)/5!

NOTE: The sum of the Pentagonal Numbers is a cube for every other value...

QUOTE (Raphie Frank+Jul 22 2011, 09:54 PM)

0 = 1 - 1 = (Penta_1 - Penta_1)
1 = 2 - 1 = (Penta_2 - Penta_1)
56 = 126 - 70 = (Penta_18 - Penta_13)
462 = 532 - 70 = (Penta_37 - Penta_13)
2002 = ?
6188 = ?

And...
2002 = 2147 - 145 = Penta_75 - Penta_19 = Binomial (14,5)
6188 = 6305 - 117 = Penta_119 - Penta_17 = Binomial (17,5)

Hopefully, there are other equivalencies for these values. Otherwise it could well be a nightmare to figure out the formula, even if true...

UPDATE:
e.g. 6188 = 8855 - 2667 = Penta_153 - Penta_84 = Binomial (17,5)

Much better...
1*2 + 0 = 2
18*2 + 1 = 37
75*2 + 3 = 153

CONJECTURE: All Binomial (n,5) are expressible as the sum or difference of two Pentagonal Numbers

- RF

Raphie Frank

23rd July 2011 - 12:26 AM

QUOTE (Raphie Frank+Jul 22 2011, 11:23 PM)

CONJECTURE: All Binomial (n,5) are expressible as the sum or difference of two Pentagonal Numbers

...in at least two ways, since all n in N can be expressed as the difference between two consecutive Pentagonal Numbers.

But what would actually be far more interesting, Jeremy, is if this proposition applies to all Binomial Coefficients. But I don't have enough data, yet, to go there...

QUOTE (Raphie Frank+Jul 10 2011, 10:03 PM)

So here's my question: How might either Euler's triangle and/or the Eulerian polynomials be related to Generalized Pentagonal Numbers in more explicit manner than Malenfant has related them to one another?

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))
http://arxiv.org/abs/1103.1585/

- RF

Raphie Frank

24th July 2011 - 07:29 PM

Notice anything, Jeremy, about the following number progression?

4-dimensional analogue of centered polygonal numbers. Also number of regions created by sides and diagonals of n-gon.
a(n)= 1 + binomial(n, 4) + binomial(n-1, 2)
1, 0, 0, 1, 4, 11, 25, 50, 91, 154, 246, 375, 550, 781, 1079, 1456, 1925, 2500, 3196, 4029, 5016, 6175, 7525, 9086, 10879, 12926, 15250, 17875, 20826, 24129, 27811, 31900, 36425, 41416, 46904, 52921, 59500, 66675, 74481, 82954, 92131
http://oeis.org/A006522

If it's not immediately clear to you, then take a look at the same progression increased by 1.
A027927 T(n,2n-4), T given by A027926
http://oeis.org/A027927

Let z = [n/3] * (3*[n/3] + 2*(n mod 3) - 1) + n mod 3 + 0^(n mod 3).
See: A096777 http://oeis.org/A096777
1, 2, 3, 5, 8, 11, 15, 20, 25, 31... etc.

Then Penta_z - 1 matches this progression from the 3rd value on.

In fact, you can use Pentagonal numbers, 7 of them to be precise, to calculate the maximal number of points, intersections, connections and areas of any n-gon.

e.g.
(100 - 92) + 70 + (40 - 12) + (92 - 1) =
(Penta_16 - Penta_15) + Penta_13 + (Penta_10 - Penta_5) + (Penta_15 - Penta_1)
-->
8 Points + 70 Intersections + 28 connections + 91 areas
= 197 = 2*99 - 1

That's the "p-i-c-a" number for an Octagon.

RELATED PROGRESSION
A000127 http://oeis.org/A000127
Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes.
1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196, 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457

Also known as Moser's problem.

And if you include the area outside the Polygon, then you only need 6 Pentagonal numbers, not 7.

(Penta_16 - Penta_15) + Penta_13 + (Penta_10 - Penta_5) + Penta_15 = 198

RELATED PROGRESSION
A059173 http://oeis.org/A059173
Maximal number of regions into which 4-space can be divided by n hyper-spheres.
1, 2, 4, 8, 16, 32, 62, 114, 198, 326, 512, 772, 1124, 1588, 2186, 2942, 3882, 5034, 6428, 8096, 10072, 12392, 15094, 18218, 21806, 25902, 30552, 35804, 41708, 48316, 55682, 63862, 72914, 82898, 93876, 105912, 119072, 133424, 149038

- RF

Raphie Frank

24th July 2011 - 07:42 PM

QUOTE (jeremyebert+Jul 21 2011, 07:24 PM)

Just so you know, there is a direct connection to the area under the circular segment with this parabolic lattice and pentagonal numbers. When the square root line y value is 3n+1 the area under it is 9*(n^2*(n+1)/2) or 9 times a Pentagonal pyramidal number.

This is an interesting observation, Jeremy. I need to think about it.

Raphie Frank

24th July 2011 - 08:19 PM

And, why, Jeremy, is the relationship to Polygons given above potentially interesting?

Excerpted from Tesselation Summary
-------------------------------------------
In 1966 Robert Berger proved that there is no algorithm that can determine whether any arbitrary set of polygons can tile the plane. This does not prove that there is no algorithm for determining whether a single polygon can tile the plane however because Berger used sets of more than one polygon. He found a way for a set of polygons to mimic a Turing machine. He then used the fact that the halting problem for Turing machines is undecidable to prove that it is also undecidable whether or not his tiles can tile the whole plane. He also gave the first example of an aperiodic set of polygons. These are sets of polygons that can tile the plane but only in such a way that the symmetry group of any of their tilings is finite. Berger's aperiodic set consisted of 20,426 polygons. By 1971, Robert Robinson had found a set consisting of only 6 tiles. In 1974, Roger Penrose discovered an aperiodic set of 2 polygons, the so-called Penrose tiles. It's unknown at this time whether there exists an aperiodic set of only one polygon.
http://www.bookrags.com/research/tesselation-wom/
-------------------------------------------

So, here's a little question for you to think about.

Are the Pentagonal Numbers, in some manner yet TBD, "Turing Complete"?

RELATED READING
"Metamath! by Gregory Chaitin

- RF

Raphie Frank

25th July 2011 - 02:14 AM

A small example of "counting two ways" as opposed to "approximating two ways."

02 + 10 + 20 + 20 + 10 = 62
26 + 15 + 92 - 70 - 01 = 62

02 + 12 + 30 + 40 + 30 = 114
51 + 35 + 155 - 126 - 1 = 114

02 + 14 + 42 + 70 + 70 = 198
92 + 70 + 247 - 220 - 1 = 198
etc.

PENTAGONALLY DERIVED FORMULA (P denotes a Pentagonal Number)
P_(z_(n)) + P_(z_(n) - 2) + P_(z_(n + 2)) - P_(z_(n + 2) - 2) - P_1
for z = [n/3] * (3*[n/3] + 2*(n mod 3) - 1) + n mod 3 + 0^(n mod 3).
See: A096777 http://oeis.org/A096777

The first set of numbers in each example above are 2* the Binomial Coefficients up to Binomial (n,4). The second set of numbers in each example are Pentagonal numbers. The sums map to the maximal number of regions obtainable by dividing a 4-D space with n-hyperspheres.

It is a trivial matter to prove, algebraically, that this relationship continues to the limit as n approaches infinity.

Divide the above sums by 2 and you get the solutions to Moser's problem.

MOSER's PROBLEM
Maximal number of regions obtained by joining n points around a circle by straight lines. Also number of regions in 4-space formed by n-1 hyperplanes.
(1*n^4 - 6*n^3 + 23*n^2 - 18*n + 24)/24
1, 2, 4, 8, 16, 31, 57, 99, 163, 256, 386, 562, 794, 1093, 1471, 1941, 2517, 3214, 4048, 5036, 6196, 7547, 9109, 10903, 12951, 15276, 17902, 20854, 24158, 27841, 31931, 36457...
http://oeis.org/A000127

Note that both expressions above, the Binomial expression and the Pentagonal Expression, involve 5 terms.

- RF

Raphie Frank

25th July 2011 - 08:34 AM

QUOTE (Raphie Frank+Jul 9 2011, 07:46 PM)

Thus...
A. C((3n + 0) ,4) = (27n^4 - 36n^3 - 18n^3 + 18n^2 + 15n^2 - 4n - 2n)/8 = -Penta_(+Penta_n - n)
B. C((3n + 1) ,4) = (27n^4 - 36n^3 + 18n^3 - 18n^2 + 15n^2 + 4n - 2n)/8 = +Penta_(+Penta_n - 0)
C. C((3n + 2) ,4) = (27n^4 + 36n^3 - 18n^3 - 18n^2 + 15n^2 - 4n + 2n)/8 = +Penta_(-Penta_n - 0)

Implied, but not stated in my most recent posts is that, knowing now what we did not know a few weeks back, we can now vastly simplify the above quoted statement.

Penta_|z - 2| = Binomial ((|n-1| + 3), 4)

z = [n/3] * (3*[n/3] + 2*(n mod 3) - 1) + n mod 3 + 0^(n mod 3).
See: A096777 http://oeis.org/A096777
1, 2, 3, 5, 8, 11, 15, 20, 25, 31... etc.

Penta_|1 - 2| = Binomial (4,4) = 1
Penta_|2 - 2| = Binomial (3,4) = 0
Penta_|3 - 2| = Binomial (4,4) = 1
Penta_|5 - 2| = Binomial (5,4) = 5
Penta_|8 - 2| = Binomial (6,4) = 15
Penta_|11 - 2| = Binomial (7,4) = 35
Penta_|15 - 2| = Binomial (8,4) = 70
Penta_|20 - 2| = Binomial (9,4) = 126
Penta_|25 - 2| = Binomial (10,4) = 210
Penta_|31 - 2| = Binomial (11,4) = 330
etc.

Generalized pentagonal numbers
Penta_n =(6n^2+6n+1)/16-(2n+1)(-1)^n/16
http://oeis.org/A001318

- RF

P.S. Note, Jeremy, that 5 - 1 = 4, 15 - 1 = 14, 35 - 5 = 30, 70 - 15 = 55, etc. Thus, while, the first difference between consecutive Binomial (n,4) is a Tetrahedral Number, the first difference between every other Binomial (n,4) is equal to the Sum of Squares. Take the first difference of every 3 Binomial (n,4) and you get (n^3 + n)/2. See: http://oeis.org/A006003

Raphie Frank

26th July 2011 - 07:30 AM

You there Jeremy?

- AC

jeremyebert

26th July 2011 - 04:40 PM

QUOTE (Raphie Frank+Jul 26 2011, 02:30 AM)

You there Jeremy?

- AC

I am here Raphie. Sometimes I feel like I'm caught somewhere in-between origami , math and science. I've been working with the wonderful Pentagonal /Binomial relationships you've shown, but I haven't been able to apply them to the model in any way as of yet. I think I would like to take some time to go over what we have found thus far. If we look at our version of the Dirichlet Divisor Sum like:

u=floor(sqrt(n))

k=1 --> u

( 2 * SUM( floor[ ( n - k^2 ) / k ] ) ) + u

compared to equivalent documented version :

( 2 * SUM( Floor[ n/k ] ) ) - u^2

the main difference is :

"( n - k^2 ) / k"

vs.

"n/k"

The "( n - k^2 ) / k" approach came directly from the locus of points "(n - k^2) / 2k" along the x axis propagating vertically up the y axis by the square root of n forming the parabolic coordinate system. I simply needed to count every time "(n - k^2) / 2k = 0 mod 0.5" from 1 to n to get the Dirichlet Divisor Sum. The sigma_0 function being equal to the count of "(n - k^2) / 2k = 0 mod 0.5". If at every "0" you SUM(k) then you have the sigma_1 function and the relation to "Euler's Pentagonal Number Theorem" and partitions.

To be honest I'm having a hard time wrapping my head around what's happening with the relationships you're showing me. I've spent so much time thinking about things from the "Congruence of squares" & " Fermat's factorization method" aspect that I haven't made it much past the 3rd row of Pascal's Triangle. I do see the n-dimensional geometric expansion of the Binomial Theorem and how it relates to exactly what I'm trying to do with a "space filling" algorithm, but I am not at all intimate with it as of yet.

I know I have much to learn and your post show me a new way of thinking about things. I'm sorry I don't have any real feed back to give you other than "Wow!" and "Thanks!" at this point.

jeremyebert

26th July 2011 - 07:48 PM

another interesting tid-bit:

k=1---->n

ABS( SUM( ABS( (n - k^2) / 2k)) / SUM( (n - k^2) / 2k) )

slowly converges to the "triangular number"/"Dirichlet Divisor Sum" ratio

T_n/D_n

at n = 1000 is accurate up to the 4th decimal place

Raphie Frank

26th July 2011 - 10:07 PM

QUOTE (jeremyebert+Jul 26 2011, 04:40 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 Frank

26th July 2011 - 10:40 PM

QUOTE (jeremyebert+Jul 26 2011, 07:48 PM)

Could you provide some sample data?

Raphie Frank

26th July 2011 - 11:02 PM

This is a great page...

Bernoulli's Triangle
http://oeis.org/wiki/Bernoulli%27s_triangle

... but, unfortunately it doesn't mention how Bernoulli's triangle is related to pi.

Let A =
A032443 Sum(binomial(2*n,i),i=0..n).
1, 3, 11, 42, 163, 638, 2510, 9908, 39203, 155382, 616666, 2449868, 9740686, 38754732, 154276028, 614429672, 2448023843, 9756737702, 38897306018, 155111585372, 618679078298, 2468152192772
http://oeis.org/A032443

Let B =
A000346 2^(2*n+1) - binomial(2*n+1,n+1).
x, 1, 5, 22, 93, 386, 1586, 6476, 26333, 106762, 431910, 1744436, 7036530, 28354132, 114159428, 459312152, 1846943453, 7423131482, 29822170718, 119766321572, 480832549478, 1929894318332
http://oeis.org/A000346

lim n--> infinity 1/n*((A_n + B_n)/(A_n - B_n))^2 = pi

e.g. for n = 21
1/21*((1929894318332 + 2468152192772)/(1929894318332 - 2468152192772))^2
=
3.179212497...

Bernoulli's Triangle is the sideways summation of Pascal's Triangle.

Bernoulli Numbers
10. The relation to the Euler numbers and pi
http://wapedia.mobi/en/Bernoulli_number?t=7.#10.

Let C_n = PP_+n
Let D_n = PP_-n
for PP denotes a Pentagonal Pyramid Number

Then...
lim n--> infinity ((C_n - D_n)/(C_n + D_n))*((A_n + B_n)/(A_n - B_n))^2 = pi

- RF

Raphie Frank

26th July 2011 - 11:33 PM

QUOTE (Raphie Frank+Jul 26 2011, 11:02 PM)

This is a great page...

Bernoulli's Triangle
http://oeis.org/wiki/Bernoulli%27s_triangle

... but, unfortunately it doesn't mention how Bernoulli's triangle is related to pi.

Nor does it mention how one can derive the following sequence, related to the Fibonacci Sequence...

A023435 Dying rabbits: a(n) = a(n-1) + a(n-2) - a(n-5).
0, 1, 1, 2, 3, 5, 7, 11, 16, 24, 35, 52, 76, 112, 164, 241, 353, 518, 759, 1113, 1631, 2391, 3504, 5136, 7527, 11032, 16168, 23696, 34728, 50897, 74593, 109322, 160219, 234813, 344135, 504355, 739168
http://oeis.org/A023435

1 = 1
1 = 1
1 + 1 = 2
1 + 2 = 3
1 + 3 + [1 = 5
1 + 4 + [2 = 7
1 + 5 + 4 + [1 = 11
1 + 6 + 7 + [2 = 16
1 + 7 + 11 + [4 + 1 = 24
1 + 8 + 16 + 8 + [2 = 35
1 + 9 + 22 + 15 + [4 + 1 = 52
1 + 10 + 29 + 26 + [8 + 2 = 76

I'll have to post the "Division of Space Square" aka "CUT(n,k) aka "k-Cake_n" so you can see where those extra terms after the brackets are coming from...

- RF

Raphie Frank

27th July 2011 - 12:51 AM

A partition number "forgery", or is there something going on here? I honestly don't know, but find the below relationship a tad "strange" to say the very least.

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181 Fibonacci = X_(n+1)
1, 1, 2, 3, 5, 7, 11, 16, 24, 35, 52, 076, 112, 164, 241, 353, 0518, 0759, 1113 Dying Rabbits = Y_(n+1)
0, 0, 0, 0, 0, 1, 02, 05, 10, 20, 37, 068, 121, 213, 369, 634, 1081, 1825, 3068 First Differences = X_(n+1) - Y_(n+1) = Z_(n+1)

Y_(n+1) - Z_(n-1)
1, 1, 2, 3, 5, 7, 11, 16, 24, 35, 52, 076, 112, 164, 241, 353, 0518 0759 1113
0, 0, 0, 0, 0, 0, 00, 01, 02, 05, 10, 020, 037, 068, 121, 213, 0369 0634 1081
---------------------------------------------------------------------------------------
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 056, 075, 096, 120, 140, 0149 0125 0032

Partition Numbers
1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 056, 077, 101, 135, 176, 0231, 297, 0385

Difference
0, 0, 0, 0, 0, 0, 00, 00, 00, 00, 00, 000, 002, 005, 015, 036, 0082, 0172, 0353
http://oeis.org/A000041

xxx = (X_n + n - 1)^2 - |y^2 - 1|(-1)^[(n+1)/2]
--------------------------------
000 = (00- 1)^2 - |0^2 - 1|
002 = (01+0)^2 + |0^2 - 1|
005 = (01+1)^2 + |0^2 - 1|
015 = (02+2)^2 - |0^2 - 1|
036 = (03+3)^2 - |1^2 - 1|
082 = (05+4)^2 + |0^2 - 1|
172 = (08+5)^2 + |2^2 - 1|
353 = (13+6)^2 - |3^2 - 1|

y = ?

No idea about y, but it comes up on OEIS in relation to both the Ramanujan Theta Function and
A075255 a(n) = n - (sum of primes factors of n (with repetition)
http://oeis.org/A075255

- RF

P.S. #1 First Differences = X_(n+1) - Y_(n+1) = Z_(n+1)
0, 0, 0, 0, 0, 1, 02, 05, 10, 20, 37, 068, 121, 213, 369, 634, 1081, 1825, 3068

20 + 37 + (13 - 2) = 68
37 + 68 + (21 - 5) = 121
68 + 121 + (34 - 10) = 213

See the addition rule and how it relates to the Fibonacci Numbers?

P.S. #2 Y_n - Y_(n-4) = The Narayana Cow Sequence

A000930 a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n-1) + a(n-3).
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491
...number of ordered partitions of n-1 into parts 1 and 2 with no two 2's adjacent. E.g. there are 6 such ways to partition 5, namely 11111, 2111, 1211, 1121, 1112, 212, so a(9) = 6.

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0...m-1. The generating function is 1/(1-x-x^m). Also a(n) = sum(binomial(n-(m-1)*i, i), i=0..n/m). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide.
http://oeis.org/A000930

jeremyebert

27th July 2011 - 01:57 AM

QUOTE (Raphie Frank+Jul 26 2011, 05:07 PM)

Thank you for the obvoius, I neeeeeeeeeed the obvious.

I'm trying to explain this to others.

Raphie Frank

27th July 2011 - 04:21 PM

QUOTE (jeremyebert+Jul 27 2011, 01:57 AM)

Thank you for the obvoius, I neeeeeeeeeed the obvious.

I'm trying to explain this to others.

Fair enough. So, then, let's re-state the 5th obvious way Pentagonal numbers relate to squares, although not all of them.

24*Penta_n + 1 = (26 - 2)Penta_n + 1 = (Penta_8 - Penta_2)Penta_n + Penta_1 is square

24*1 + 1 = 5^2
24*2 + 1 = 7^2
24*5 + 1 = 11^2
24*7 + 1 = 13^2
24*12 + 1 = 17^2
24*15 + 1 = 19^2
24*22 + 1 = 23^2

etc.

The prime numbers squared are a subset of the set of elements constructed thusly.

Also...
+0 + 1 - 2 - 5 + 7 + 12 - 15 - 22 + 26 + 35 - 40 - 51 + 57 + 70 - 77 - 92
=
+0, +1, -1, -6, +1, +13, -2, - 24, +2, + 37, -3, -54, +3, + 73, -4, -96

See the pattern? And see how it relates to the above?

Delta (1, 6, 13, 24, 37, 54, 73, 96...) = 5, 7, 11, 13, 17, 19, 23...

The prime numbers are a subset of this sequence.

- RF

P.S. That 26 is the number of dimensions in Bosonic String Theory, and that 26 - 2 is the number of associated dimensions in Euclidean Space is most surely just "coincidence." Cicadas, quasi-crystals and Zipfian distributions notwithstanding, and without any regard for Euler's Gamma Function, we all "know" that prime numbers and pentagons have nothing to do with physical, social or cognitive spaces. That's pretty damn obvious too.

jeremyebert

27th July 2011 - 06:28 PM

QUOTE (Raphie Frank+Jul 25 2011, 03:34 AM)

I seem to have another link on this relationship.

m = ((n+1)*(n+2)/6) related to http://oeis.org/A001840

01
02
03.333
05
07
09.333
12
15
18.333
22
26
30.333
35
40
45.333
51
57

if m is an integer then m is itself a generalized pentagonal number

NON-GEN-PENTA_n = n(3*n -1)/2

m(3*m - 1)/2 | = Binom(n+4,4)

What do you think?

jeremyebert

27th July 2011 - 06:59 PM

QUOTE (Raphie Frank+Jul 27 2011, 11:21 AM)

That 26 is the number of dimensions in Bosonic String Theory, and that 26 - 2 is the number of associated dimensions in Euclidean Space is most surely just "coincidence." Cicadas, quasi-crystals and Zipfian distributions notwithstanding, and without any regard for Euler's Gamma Function, we all "know" that prime numbers and pentagons have nothing to do with physical, social or cognitive spaces. That's pretty damn obvious too.

rpenner

27th July 2011 - 07:11 PM

QUOTE (Raphie Frank+Jun 12 2011, 09:22 AM)

Purely as an informational point, Kino, p^2 - 1, for p>3 is always divisible by 24. In other words...

p^2 - 1 == 0 (modulo 24); for p> 3

And also...

(p^2 - 1)/24 is Pentagonal; for p> 3

The latter point follows rather obviously from the fact that all primes > 3 are congruent to either -1 or 1 modulo 6, echoing the form of a generalized Pentagonal Number.

- RF

As Kino later points out, there is no value in searching for primes using the first test since it merely asserts p = 6k ± 1 which is to say neither 2 nor 3 is a factor of p, which is only the uninteresting part of the definition of primes larger than 3.It also works for an infinite number of non-primes {1, 25, 35, 49, 65, 77, 85, 91, ... }

But is it a pentagonal number?

p = 6k ± 1 ⇒ (p^2 - 1)/24 = (p + 1)(p - 1)/24 = (6k + 1 ± 1)(6k - 1 ± 1)/24 = (36k^2 ± 12 k)/24 = (3k±1)k/2

As Kino points out, you can invert the formula Poly(5,n) = (3 n^2 - n )/2

For m > 1, Poly(m,n) = [(m-2) n^2 + (4-m) n]/2
For m > 2, IPoly(m,x) = [ m - 4 + √(m^2 + 8(m - 2)(x-1)) ] / ( 2 m - 4 )
For m > 2, n >= 1 IPoly(m, Poly(m,n)) = n (even if m and n aren't integers)
For n > 1/4, IPoly(3, Poly(6,n)) = 2n - 1 ( which is a constructive proof that all hexagonal numbers are triangular numbers, which is a theorem given in Conway & Guy)
From n > 1/6, IPoly(3, 3 Poly(5,n)) = 3n - 1 (which is a constructive proof that three times a pentagonal number is a triangular number which is a theorem given in Conway & Guy)

And this inverse IPoly(5, x) = (1 + √(1 + 24 x))/6 when applied to (p^2 - 1)/24 when p = 6k ± 1 only gives an integer when p is in the form 6k-1 regardless if it is prime or not.

Specifically, it fails when p = 7, 13, 19 ... because 2, 7, 15, etc are not pentagonal numbers.
and so Raphie, when not cribbing from the materials of others, is shown to be trivial and unreliable once again.

The main point to both of you, is
1) stop cribbing from others -- attribute your work properly
2) stop talking about relationships when you really mean a "non-empty intersection of sets"
3) start writing proofs

jeremyebert

27th July 2011 - 07:22 PM

QUOTE (jeremyebert+Jul 27 2011, 01:28 PM)

I find it interesting that

1.33333 = -1
2.33333 = -2
3.33333 = -3
4.33333 = -4
.......

for the n(3n - 1)/2 formula

generating Generalized pentagonal numbers from either sequence

+1, -1, +2, -2, +3, -3, +4, -4, +5, -5, +6, -6.....

1, 1.33, 2, 2.33, 3, 3.33, 4, 4.33, 5, 5.33, 6, 6.33......

jeremyebert

27th July 2011 - 07:55 PM

QUOTE (rpenner+Jul 27 2011, 02:11 PM)

Please bear with me RPenner...

Is the Poly function shown this? http://en.wikipedia.org/wiki/Polygonal_number

If so what is the IPoly function?

rpenner

27th July 2011 - 08:43 PM

The inverse function, similar to the last equation in the section marked Formulae on that same Wikipedia page.

jeremyebert

27th July 2011 - 08:58 PM

QUOTE (rpenner+Jul 27 2011, 03:43 PM)

The inverse function, similar to the last equation in the section marked Formulae on that same Wikipedia page.

I see the Poly functoin in the section. What is the formula for its inverse?

jeremyebert

27th July 2011 - 09:08 PM

QUOTE (jeremyebert+Jul 27 2011, 03:58 PM)

I see the Poly functoin in the section. What is the formula for its inverse?

I think I see a very possible connection from:

"The problem of determining, given two such sets, all numbers that belong to both can be solved by reducing the problem to Pell's equation" - WIKI

http://en.wikipedia.org/wiki/Pell%27s_equation

"These solutions yield good rational approximations of the form x/y to the square root of n."

to my model.

Thanks Rpenner for showing me this, it gives me much to think about.

Raphie Frank

28th July 2011 - 01:14 AM

QUOTE (Raphie Frank+Jul 27 2011, 12:51 AM)

P.S. #2 Y_n - Y_(n-4) = The Narayana Cow Sequence

A000930 a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n-1) + a(n-3).
1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491
[i]...number of ordered partitions of n-1 into parts 1 and 2 with no two 2's adjacent. E.g. there are 6 such ways to partition 5, namely 11111, 2111, 1211, 1121, 1112, 212, so a(9) = 6.
http://oeis.org/A000930

The formula given here is INCORRECT.

The Dying Rabbit Sequence and the Narayana Cow Sequence are definitely related, which I can show at another point in time, but this formula is flat out wrong. So disregard it.

Wanted to get that out there before RPenner caught the error, which, inevitably, he would have done. Thankfully for me, however, he is caught up arguing trivial points based on language usage, such as that Generalized Pentagonal Numbers are not Pentagonal.

- RF

Raphie Frank

28th July 2011 - 01:18 AM

QUOTE (rpenner+Jul 27 2011, 07:11 PM)

Specifically, it fails when p = 7, 13, 19 ... because 2, 7, 15, etc are not pentagonal numbers.
and so Raphie, when not cribbing from the materials of others, is shown to be trivial and unreliable once again.

2, 7 and 15 are indeed GENERALIZED Pentagonal Numbers which is what Jeremy and I have been discussing. And GENERALIZED Pentagonal Numbers are the basis of Euler's summing technique, Euler's Pentagonal Number Theorem, and crop up in Malenfant's recent paper, Ono's recent paper, etc.

0, 1, 2, 5, 7, 12, 15, 22, 30, 35, 40, 51, 57. etc...

To be clear, as I am notating...

Penta_n refers to a Generalized Pentagonal Number
+ Penta_n refers to a "positive" Pentagonal Number (which is what you are referring to) [aka "First Pentagonal Numbers"]
- Penta_n refers to a "negative" Pentagonal Number (which includes 2, 7 and 15...) [aka "Second Pentagonal Numbers"]

You are flat out wrong here, RPenner, and as for "cribbing" material, that is just about the silliest statement you have ever made upon this forum. "And so, RPenner, when not..." No. I won't go there.

QUOTE (Raphie Frank+Jun 22 2011, 04:55 AM)

P.S. If I lack clarity in some manner in terms of this question, or have in some manner mis-stated, then I ask you only to ask for clarification before engaging in any "moral mathematical" tirades or crusades.

Second pentagonal numbers: n * (3*n + 1) / 2.
0, 2, 7, 15, 26, 40, 57, 77, 100, 126, 155, 187, 222, 260, 301, 345, 392, 442, 495, 551, 610, 672, 737, 805, 876, 950, 1027, 1107, 1190, 1276, 1365, 1457, 1552, 1650, 1751, 1855, 1962, 2072, 2185, 2301, 2420, 2542, 2667, 2795, 2926, 3060, 3197, 3337, 3480
http://oeis.org/A005449

Generalized pentagonal numbers: n(3n-1)/2, n=0, +- 1, +- 2,....
0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, 92, 100, 117, 126, 145, 155, 176, 187, 210, 222, 247, 260, 287, 301, 330, 345, 376, 392, 425, 442, 477, 495, 532, 551, 590, 610, 651, 672, 715, 737, 782, 805, 852, 876, 925, 950, 1001, 1027, 1080, 1107, 1162, 1190, 1247, 1276, 1335
http://oeis.org/A001318

Best,
Raphie

P.S. As a refresher "Sum of divisors" can refer to SUM(sigma_0(n)) or (sigma_1(n)). Typically, if there is any doubt as to one's intended usage, a reasonable person will consider the context in which the term is being used. Especially when the interpreter knows he or she is discussing with one who is self-taught.

Raphie Frank

28th July 2011 - 02:16 AM

QUOTE (Raphie Frank+Jul 26 2011, 11:02 PM)

lim n--> infinity 1/n*((A_n + B_n)/(A_n - B_n))^2 = pi

e.g. for n = 21
1/21*((1929894318332 + 2468152192772)/(1929894318332 - 2468152192772))^2
=
3.179212497...

QUOTE (Raphie Frank+Jul 26 2011, 11:02 PM)

Let A =
A032443 Sum(binomial(2*n,i),i=0..n).
1, 3, 11, 42, 163, 638, 2510, 9908, 39203, 155382, 616666, 2449868, 9740686, 38754732, 154276028, 614429672, 2448023843, 9756737702, 38897306018, 155111585372, 618679078298, 2468152192772
http://oeis.org/A032443

Let B =
A000346 2^(2*n+1) - binomial(2*n+1,n+1).
x, 1, 5, 22, 93, 386, 1586, 6476, 26333, 106762, 431910, 1744436, 7036530, 28354132, 114159428, 459312152, 1846943453, 7423131482, 29822170718, 119766321572, 480832549478, 1929894318332
http://oeis.org/A000346

- RF

P.S. Jeremy, as with the formula I showed you for Euler's Number, it's a very slowly converging series. You can get there much faster, in approximation terms, by inserting the Fibonacci Series (F_n) into the formula 5*arccos ((F_(n+1)/2*F_(n)).

Raphie Frank

28th July 2011 - 02:53 AM

QUOTE (Raphie Frank+Jul 28 2011, 01:18 AM)

P.S. As a refresher "Sum of divisors" can refer to SUM(sigma_0(n)) or (sigma_1(n)). Typically, if there is any doubt as to one's intended usage, a reasonable person will consider the context in which the term is being used. Especially when the interpreter knows he or she is discussing with one who is self-taught.

Recognizing the potential for confusion, Jeremy and I, without even discussing it, have adopted the term "Divisor Summatory Function" when referring to D(n), otherwise known as the Dirichlet Divisor Sum = SUM (sigma_0(n)) = SUM (d(n)).

As for [(p^2 - 1)/24] is Pentagonal. That is far too elementary for either of us to feel the need for clarification as the maths involved are Junior High School level.

Brackets, by the way, typically indicate the floor function, although it is perfectly acceptable to use parentheses instead if one states that the floor function is being used.

e.g. floor ((p^2 - 1)/24) is Pentagonal.

- RF

P.S. Oops. "p" refers here to "prime number," RPenner. Not sure that's clear to you, since p(n) refers to partition number as opposed to p_n which denotes the n-th prime, indexed from 1, whereas the partition numbers are indexed from 0. To avoid confusion, I notate the n-th partition number "par(n)."

jeremyebert

28th July 2011 - 01:24 PM

QUOTE (Raphie Frank+Jul 27 2011, 09:16 PM)

Incidentally, Jeremy, the above quoted statement is "elementary," assuming one knows the following identity:

lim n--> infinity sqrt (n*pi) = 4^n/Binomial (2n, n)

The difference is that one formula relates pi to powers of 4 and the Central Binomial Coefficients while the other relates it to the Geometric Division of Space and happens to be a heck of a lot more elegant if you ask me.

lim n--> infinity sqrt (n*pi) = ( A + B )/( A - B )

- RF

P.S. Jeremy, as with the formula I showed you for Euler's Number, it's a very slowly converging series. You can get there much faster, in approximation terms, by inserting the Fibonacci Series (F_n) into the formula 5*arccos ((F_(n+1)/2*F_(n)).

"lim n--> infinity sqrt (n*pi) = ( A + B )/( A - B )"

elegant seems to fall short if you ask me.

jeremyebert

28th July 2011 - 02:14 PM

QUOTE (Raphie Frank+Jul 26 2011, 07:51 PM)

a model of n - sopfr(n) with relation to these sets as you've shown would be very interesting. the more I learn about primes the more I realize the unification that comes from them.

Raphie Frank

28th July 2011 - 02:21 PM

Jeremy, you might want to go back and edit out from your previous post the reference to the Narayana Cow Sequence (P.S. #2) as the formula I provided was incorrect, as I noted in another post.

As for me, the more I learn about Pentagonal Numbers, + and -, the more I realize the unification that comes from them.

- RF

Raphie Frank

28th July 2011 - 03:08 PM

QUOTE (jeremyebert+Jul 28 2011, 01:24 PM)

"lim n--> infinity sqrt (n*pi) = ( A + B )/( A - B )"

elegant seems to fall short if you ask me.

The "cool" part for me, Jeremy, is that the "n" term can also be expressed ( A + B )/( A - B ) = n^3/n^2 = n, where n^3 is equal to (+PP_n) + (-PP_n) and n^2 is equal to (+PP_n) - (-PP_n), for +PP_n the summation of the +Penta Numbers (Penta_(2n + 1)) and -PP_n the summation of the -Penta numbers (Penta_(2n)). In other words "+Pentagonal Pyramid Numbers" and "-Pentagonal Pyramid Numbers."

Typically, for the sake of clarity, I will notate +PP_n as PP_n and -PP_n as PP_-n. But they are interchangeable.

You could also notate: n*T_n and n*T_-n

e.g.
4*T_-4 = 24 = PP_-4
4*T_4 = 40 = PP_4

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

Where, as I know I need not specify, but will, T_n denotes the form (n^2 + n)/2, which is a Triangular Number.

- RF

jeremyebert

28th July 2011 - 04:10 PM

QUOTE (Raphie Frank+Jul 28 2011, 09:21 AM)

missed the window

jeremyebert

28th July 2011 - 04:48 PM

QUOTE (Raphie Frank+Jul 26 2011, 05:40 PM)

another interesting tid-bit:

k=1---->n

ABS( SUM( ABS( (n - k^2) / 2k)) / SUM( (n - k^2) / 2k) )

slowly converges to the "triangular number"/"Dirichlet Divisor Sum" ratio

T_n/D_n

Could you provide some sample data?

my idea with this is:

12^(1/2) = 3.4641016151

Divisors = 001, 002, 003, 004, 006, 012

k--->n = 001, 002, 003, 004, xx5, 006, xx7, xx8, xx9, x10, x11, 012
(n-k^2)/2k = 5.5, 2.0, 0.5,-0.5,-1.3,-2.0,-2.6,-3.2,-3.8,-4.4,-4.9,-5.5

SUM((n-k^2)/2k) = -20.38073593

because the divisor symmetry of n sums to 0 with this function, the residue should show some relation to the non-divisors (Cicada Number) of n, hence the T_n - C_n = D_n relationship we have seen.

T_n = Triangular Number http://oeis.org/A000217
C_n = Cicada Number http://oeis.org/A161664
D_n = Dirichlet Divisor Sum http://oeis.org/A006218

I will work on providing some data.

rpenner

28th July 2011 - 07:45 PM

QUOTE (jeremyebert+Jul 28 2011, 04:48 PM)

show some relation to the non-divisors (Cicada Number) of n, hence the T_n - C_n = D_n relationship we have seen.

If you go to the page for the sequence: http://oeis.org/A161664

Already the "formula" gives:
A161664(n) = a(n) = A000217(n)-A006218(n). or in your notation:
C_n = T_n - D_n
which is equivalent to saying
D_n = T_n - C_n

No data required.

The definition of http://oeis.org/A161664 as a running sum of http://oeis.org/A049820 makes this obvious.

And if that didn't do it for you, there's a geometric diagram which shows C_n is just the sum of the -'s and |'s in http://oeis.org/A161664/a161664.pdf

jeremyebert

28th July 2011 - 08:08 PM

QUOTE (rpenner+Jul 28 2011, 02:45 PM)

Yes, Rpenner. I think I posted that pdf in this thread already. I really like the geometric layout that it shows.

I am trying to equate it to

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

because I've used the

u=FLOOR(sqrt(n))
k---> u
(n-k^2)/k

part in a version of the "Dirichlet Divisor Sum" I posted.

Also it forms the parabolic coordinate system that I'm so obsessed with.

rpenner

28th July 2011 - 10:20 PM

Your sum = (n/2) (H_n - (1/2) T_n ) = (n/4)(2 H_n - n - 1 )
Where H_n = Sum(k=1..n) 1/k

Proof: sum(k=1..n) (n-k^2)/(2k)
= sum(k=1..n) n/(2k) - sum(k=1..n) k^2/(2k)
= (n/2) sum(k=1..n) 1/k - (1/2) sum(k=1..n) k
= (n/4) 2 H_n - (1/2)T_n
= (n/4) 2 H_n - (n/4)(n+1)
= (n/4) (2 H_n - n - 1 )

Similarly when u = floor(sqrt(n))
sum(k=1..u) (n-k^2)/(2k) = ( n H_u - T_u )/2 = ( 2 n H_u - u^2 - u)/4

Here's the first 100 numbers:
0, 1/2, 1, 3/2, 9/4, 3, 15/4, 9/2, 21/4, 37/6, 85/12, 8, 107/12, 59/6, 43/4, 35/3, 305/24, 55/4, 355/24, 95/6, 135/8, 215/12, 455/24, 20, 505/24, 1331/60, 933/40, 367/15, 3073/120, 107/4, 3347/120, 871/30, 1207/40, 1879/60, 779/24, 168/5, 1393/40, 721/20, 1491/40, 77/2, 1589/40, 819/20, 1687/40, 217/5, 357/8, 917/20, 1883/40, 483/10, 1981/40, 1423/28, 14593/280, 3739/70, 15319/280, 7841/140, 3209/56, 293/5, 16771/280, 8567/140, 17497/280, 893/14, 18223/280, 9293/140, 2707/40, 2414/35, 7877/112, 20073/280, 40907/560, 10417/140, 42429/560, 617/8, 43951/560, 5589/70, 45473/560, 23117/280, 9399/112, 11939/140, 6931/80, 24639/280, 50039/560, 635/7, 51561/560, 235589/2520, 478307/5040, 5779/60, 98513/1008, 249847/2520, 168941/1680, 32122/315, 521081/5040, 5869/56, 76477/720, 135617/1260, 183199/1680, 278363/2520, 112771/1008, 23791/210, 578113/5040, 41803/360, 65819/560, 29975/252

It has long-term growth like (n/4) ln n
Or in detail, it is approximately: ( ln n + 2 (Euler-Mascheroni Constant) - 1)(n/4) - 1/24

At n=100, the approximation is 118.94837...
and the actual value is 118.94841...

at n=10000, the approximation is 23411.887587781...
and the actual value is 23411.8875881981....

naturally, the approximation works best when n is a perfect square and large.

Raphie Frank

28th July 2011 - 11:12 PM

Thank you RPenner. This is really great information.

jeremyebert

29th July 2011 - 05:11 PM

QUOTE (rpenner+Jul 28 2011, 05:20 PM)

YOWZERS!! Thank you RPenner! A direct relation to the Euler–Mascheroni constant. I've been trying to see how that works for a while now. My mind is currently blown.

jeremyebert

29th July 2011 - 05:15 PM

I can definitly see the link to the geometry as well. http://upload.wikimedia.org/wikipedia/comm...ma-area.svg.png

jeremyebert

29th July 2011 - 06:22 PM

Summing just the divisors gives you this sequence:

http://oeis.org/A079667

jeremyebert

29th July 2011 - 06:26 PM

QUOTE (jeremyebert+Jul 29 2011, 01:22 PM)

Summing just the divisors gives you this sequence:

http://oeis.org/A079667

Which is silimar to this sequence:

http://oeis.org/A006874

jeremyebert

29th July 2011 - 06:34 PM

QUOTE (jeremyebert+Jul 29 2011, 01:26 PM)

Which is silimar to this sequence:

http://oeis.org/A006874

and
http://www.mrob.com/pub/muency/muatom.html

Raphie Frank

29th July 2011 - 08:00 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}

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

- RF

Raphie Frank

29th July 2011 - 08:10 PM

QUOTE (jeremyebert+Jul 29 2011, 06:26 PM)

Which is silimar to this sequence:

http://oeis.org/A006874

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

PhysOrg scientific forums are totally dedicated to science, physics, and technology. Besides topical forums such as nanotechnology, quantum physics, silicon and III-V technology, applied physics, materials, space and others, you can also join our news and publications discussions. We also provide an off-topic forum category. If you need specific help on a scientific problem or have a question related to physics or technology, visit the PhysOrg Forums. Here you’ll find experts from various fields online every day.

To quit out of "lo-fi" mode and return to the regular forums, please click here.