Observation: A Prime / Mersenne / (Ramanujan) Triangular Number Convolution

  • Thread starter Raphie
  • Start date
  • #1
Raphie
151
0
for...

p'_n = {1 Union Prime Numbers}
M_n = n-th Mersenne Number (2^n - 1)
T_n = n-th Triangular Number (n^2 + n)/2

x = {0,1,2,3,13} --> F_(0, 1/2, 3, 4, 7) for F_n = n-th Fibonacci Number

Then...

((p'_x*p'_2x)*(M_x - (T_x - 1))) / ((T_(M_x) - T_(T_x - 1)) is in N

EXPANSION
((1*1)*(0 + 1))/((0^2 + 0)/2 - (-1^2 -1)/2) = 1
((2*3)*(1 - 0))/((1^2 + 1)/2 - (0^2 + 0)/2) = 6
((3*7)*(3 - 2))/((3^2 + 3)/2 - (2^2 + 2)/2) = 7
((5*13)*(7 - 5)) / ((7^2 + 7)/2 - (5^2 + 5)/2) = 10
((41*101)*(8191 - 90))/((8191^2 + 8191)/2 - (90^2 + 90)/2) = 1

|{-1, 0, 2, 5, 90}| = 1, 0, 2, 5, 90

... where 1, 0, 2, 5, & 90 is the complete set of indices associated with the Ramanujan-Nagell Triangular Numbers...

T_01 = 1
T_00 = 0
T_02 = 3
T_05 = 15
T_90 = 4095

I'm pretty positive these are the only integers for which the above formula holds...

e.g.

((43*135)*(16383 - 104)) / ((16383^2 +16383)/2 - (104^2 + 104)/2) = 645/916 --> .70
((47*176)*(32767 - 119)) / ((32767^2 + 32767)/2 - (119^2 + 119)/2) = 16544/32887 --> .50
((53*231)*(65535 - 135)) / ((65535^2 + 65535)/2 - (135^2 + 135)/2) = 24486/65671 --> .37
((59*297)*(131071 - 152)) / ((131071^2 + 131071)/2 - (152^2 + 152)/2) = 17523/65612 --> 0.26
((61*385)*(262143 - 170)) / ((262143^2 + 262143)/2 - (170^2 + 170)/2) = 23485/131157 --> .18

For the special cases of x = 0, 1, 2 & 13, where par_n denotes "partition #," then the following statement also holds...

((p'_x*par_x)*(M_x - (T_x - 1))) / ((T_(M_x) - T_(T_x - 1)) is in N

Worth noting is the relationship between the following numbers and Multiply Perfect Numbers:

((0^2 + 0)/2 = 0; sigma (0) = 0* (n-fold Perfect)**
((1^2 + 1)/2 = 1; sigma (1) = 1 (1-fold Perfect)
((3^2 + 3)/2 = 6; sigma (6) = 12 (2-fold Perfect)
((7^2 + 7)/2 = 28; sigma (28) = 56 (2-fold Perfect)
((8191^2 + 8191)/2 = 33550336; sigma (33550336) = 67100672 (2-fold Perfect)

* Assumes divisors of an integer must divide that integer and be less than or equal to that integer...
** I don't know if 0 is typically considered "Multiply-Perfect," but I see no reason for it not to be based upon the above definition.

Best,
Raphie

P.S. The above is a "somewhat" accidental observation that followed from exploratory investigations into the following (dimensionless) numerical equivalency for the Josephson Constant Derivation of Planck's Constant where 67092479/8191 = Carol_13/Mersenne_13 = C_13/M_13 for Carol Numbers = M_n^2 - 2, and 9.10938215*10^-31 is the mass of an electron.

sqrt ((4*pi^2*9109.38215)/(67092479/8191) ~ 6.62606776
Related Link: Planck's Constant (Determination) http://en.wikipedia.org/wiki/Planck_constant#Determination
 
Last edited:

Answers and Replies

  • #2
CRGreathouse
Science Advisor
Homework Helper
2,842
0
The modulus grows so quickly that it is easy to prove that this doesn't happen for x > 13.

My calculations disagree with yours on x = 0.
 
  • #3
Raphie
151
0
My calculations disagree with yours on x = 0.

How are you calculating at 0? It's entirely possible I made a physical typo or "mind typo" somewhere. (I'm rather known in certain quarters for those...) :-)

Best,
Raphie
 
  • #4
Raphie
151
0
My calculations disagree with yours on x = 0.

Here is the typo...

((1*1)*(0 + 1))/((0^2 + 0)/2 - ((-1)^2 -1)/2) = Undefined
((1*1)*(0 + 1))/((0^2 + 0)/2 - (-(1)^2 -1)/2) = 1

Thank you for the catch, CRGreathouse. Obviously T_-1 = 0. The basic principle of the formula can still work for all values, but only if one uses factorials in the denominator. Thus...

n!/|n -1|! replaces x in the denominator (for n = ((T_(M_x) - T_(T_x - 1)))

... a mathematical "part of speech" I use frequently (and originally used in a formula related to the above that I call "The Chess King Sequence").

And why do I use n!/|n-1|! ?

EXAMPLE:

For y = Ramanujan-Nagell Triangular # *2
For x = sigma (2^n -1)
K_x = x-th Maximal Kissing Number

Then...

y*x = K_x
-------------
0 * 0 = 0 = K_0
2 * 1 = 2 = K_1
6 * 4 = 24 = K_4
30 * 8 = 240 = K_8
8190 * 24 = 196560 = K_24

But how to state in the following form?

K_x/x = y

It only works if I adopt some "kludge" such as n!/|n - 1|! to handle the special case of x = 0.

Best,
Raphie
 
Last edited:
  • #5
CRGreathouse
Science Advisor
Homework Helper
2,842
0
It only works if I adopt some "kludge" such as n!/|n - 1|! to handle the special case of x = 0.

That doesn't work, since (-1)! is undefined.
 
  • #6
Raphie
151
0
That doesn't work, since (-1)! is undefined.

That is why I present it as an absolute value... |n-1|!

|0 - 1|! = 1

It's a "kludge" (aka "a workaround"), CRGreathouse. Nothing more and nothing less. The purpose of the kludge? To preserve symmetry. Not mathematical symmetry, mind you, but (forgive the philosophical turn here...) cognitive symmetry.

a*b = x
x/b = a

... "wrong" for b = 0, a mathematical "fact." I know that.

But practically speaking, it gets in the way of any "number mappings" that start at 0 rather than 1. Which, in large part, is also why I include 1 as the 0-th prime (p'_0). It provides a reference point for comparison with other integer progressions.

- RF
 
  • #7
CRGreathouse
Science Advisor
Homework Helper
2,842
0
That is why I present it as an absolute value... |n-1|!

|0 - 1|! = 1

It's a "kludge" (aka "a workaround"), CRGreathouse. Nothing more and nothing less. The purpose of the kludge? To preserve symmetry. Not mathematical symmetry, mind you, but (forgive the philosophical turn here...) cognitive symmetry.

Oh, I missed that. It does make me cringe, though. If it makes you happy, so much the better I suppose.
 
  • #8
Raphie
151
0
Oh, I missed that. It does make me cringe, though. If it makes you happy, so much the better I suppose.

You just made me laugh out loud, CRG, so I guess I can accept that. :smile:

In general, thank you for all the feedback you have given me. It does not go unnoticed or unappreciated.

- RF
 
  • #9
Raphie
151
0
Hi CRGreathouse. Don't know if you've been checking in at all, but I thought I'd make mention of the fact that I have cleaned up my initial statement just a bit. It might still make one "cringe," but it does accurately "capture" all the indices associated with the Ramanujan-Nagell Triangular Numbers, and has led to some other observations, a few of which are mentioned here:

OBSERVATION: The #31, The Golden Scale, Prime Counting Function & Partition Numbers
https://www.physicsforums.com/showthread.php?t=469982

A Prime / Mersenne / (Ramanujan) Triangular Number Convolution
https://www.physicsforums.com/showthread.php?t=452148

For p'_x denotes an Integer| 0 < d(n) < 3, M_x a Mersenne Number and T_n a Triangular Number, then...

((p'_x * p'_2x) * (M_x - |T_x - 1|)) / ((T_(M_x) - T_|T_x - 1|) is in N

for |T_x - 1| = 1, 0, 2, 5, 90 and x = 0, 1, 2, 3, 13

EXPANSION
((01 * 001) * (0000 - 01)) / ((0000^2 + 0000)/2 - (01^2 + 01)/2) = 01
((02 * 003) * (0001 - 00)) / ((0001^2 + 0001)/2 - (00^2 + 00)/2) = 06
((03 * 007) * (0003 - 02)) / ((0003^2 + 0003)/2 - (02^2 + 02)/2) = 07
((05 * 013) * (0007 - 05)) / ((0007^2 + 0007)/2 - (05^2 + 05)/2) = 10
((41 * 101) * (8191 - 90)) / ((8191^2 + 8191)/2 - (90^2 + 90)/2) = 01
For quick verification, anyone interested can just copy and paste these equations into the compute box of Wolfram Alpha at http://www3.wolframalpha.com/

0, 1, 2, 5, 90 are the index numbers associated with the Ramanujan-Nagell Triangular Numbers 0, 1, 3, 15, 4095, which, as Î have previously noted, map, whether coincidentally or no, in 1:1 manner with Maximal Sphere Packings for dimension sigma (M_n(mod 5)) = 0, 1, 4, 8, 24 by the formula 2*T_z*sigma (M_n(mod 5)), for T_z denotes a Ramanujan-Nagell Triangular Number and sigma(n) the sum of divisor function.

2* 0 * 0 = 0 = K_0
2* 1 * 1 = 2 = K_1
2* 3 * 4 = 24 = K_4
2* 15 * 8 = 240 = K_8
2* 4095 * 24 = 196560 = K_24

RELATED PAPER
Kissing Numbers, Sphere Packings, and Some Unexpected Proofs
Florian Pfender & Günter M. Ziegler (2004)
http://mathdl.maa.org/mathDL/22/?pa=content&sa=viewDocument&nodeId=3065

A couple examples of observations I believe to be unique utilizing the first and last x values associated with the above equation form where...
((p'_x * p'_2x) * (M_x - |T_x - 1|)) / ((T_(M_x) - T_|T_x - 1|) = 1
(indicative of an elliptic curve of some form?)

======================================
T_13 + T_06 = par_13 + par_06 = 91 + 21 = 101 + 11 = 112
T_00 + T_12 = par_00 + par_12 = 00 + 78 = 001 + 77 = 078

0 + pi (31 - 11) + phi (31 - 11) + (31 - 11) = 0 + 8 + 08 + 20 = 36
0 + pi (31 - 12) + phi (31 - 12) + (31 - 12) = 0 + 8 + 18 + 19 = 45
0 + pi (31 - 13) + phi (31 - 13) + (31 - 13) = 0 + 7 + 06 + 18 = 31
36 + 45 + 31 = 112

0 + pi (00 + 11) + phi (00 + 11) + (00 + 11) = 0 + 5 + 10 + 11 = 26
0 + pi (00 + 12) + phi (00 + 12) + (00 + 12) = 0 + 5 + 04 + 12 = 21
0 + pi (00 + 13) + phi (00 + 13) + (00 + 13) = 0 + 6 + 12 + 13 = 31
26 + 21 + 31 = 78

T_(13 + 6) = T_19 = 112 + 78 = 190
T_(0 + 12) = T_12 = 000 + 78 = 078

p_(par_6 + par_7) = par (6 + 7) = par_(0+13) = p_26 = 101
======================================

0 + 6 + 12 + 13 = 31 --> 00 + pi (00 + 13) + phi (00 + 13) + 13, and 31 is the largest prime associated with partition numbers of Hausdorff Dimension 1 as per Ken Ono et al's recent mathematical findings. 13 is the smallest prime.

Metaphor: All the matters is the beginning and the end of the shot (a film industry truism...) In other words, the metaphor being applied here is that the beginning and end of the shot equate in some manner with "lattice points" of human perception.

Oh, and by the way, in relation to the number 31 (= 31 +/- 00)...

T_00 + p'_00 - par_00 = 00 + 01 - 001 = 00
T_13 + p'_13 - par_13 = 91 + 41 - 101 = 31

Best,
RF

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

P.S. Also, 45 is a Sophie Germain Triangular Number (2x + 1 is triangular) partnered with 91 (and when summed, 91 + 45 equals another triangular number, T_9 + T_13 = T_(7+9) = T_16 = 136):

112 == sigma (91) = sigma (T_13) = sigma (2*(par_7 + par_9) + 1) = sigma (2*(15 + 30) + 1)
078 == sigma (45) = sigma (T_09) = sigma (1*(par_7 + par_9) + 0) = sigma (1*(15 + 30) + 0)

A Related Conjecture:
Sophie Germain Triangles & x | 2y^2 + 2y - 3 = z^2
https://www.physicsforums.com/showthread.php?t=462793
(sqrt (16x + 9) - 1)/2 = y | 2y^2 + 2y - 3 = z^2
for x a Sophie Germain Triangular Number

e.g.
(sqrt (16*45 + 9) - 1)/2 = (0 + 13) = y
(2*13^2 + 2*13 - 3 = (6 + 13)^2 = 19^2 = z^2

Also related is the form: (n^2 - 2*a^2 - 7)/2 = 0
e.g. (13^2 - 2*(9)^2 - 7)/2 = 0

P.P.S. Would it were that one were not so discouraged by modern day sensibilities to note, for instance, that the above conjecture and observations are indirectly derivative of number mappings such as the following: (C (31, 1) + C (31, 2) + C (31, 3) + C (31, 4))*(totient (pi (31)))^-(pi (31)) == 3.6456*10^-11, the value of a number quite central to modern day physics, although derived by a "numerologist," Johann Balmer.
 
Last edited:

Suggested for: Observation: A Prime / Mersenne / (Ramanujan) Triangular Number Convolution

Replies
2
Views
152
Replies
5
Views
560
  • Last Post
Replies
1
Views
292
Replies
3
Views
618
  • Last Post
Replies
4
Views
782
  • Last Post
Replies
1
Views
567
  • Last Post
Replies
1
Views
418
Replies
1
Views
860
Replies
6
Views
684
Replies
7
Views
545
Top