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Observation: A Prime / Mersenne / (Ramanujan) Triangular Number Convolution 
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#1
Nov2910, 12:00 PM

P: 153

for...
p'_n = {1 Union Prime Numbers} M_n = nth Mersenne Number (2^n  1) T_n = nth Triangular Number (n^2 + n)/2 x = {0,1,2,3,13} > F_(0, 1/2, 3, 4, 7) for F_n = nth 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 RamanujanNagell 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* (nfold Perfect)** ((1^2 + 1)/2 = 1; sigma (1) = 1 (1fold Perfect) ((3^2 + 3)/2 = 6; sigma (6) = 12 (2fold Perfect) ((7^2 + 7)/2 = 28; sigma (28) = 56 (2fold Perfect) ((8191^2 + 8191)/2 = 33550336; sigma (33550336) = 67100672 (2fold 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 "MultiplyPerfect," 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_...#Determination 


#2
Nov2910, 12:13 PM

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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
Nov2910, 12:29 PM

P: 153

Best, Raphie 


#4
Nov3010, 09:45 AM

P: 153

Observation: A Prime / Mersenne / (Ramanujan) Triangular Number Convolution
((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!/n1! ? EXAMPLE: For y = RamanujanNagell Triangular # *2 For x = sigma (2^n 1) K_x = xth 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 


#5
Nov3010, 10:34 PM

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#6
Dec110, 12:58 AM

P: 153

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 0th prime (p'_0). It provides a reference point for comparison with other integer progressions.  RF 


#7
Dec110, 01:05 AM

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#8
Dec110, 01:09 AM

P: 153

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


#9
Feb911, 08:33 PM

P: 153

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 RamanujanNagell 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 http://www.physicsforums.com/showthread.php?t=469982 ((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 http://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. 


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