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
Quote by Raphie
A Prime / Mersenne / (Ramanujan) Triangular Number Convolution
http://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 RamanujanNagell 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 RamanujanNagell 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=...nt&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
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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.