Goongyae said:
The amazing thing is d(n) can be computed recursively:
d(n) = d(n-1)+d(n-2)-d(n-5)-d(n-7)+d(n-12)+d(n-15)-...
except if you get d(0) on the right hand-side, you must put in the number n itself.
Those numbers up above there are all generalized Pentagonal numbers, so I'm wondering how the identity you posted relates to Euler's Pentagonal number theorem:
p(n-0)-p(n-1)-p(n-2)+p(n-5)+p(n-7)-p(n-12)-p(n-15)++-- =0^n
where p(n) is the partition function
http://oeis.org/A001318
Multiply any generalized Pentagonal number by 3, by the way, and you get triangular numbers of the form T_(3n + 2) and T_(3n+3) e.g. 3, 6, 15, 21, 36, 45. Multiply by 24 and add 1, and you get square numbers, including a nice little straight run of prime squares from 5 --> 23, not surprising since all of those square numbers take the form (6n + 1)^2 or (6n - 1)^2...
24*0 + 1 = 1^2
---------------------
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
---------------------
24*26 + 1 = 25^2
In other words, all primes greater than 3 are constructible in the following manner:
sqrt (24*(Generalized Pentagonal Number_n) + 1), a fact which might naturally lead one to ask: Is this in some manner related to the Dedekind eta function? Conversely, working backwards, then
(p^2 - 1)/24 for p > 3, will always be a Generalized Pentagonal Number
Also, via summation, the Pentagonal Numbers can be quite simply related to both Cubes and Squares. e.g. 4^3 - 40 = 24, for 40 = 1 + 5 + 12 + 22 and 24 = 0 + 2 + 7 + 15. The difference between 24 and 40 = 4^2.
UPDATE: via Wolfram Mathworld
Letting x_i be the set of numbers relatively prime to 6, the generalized pentagonal numbers are given by (x_i^2-1)/24. Also, letting y_i be the subset of the x_i for which x_i=5 (mod 6), the usual pentagonal numbers are given by (y_i^2-1)/24 (D. Terr, pers. comm., May 20, 2004).
http://mathworld.wolfram.com/PentagonalNumber.html