0,3

Reinhard Zumkeller, Table of n, a(n) for n = 0..5000

Sylvie Corteel, Carla D. Savage, Herbert S. Wilf, Doron Zeilberger, A pentagonal number sieve, J. Combin. Theory Ser. A 82 (1998), no. 2, 186-192.

Eric Weisstein's World of Mathematics, Pentagonal Number Theorem

Wikipedia, Pentagonal number theorem

a(n) = p(n)^3 - p(n-k(1))^3 - p(n-k(2))^3 + p(n-k(3))^3 + p(n-k(4))^3 - p(n-k(5))^3 - ..., with p=A000041 and k=A001318, see Wilf link: p. 2, (3).

G.f.: Sum[p(n)^3*x^n]/Sum[p(n)*x^n], with p(n)=number of partitions of n. - Vaclav Kotesovec, Jul 04 2016

a(n) ~ 2^(3/2) * exp(4*Pi*sqrt(n/3)) / (729 * 3^(1/4) * n^(11/4)). - Vaclav Kotesovec, May 20 2018

a(3) = 18 because of the 18 triples of partitions of 3: (3,3,21), (3,3,111), (3,21,3), (3,21,21), (3,21,111), (3,111,3), (3,111,21), (3,111,111), (21,3,3), (21,3,21), (21,3,111), (21,21,3), (21,111,3), (111,3,3), (111,3,21), (111,3,111), (111,21,3) and (111,111,3);

a(3) = A000041(3-A001318(0))^3 - A000041(3-A001318(1))^3 - A000041(3-A001318(2))^3 = 3^3 - 2^3 - 1^3 = 27 - 8 - 1 = 18.

Table[Sum[(Cos[Pi*j/2] - Sin[Pi*j/2]) * PartitionsP[n - ((6*j^2 + 6*j + 1)/16 - (2*j + 1)*(-1)^j/16)]^3, {j, 0, Floor[Sqrt[8*n/3]]}], {n, 0, 30}] (* Vaclav Kotesovec, Jul 04 2016 *)

nmax = 50; CoefficientList[Series[Sum[PartitionsP[k]^3*x^k, {k, 0, nmax}] / Sum[PartitionsP[k]*x^k, {k, 0, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 04 2016 *)

(Haskell)

a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row

Cf. A000041, A133042, A001318, A087960, A260672, A054440, A304873, A304878.

Sequence in context: A188119 A239420 A219590 * A279260 A379178 A371987

Adjacent sequences: A260661 A260662 A260663 * A260665 A260666 A260667

Reinhard Zumkeller, Nov 15 2015

approved