A260664 Number of ordered triples of partitions of n with no common parts.
1, 0, 6, 18, 90, 192, 864, 1710, 5970, 13110, 36810, 75984, 210546, 410130, 1003908, 2045808, 4616730, 8950176, 19746720, 37297710, 78247344, 147410640, 294299424, 543058032, 1067679540, 1925323308, 3653769792, 6555529158, 12129597486, 21348640230
Offset: 0
Keywords
Examples
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.
Links
- 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
Programs
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Haskell
a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row
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Mathematica
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 *)
Formula
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
Comments