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A260664 Number of ordered triples of partitions of n with no common parts.

%I A260664 #32 Feb 16 2025 08:33:26
%S A260664 1,0,6,18,90,192,864,1710,5970,13110,36810,75984,210546,410130,
%T A260664 1003908,2045808,4616730,8950176,19746720,37297710,78247344,147410640,
%U A260664 294299424,543058032,1067679540,1925323308,3653769792,6555529158,12129597486,21348640230
%N A260664 Number of ordered triples of partitions of n with no common parts.
%H A260664 Reinhard Zumkeller, <a href="/A260664/b260664.txt">Table of n, a(n) for n = 0..5000</a>
%H A260664 Sylvie Corteel, Carla D. Savage, Herbert S. Wilf, Doron Zeilberger, <a href="https://doi.org/10.1006/jcta.1997.2846">A pentagonal number sieve</a>, J. Combin. Theory Ser. A 82 (1998), no. 2, 186-192.
%H A260664 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PentagonalNumberTheorem.html">Pentagonal Number Theorem</a>
%H A260664 Wikipedia, <a href="http://www.wikipedia.org/wiki/Pentagonal_number_theorem">Pentagonal number theorem</a>
%F A260664 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).
%F A260664 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
%F A260664 a(n) ~ 2^(3/2) * exp(4*Pi*sqrt(n/3)) / (729 * 3^(1/4) * n^(11/4)). - _Vaclav Kotesovec_, May 20 2018
%e A260664 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);
%e A260664 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.
%t A260664 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 *)
%t A260664 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 *)
%o A260664 (Haskell)
%o A260664 a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row
%Y A260664 Cf. A000041, A133042, A001318, A087960, A260672, A054440, A304873, A304878.
%K A260664 nonn
%O A260664 0,3
%A A260664 _Reinhard Zumkeller_, Nov 15 2015