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A260664
Number of ordered triples of partitions of n with no common parts.
11
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
0,3
LINKS
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
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
EXAMPLE
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.
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 *)
PROG
(Haskell)
a260664 = sum . zipWith (*) a087960_list . map a133042 . a260672_row
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Nov 15 2015
STATUS
approved