A256310 Number of partitions of 2n into exactly 6 parts.

0, 0, 0, 1, 2, 5, 11, 20, 35, 58, 90, 136, 199, 282, 391, 532, 709, 931, 1206, 1540, 1945, 2432, 3009, 3692, 4494, 5427, 6510, 7760, 9192, 10829, 12692, 14800, 17180, 19858, 22856, 26207, 29941, 34085, 38677, 43752, 49342, 55491, 62239, 69624, 77695, 86499

Offset

0,5

Comments

The number of partitions of 2*(n-3) into at most 6 parts. - Colin Barker, Mar 31 2015

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,0,0,-3,1,1,0,0,1,1,-3,0,0,2,-1).

Formula

G.f.: x^3*(x^4+x^3+x^2+1) / ((x-1)^6*(x+1)*(x^2+x+1)^2*(x^4+x^3+x^2+x+1)).

a(n) = 2*a(n-1) - 3*a(n-4) + a(n-5) + a(n-6) + a(n-9) + a(n-10) - 3*a(n-11) + 2*a(n-14) - a(n-15). - Wesley Ivan Hurt, Jul 28 2022

Example

For n=4 the 2 partitions of 2*4 = 8 are [1,1,1,1,1,3] and [1,1,1,1,2,2].

Mathematica

CoefficientList[Series[x^3 (x^4 + x^3 + x^2 + 1) / ((x - 1)^6 (x + 1) (x^2 + x + 1)^2 (x^4 + x^3 + x^2 + x + 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 22 2015 *)

Prog

(PARI) concat(0, vector(40, n, k=0; forpart(p=2*n, k++, , [6, 6]); k))
(PARI) concat([0, 0, 0], Vec(x^3*(x^4+x^3+x^2+1)/((x-1)^6*(x+1)*(x^2+x+1)^2*(x^4+x^3+x^2+x+1)) + O(x^100)))

Crossrefs

Cf. Similar sequences: A000212 (3 parts), A001477 (2 parts), A014126 (4 parts), A256309 (5 parts).

Sequence in context: A038377 A261227 A022908 * A026390 A005575 A328670

Adjacent sequences: A256307 A256308 A256309 * A256311 A256312 A256313

Keyword

nonn,easy

Author

Colin Barker, Mar 22 2015

Status

approved