A256313 Number of partitions of 3n into exactly 4 parts.

0, 0, 2, 6, 15, 27, 47, 72, 108, 150, 206, 270, 351, 441, 551, 672, 816, 972, 1154, 1350, 1575, 1815, 2087, 2376, 2700, 3042, 3422, 3822, 4263, 4725, 5231, 5760, 6336, 6936, 7586, 8262, 8991, 9747, 10559, 11400, 12300, 13230, 14222, 15246, 16335, 17457

Offset

0,3

Links

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

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

Formula

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

a(n) = (6*n^3+6*n^2-3*n-5+(3*n+1)*(-1)^n+2*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/32. - Luce ETIENNE, Feb 17 2017

Example

For n=3 the 6 partitions of 3*3 = 9 are [1,1,1,6], [1,1,2,5], [1,1,3,4], [1,2,2,4], [1,2,3,3] and [2,2,2,3].

Mathematica

LinearRecurrence[{2, 0, -2, 2, -2, 0, 2, -1}, {0, 0, 2, 6, 15, 27, 47, 72}, 60] (* Harvey P. Dale, Jul 18 2021 *)

Prog

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

Crossrefs

Cf. A077043, A256314, A256315.

Sequence in context: A293402 A192691 A360405 * A138621 A163061 A331773

Adjacent sequences: A256310 A256311 A256312 * A256314 A256315 A256316

Keyword

nonn,easy

Author

Colin Barker, Mar 23 2015

Status

approved