A277229 Convolution of the odd-indexed triangular numbers (A000384(n+1)) and the squares (A000290).

0, 1, 10, 48, 158, 413, 924, 1848, 3396, 5841, 9526, 14872, 22386, 32669, 46424, 64464, 87720, 117249, 154242, 200032, 256102, 324093, 405812, 503240, 618540, 754065, 912366, 1096200, 1308538, 1552573, 1831728, 2149664, 2510288, 2917761, 3376506, 3891216

Offset: 0

Author

Wolfdieter Lang, Oct 20 2016

Keywords

• nonn
• easy

Comments

This sequence was originally proposed in a comment on A071238 by J. M. Bergot as giving the first differences. Therefore, a(n) gives the partial sums of A071238.

Links

• Colin Barker, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A000217, A000290, A000384, A071238.

Programs

Mathematica

Table[n (n + 1) (n + 2) (4 n^2 + 3 n + 3)/60, {n, 0, 40}] (* Bruno Berselli, Oct 21 2016 *)

PARI

concat(0, Vec(x*((1+x)*(1+3*x))/(1-x)^6 + O(x^50))) \\ Colin Barker, Oct 21 2016

Formula

O.g.f.: x*(1 + x)*(1 + 3*x)/(1 - x)^6 = ((1 + 3*x)/(1 - x)^3)*(x*(1 + x)/(1 - x)^3).

a(n) = Sum_{k=0..n} A000384(n+1-k)*A000290(k).

a(n) = binomial(n+2, 3)*(4*n^2 + 3*n + 3)/10 = n*(n + 1)*(n + 2)*(4*n^2 + 3*n + 3)/60.

a(n) = Sum_{k=0..n} A071238(k).

a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>5. - Colin Barker, Oct 21 2016