A277228 Convolution of the even-indexed triangular numbers (A014105) and the squares (A000290).

0, 0, 3, 22, 88, 258, 623, 1316, 2520, 4476, 7491, 11946, 18304, 27118, 39039, 54824, 75344, 101592, 134691, 175902, 226632, 288442, 363055, 452364, 558440, 683540, 830115, 1000818, 1198512, 1426278, 1687423, 1985488, 2324256, 2707760, 3140291, 3626406

Offset: 0

Author

Wolfdieter Lang, Oct 20 2016

Keywords

• nonn
• easy

Comments

This sequence was originally proposed in a comment on A071245 by J. M. Bergot as giving the first differences. Therefore, a(n) gives the partial sums of A071245.

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, A014105, A071245.

Programs

Magma

[Binomial(n+1, 3)*(4*n^2 +5*n +4)/10: n in [0..40]]; // G. C. Greubel, Oct 22 2018

Mathematica

Table[(n - 1) n (n + 1) (4 n^2 + 5 n + 4)/60, {n, 0, 40}] (* Bruno Berselli, Oct 21 2016 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,0,3,22,88,258},40] (* Harvey P. Dale, Jun 04 2023 *)

PARI

concat(vector(2), Vec(x^2*(1+x)*(3+x)/(1-x)^6 + O(x^50))) \\ Colin Barker, Oct 21 2016

Formula

O.g.f.: x^2*(1 + x)*(3 + x)/(1 - x)^6 = (x*(3 + x)/(1 - x)^3)*(x*(1 + x)/(1 - x)^3).

a(n) = Sum_{k=0..n} A014105(n-k)*A000290(k).

a(n) = binomial(n+1, 3)*(4*n^2 + 5*n + 4)/10 = (n - 1)*n*(n + 1)*(4*n^2 + 5*n + 4)/60.

a(n) = Sum_{k=0..n} A071245(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