A110450 a(n) = n*(n+1)*(n^2+n+1)/2.

0, 3, 21, 78, 210, 465, 903, 1596, 2628, 4095, 6105, 8778, 12246, 16653, 22155, 28920, 37128, 46971, 58653, 72390, 88410, 106953, 128271, 152628, 180300, 211575, 246753, 286146, 330078, 378885, 432915, 492528, 558096, 630003, 708645, 794430

Offset: 0

Author

Reinhard Zumkeller, Jul 21 2005

Keywords

• nonn
• easy

Comments

This sequence is related to A085461 by 3*A085461(n) = n*a(n) - Sum_{i=0..n-1} a(i) for n>0. - Bruno Berselli, Dec 27 2010

Subsequence of the triangular numbers A000217, see formulas below. - David James Sycamore, Jul 31 2018

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000217, A002378, A061317, A085461, A110449, A014105.

Programs

GAP

List([0..40],n->n*(n+1)*(n^2+n+1)/2); # Muniru A Asiru, Aug 02 2018

Magma

[n*(n+1)*(n^2+n+1)/2: n in [0..40]]; // Vincenzo Librandi, Dec 26 2010

Maple

A110450:=n->n*(n+1)*(n^2+n+1)/2; seq(A110450(k), k=0..50); # Wesley Ivan Hurt, Sep 27 2013

Mathematica

Table[n (n + 1) (n^2 + n + 1)/2, {n, 0, 100}] (* Wesley Ivan Hurt, Sep 27 2013 *)
CoefficientList[Series[-3 x (x^2 + 2 x + 1)/(x - 1)^5, {x, 0, 36}], x] (* or *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 3, 21, 78, 210}, 36] (* Robert G. Wilson v, Jul 31 2018 *)

PARI

a(n)=n*(n+1)*(n^2+n+1)/2 \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = Sum_{k=0..n} A110449(n,k), sums of rows in triangle A110449.

From Bruno Berselli, Dec 27 2010: (Start)

G.f.: 3*x*(1 + x)^2/(1 - x)^5.

a(n) = A014105(A000217(n)). (End)

a(n) = Sum_{i=1..n*(n+1)} i. - Wesley Ivan Hurt, Sep 27 2013

a(n) = Sum_{i=0..n} i*(2*i^2+1), and these are the partial sums of A061317. - Bruno Berselli, Feb 09 2017

a(n) = t(n,t(n,A000217(n))), where t(n,k) = n*(n+1)/2 + k*n and k=0. - Bruno Berselli, Feb 28 2017

E.g.f.: (x/2)*(6 + 15*x + 8*x^2 + x^3)*exp(x). - G. C. Greubel, Aug 24 2017

a(n) = A000217(n*(n+1)). - David James Sycamore, Jul 31 2018

a(n) = A000217(2*A000217(n)) = A000217(A002378(n)). - Alois P. Heinz, Jul 31 2018

From R. J. Mathar, Mar 23 2021: (Start)

a(n) = A002378(n)+A062392(n).

a(n) = 3*A006325(n+1). (End)

Sum_{n>=1} 1/a(n) = 4 - 2*Pi*tanh(sqrt(3)*Pi/2)/sqrt(3). - Amiram Eldar, May 10 2025