A116689 Partial sums of dodecahedral numbers (A006566).

0, 1, 21, 105, 325, 780, 1596, 2926, 4950, 7875, 11935, 17391, 24531, 33670, 45150, 59340, 76636, 97461, 122265, 151525, 185745, 225456, 271216, 323610, 383250, 450775, 526851, 612171, 707455, 813450, 930930, 1060696, 1203576, 1360425

Offset: 0

Author

Jonathan Vos Post, Mar 15 2006

Keywords

• nonn
• easy

Comments

Geometrically, the partial sums of dodecahedral numbers may be interpreted as 4-dimensional dodecahedral hyperpyramidal numbers. It is somewhat surprising that this is (with proper offset) the triangular number of the "second pentagonal numbers, minus 1."

Also, the sequence is related to A004188 by a(n) = n*A004188(n)-sum(A004188(i), i=0..n-1). [Bruno Berselli, Apr 05 2012]

Links

• Bruno Berselli, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000217, A001844, A004188, A005449, A006566, A095794.

Programs

Maple

A116689:=n->binomial(n*(3*n+1)/2,2); seq(A116689(k), k=0..100); # Wesley Ivan Hurt, Oct 06 2013

Mathematica

Table[Binomial[n(3n+1)/2,2], {n,0,100}] (* Wesley Ivan Hurt, Oct 06 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{0,1,21,105,325},40] (* Harvey P. Dale, Apr 01 2018 *)

Formula

a(n) = Sum_{i = 0..n} A006566(i).

a(n) = Sum_{i = 0..n} i*(3*i-1)*(3*i-2)/2.

a(n+1) = A000217(A095794(n)).

a(n+1) = A000217(A005449(n) - 1).

a(n+1) = A000217(n*(3n+1)/2-1).

a(n+1) = A000217(A001844(n) - A000217(n+1) - 1).

a(n) = n*(9*n^3+6*n^2-5*n-2)/8. G.f.: x*(1+16*x+10*x^2)/(1-x)^5. [Colin Barker, Apr 04 2012]

a(n) = binomial(A005449(n), 2). - Wesley Ivan Hurt, Oct 06 2013

From Peter Bala, Sep 03 2023: (Start)

a(n) = n*(n + 1)*(3*n + 1)*(3*n - 2)/8.

a(n) = Sum_{0 <= i <= j <= n-1} (3*i + 1)*(3*j + 1). Cf. A024212 (End)