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

0,3

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).

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)

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 *)

Crossrefs

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

Sequence in context: A068142 A126229 A060537 * A235876 A279763 A068986

Adjacent sequences: A116686 A116687 A116688 * A116690 A116691 A116692

Keyword

nonn,easy

Author

Jonathan Vos Post, Mar 15 2006

Status

approved