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A116689
Partial sums of dodecahedral numbers (A006566).
4
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]
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
KEYWORD
nonn,easy
AUTHOR
Jonathan Vos Post, Mar 15 2006
STATUS
approved