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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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified December 9 19:36 EST 2022. Contains 358703 sequences. (Running on oeis4.)