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%I #26 Sep 05 2023 08:15:25
%S 0,1,21,105,325,780,1596,2926,4950,7875,11935,17391,24531,33670,45150,
%T 59340,76636,97461,122265,151525,185745,225456,271216,323610,383250,
%U 450775,526851,612171,707455,813450,930930,1060696,1203576,1360425
%N Partial sums of dodecahedral numbers (A006566).
%C 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."
%C 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]
%H Bruno Berselli, <a href="/A116689/b116689.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F a(n) = Sum_{i = 0..n} A006566(i).
%F a(n) = Sum_{i = 0..n} i*(3*i-1)*(3*i-2)/2.
%F a(n+1) = A000217(A095794(n)).
%F a(n+1) = A000217(A005449(n) - 1).
%F a(n+1) = A000217(n*(3n+1)/2-1).
%F a(n+1) = A000217(A001844(n) - A000217(n+1) - 1).
%F 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]
%F a(n) = binomial(A005449(n), 2). - _Wesley Ivan Hurt_, Oct 06 2013
%F From _Peter Bala_, Sep 03 2023: (Start)
%F a(n) = n*(n + 1)*(3*n + 1)*(3*n - 2)/8.
%F a(n) = Sum_{0 <= i <= j <= n-1} (3*i + 1)*(3*j + 1). Cf. A024212 (End)
%p A116689:=n->binomial(n*(3*n+1)/2,2); seq(A116689(k), k=0..100); # _Wesley Ivan Hurt_, Oct 06 2013
%t Table[Binomial[n(3n+1)/2,2], {n,0,100}] (* _Wesley Ivan Hurt_, Oct 06 2013 *)
%t LinearRecurrence[{5,-10,10,-5,1},{0,1,21,105,325},40] (* _Harvey P. Dale_, Apr 01 2018 *)
%Y Cf. A000217, A001844, A004188, A005449, A006566, A095794.
%K nonn,easy
%O 0,3
%A _Jonathan Vos Post_, Mar 15 2006
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