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%I #65 Dec 11 2024 06:10:22
%S 1,11,48,140,325,651,1176,1968,3105,4675,6776,9516,13013,17395,22800,
%T 29376,37281,46683,57760,70700,85701,102971,122728,145200,170625,
%U 199251,231336,267148,306965,351075,399776,453376,512193,576555,646800,723276,806341
%N Partial sums of cupolar numbers (1/3)*(n+1)*(5*n^2+7*n+3) (A096000).
%C Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, p and q. - _Wesley Ivan Hurt_, Apr 15 2018
%H G. C. Greubel, <a href="/A117066/b117066.txt">Table of n, a(n) for n = 1..5000</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=1..n} A096000(i).
%F a(n) = Sum_{i=1..n} (1/3)*(i+1)*(5*i^2+7*i+3).
%F a(n) = Sum_{i=1..n} (1/2)*(Q(i) + 3*i^2 + 3*i + 1), where Q(i) are the cuboctahedral numbers (A005902).
%F a(n) = Sum_{i=0..n} A073254(n,i)*i. - _Peter Luschny_, Oct 29 2011
%F G.f.: x*(1+6*x+3*x^2) / (1-x)^5. - _Colin Barker_, May 08 2013
%F 9*a(n) = Sum_{i=0..n} (n+i)^3, see Maple code by _Zerinvary Lajos_. - _Bruno Berselli_, Apr 01 2014
%F a(n) = n^2*(n+1)*(5*n+1)/12. - _Vaclav Kotesovec_, Jan 03 2017
%F E.g.f.: (x/12)*(12 + 54*x + 36*x^2 + 5*x^3)*exp(x). - _G. C. Greubel_, Jul 19 2017
%F Another identity: ..., a(4) = (1/2)*(7*(2+4+6+8)+5*(4+6+8)+3*(6+8)+1*(8)) = 140, a(5) = (1/2)*(9*(2+4+6+8+10)+7*(4+6+8+10)+5*(6+8+10)+3*(8+10)+1*(10)) = 325, ... - _J. M. Bergot_, Aug 25 2022
%e For n=6, 9*a(6) = 6^3 + 7^3 + 8^3 + 9^3 + 10^3 + 11^3 +12^3 = 9*651. - _Bruno Berselli_, Apr 01 2014
%p a:=n->sum ((n+j)^3, j=0..n): seq(a(n)/9, n=1..40);# _Zerinvary Lajos_, Dec 17 2008
%t Table[Sum[n i (n + i), {i, 0, n}]/2, {n, 40}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 03 2011 *)
%t Accumulate[Table[((n+1)(5n^2+7n+3))/3,{n,0,50}]] (* or *) LinearRecurrence[{5,-10,10,-5,1},{1,11,48,140,325},50] (* _Harvey P. Dale_, Jan 03 2024 *)
%o (PARI) my(x='x+O('x^40)); Vec(x*(3*x^2+6*x+1)/(1-x)^5) \\ _G. C. Greubel_, Jul 19 2017
%o (PARI) a(n) = n^2*(n+1)*(5*n+1)/12; \\ _Altug Alkan_, Apr 16 2018
%o (Magma) [n^2*(n+1)*(5*n+1)/12: n in [1..40]]; // _Vincenzo Librandi_, Apr 16 2018
%o (Sage) [n^2*(n+1)*(5*n+1)/12 for n in (1..40)] # _G. C. Greubel_, Jul 05 2019
%o (GAP) List([1..40], n-> n^2*(n+1)*(5*n+1)/12); # _G. C. Greubel_, Jul 05 2019
%Y Cf. A005902, A096000.
%K nonn,easy
%O 1,2
%A _Jonathan Vos Post_, Apr 17 2006
%E Terms corrected by _Colin Barker_, May 08 2013