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%I #36 Jan 13 2024 00:29:01
%S 1,10,42,120,275,546,980,1632,2565,3850,5566,7800,10647,14210,18600,
%T 23936,30345,37962,46930,57400,69531,83490,99452,117600,138125,161226,
%U 187110,215992,248095,283650,322896,366080,413457,465290,521850,583416,650275,722722
%N a(n) = (n+1)^2*(n+2)*(2*n+3)/6.
%C Kekulé numbers for certain benzenoids.
%D S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 232, # 44).
%H G. C. Greubel, <a href="/A108678/b108678.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 G.f.: (1 + 5*x + 2*x^2)/(1-x)^5.
%F a(n) = A098077(n+1)/2. - _Alexander Adamchuk_, Apr 12 2006
%F From _Amiram Eldar_, May 31 2022: (Start)
%F Sum_{n>=0} 1/a(n) = Pi^2 + 48*log(2) - 42.
%F Sum_{n>=0} (-1)^n/a(n) = Pi^2/2 - 12*Pi - 12*log(2) + 42. (End)
%F From _G. C. Greubel_, Apr 09 2023: (Start)
%F a(n) = (1/3)*binomial(n+2, 2)*binomial(2*n+3, 2).
%F a(n) = (1/3)*A000217(n+1)*A014105(n+1)
%F a(n) = (1/8)*A100431(n).
%F E.g.f.: (1/6)*(6 + 54*x + 69*x^2 + 23*x^3 + 2*x^4)*exp(x). (End)
%F a(n) = (n+1)*A000330(n+1). - _Olivier Gérard_, Jan 13 2024
%p a:=n->(n+1)^2*(n+2)*(2*n+3)/6: seq(a(n),n=0..42);
%p a:=n->sum(n*j^2, j=1..n): seq(a(n), n=1..36); # _Zerinvary Lajos_, Apr 29 2007
%t Table[(n+1)^2*(n+2)(2n+3)/6,{n,0,100}] (* _Vladimir Joseph Stephan Orlovsky_, Jun 03 2011 *)
%o (Magma) [(n+1)^2*(n+2)*(2*n+3)/6: n in [0..60]]; // _G. C. Greubel_, Apr 09 2023
%o (SageMath) [(n+1)^2*(n+2)*(2*n+3)/6 for n in range(61)] # _G. C. Greubel_, Apr 09 2023
%Y Cf. A000217, A014105, A098077, A100431.
%K nonn,easy
%O 0,2
%A _Emeric Deutsch_, Jun 17 2005
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