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a(n) = 4*(n+1)^2*(n+3)^2*(5*n^2 + 20*n + 12).
0

%I #13 Jun 02 2022 14:50:20

%S 432,9472,64800,269568,842800,2184192,4953312,10163200,19288368,

%T 34387200,58238752,94493952,147841200,224186368,330847200,476762112,

%U 672713392,931564800,1268513568,1701356800,2250772272,2940613632,3798220000,4854739968,6145470000,7710207232

%N a(n) = 4*(n+1)^2*(n+3)^2*(5*n^2 + 20*n + 12).

%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. 311).

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F G.f.: 16*(27 + 403*x + 473*x^2 - 15*x^3 + 14*x^4 - 2*x^5)/(1-x)^7.

%F From _Amiram Eldar_, Jun 02 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 197/864 - Pi^2/144 - (5/144)*sqrt(5/2)*Pi*cot(2*sqrt(2/5)*Pi).

%F Sum_{n>=0} (-1)^n/a(n) = -85/432 - Pi^2/288 - (5/144)*sqrt(5/2)*Pi*cosec(2*sqrt(2/5)*Pi). (End)

%p a:=n->4*(n+1)^2*(n+3)^2*(5*n^2+20*n+12): seq(a(n),n=0..28);

%t Table[4*(n + 1)^2*(n + 3)^2*(5*n^2 + 20*n + 12), {n, 0, 30}] (* _Amiram Eldar_, Jun 02 2022 *)

%K nonn

%O 0,1

%A _Emeric Deutsch_, Jun 19 2005