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a(n) = (n+1)*(n+2)^2*(n+3)*(n+4)*(3*n+5)/240.
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%I #22 Jun 02 2022 14:50:23

%S 1,12,66,245,714,1764,3864,7722,14355,25168,42042,67431,104468,157080,

%T 230112,329460,462213,636804,863170,1152921,1519518,1978460,2547480,

%U 3246750,4099095,5130216,6368922,7847371,9601320,11670384,14098304,16933224,20227977

%N a(n) = (n+1)*(n+2)^2*(n+3)*(n+4)*(3*n+5)/240.

%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 (pp. 167-169, Table 10.5/II/3).

%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.: (1 + 5*x + 3*x^2)/(1-x)^7.

%F a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7). - _Wesley Ivan Hurt_, May 03 2015

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

%F Sum_{n>=0} 1/a(n) = 405*sqrt(3)*Pi/7 + 20*Pi^2 - 3645*log(3)/7 + 1280/21.

%F Sum_{n>=0} (-1)^n/a(n) = 810*sqrt(3)*Pi/7 - 10*Pi^2 - 4160*log(2)/7 - 2480/21. (End)

%p a:=n->(n+1)*(n+2)^2*(n+3)*(n+4)*(3*n+5)/240: seq(a(n),n=0..35);

%t CoefficientList[Series[(1+5x+3x^2)/(1-x)^7,{x,0,40}],x] (* _Harvey P. Dale_, Feb 19 2011 *)

%t Table[(n + 1) (n + 2)^2 (n + 3) (n + 4) (3 n + 5) / 240, {n, 0, 50}] (* _Vincenzo Librandi_, May 03 2015 *)

%o (Magma) [(n+1)*(n+2)^2*(n+3)*(n+4)*(3*n+5)/240 : n in [0..50]]; // _Wesley Ivan Hurt_, May 03 2015

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005