%I #12 Jun 13 2015 00:51:50
%S 1,31,371,2646,13524,54684,185724,551034,1467609,3578575,8107099,
%T 17257604,34826064,67098864,124140528,221594796,383151321,643861911,
%U 1054526011,1687405258,2643571700,4062243900,6132519900,9107976150,13324667265,19223133111,27375097491
%N a(n) = (n+1)(n+2)^2*(n+3)^2*(n+4)^2*(n+5)(3n^2 + 13n + 15)/43200.
%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. 229).
%H T. D. Noe, <a href="/A107941/b107941.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).
%F G.f.: (1+20*x+85*x^2+105*x^3+38*x^4+3*x^5)/(1-x)^11. - _Colin Barker_, Sep 18 2012
%p a:=n->(1/43200)*(n+1)*(n+2)^2*(n+3)^2*(n+4)^2*(n+5)*(3*n^2+13*n+15): seq(a(n),n=0..28);
%K nonn,easy
%O 0,2
%A _Emeric Deutsch_, Jun 12 2005