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a(n) = (n+1)(n+2)^2*(n+3)(7n^2 + 23n + 20)/240.
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%I #10 Jul 22 2022 08:52:06

%S 1,15,94,380,1176,3038,6888,14148,26895,48037,81510,132496,207662,

%T 315420,466208,672792,950589,1318011,1796830,2412564,3194884,4178042,

%U 5401320,6909500,8753355,10990161,13684230,16907464,20739930,25270456

%N a(n) = (n+1)(n+2)^2*(n+3)(7n^2 + 23n + 20)/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 (p. 168).

%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.: -(2*x^3+10*x^2+8*x+1)/(x-1)^7. - _Alois P. Heinz_, Feb 27 2015

%p a:=n->(n+1)*(n+2)^2*(n+3)*(7*n^2+23*n+20)/240: seq(a(n),n=0..33);

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005