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

%I #12 Jul 22 2022 09:33:30

%S 1,144,1485,7168,23625,61776,138229,276480,508113,874000,1425501,

%T 2225664,3350425,4889808,6949125,9650176,13132449,17554320,23094253,

%U 29952000,38349801,48533584,60774165,75368448,92640625,112943376

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

%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. 310).

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F G.f.: (1 + 138z + 636z^2 + 398z^3 + 27z^4)/(1-z)^6.

%p a:=n->(n+1)^3*(2*n+1)*(5*n+1): seq(a(n), n=0..30);

%t Table[(2n + 1)(5n + 1)(n + 1)^3, {n, 0, 30}] (* _Wesley Ivan Hurt_, Jan 24 2014 *)

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Jun 19 2005