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a(n) = (n+1)(n+2)^2*(n+3)^2*(n+4)(7n^2 + 23n + 20)/2880.
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%I #12 Aug 16 2021 14:01:09

%S 1,25,235,1330,5488,18228,51660,129690,295845,624481,1236235,2318680,

%T 4153240,7149520,11888304,19174572,30101985,46130385,69177955,

%U 101729782,146964664,208902100,292571500,404205750,551461365,743667561,992106675

%N a(n) = (n+1)(n+2)^2*(n+3)^2*(n+4)(7n^2 + 23n + 20)/2880.

%C Kekulé numbers for certain benzenoids.

%H Colin Barker, <a href="/A108178/b108178.txt">Table of n, a(n) for n = 0..1000</a>

%H S. J. Cyvin and I. Gutman, <a href="https://doi.org/10.1007/978-3-662-00892-8">Kekulé structures in benzenoid hydrocarbons</a>, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 20).

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%F From _Colin Barker_, Apr 22 2020: (Start)

%F G.f.: (1 + 16*x + 46*x^2 + 31*x^3 + 4*x^4) / (1 - x)^9.

%F a(n) = 9*a(n-1) - 36*a(n-2) + 84*a(n-3) - 126*a(n-4) + 126*a(n-5) - 84*a(n-6) + 36*a(n-7) - 9*a(n-8) + a(n-9) for n>8.

%F (End)

%p a:=(n+1)*(n+2)^2*(n+3)^2*(n+4)*(7*n^2+23*n+20)/2880: seq(a(n),n=0..30);

%t Table[(n+1)(n+2)^2(n+3)^2(n+4)(7n^2+23n+20)/2880,{n,0,50}] (* or *) LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,25,235,1330,5488,18228,51660,129690,295845},50] (* _Harvey P. Dale_, Aug 16 2021 *)

%o (PARI) Vec((1 + 16*x + 46*x^2 + 31*x^3 + 4*x^4) / (1 - x)^9 + O(x^30)) \\ _Colin Barker_, Apr 22 2020

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Jun 13 2005