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a(n) = (n+1)(n+2)^3*(n+3)^2*(n+4)(n^2 + 4n + 5)/1440.
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%I #12 Apr 23 2020 07:39:40

%S 1,30,340,2275,10878,41160,131040,365310,916575,2110966,4528524,

%T 9150505,17568460,32272800,57041664,97454268,161556525,260710590,

%U 410664100,632879247,956166442,1418672200,2070276000,2975456250,4216691115

%N a(n) = (n+1)(n+2)^3*(n+3)^2*(n+4)(n^2 + 4n + 5)/1440.

%C Kekulé numbers for certain benzenoids.

%H Colin Barker, <a href="/A107967/b107967.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).

%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

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

%F G.f.: (1 + 20*x + 85*x^2 + 105*x^3 + 38*x^4 + 3*x^5) / (1 - x)^10.

%F a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10) for n>9.

%F (End)

%p a:=n->(1/1440)*(n+1)*(n+2)^3*(n+3)^2*(n+4)*(n^2+4*n+5): seq(a(n),n=0..30);

%o (PARI) Vec((1 + 20*x + 85*x^2 + 105*x^3 + 38*x^4 + 3*x^5) / (1 - x)^10 + O(x^30)) \\ _Colin Barker_, Apr 22 2020

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Jun 12 2005