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a(n) = 4*(n+1)^2*(3*n+1)^2*(12*n^2+20*n+5).
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%I #15 Feb 22 2020 20:41:45

%S 20,9472,164052,1107200,4681300,14929920,39411092,90821632,188932500,

%T 362835200,653499220,1116640512,1825901012,2876339200,4388231700,

%U 6511185920,9428563732,13362216192,18577529300,25388780800,34164808020

%N a(n) = 4*(n+1)^2*(3*n+1)^2*(12*n^2+20*n+5).

%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. 311).

%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.: 4*(5 + 2333*z + 24542*z^2 + 39262*z^3 + 11293*z^4 + 325*z^5)/(1-z)^7.

%p a:=n->4*(n+1)^2*(3*n+1)^2*(12*n^2+20*n+5): seq(a(n),n=0..24);

%t Table[4(n+1)^2(3n+1)^2(12n^2+20n+5),{n,0,30}] (* _Harvey P. Dale_, Jun 14 2019 *)

%K nonn,easy

%O 0,1

%A _Emeric Deutsch_, Jun 19 2005