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a(n) = (n+1)(n+2)(n+3)(9n^2 + 26n + 20)/120.
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%I #17 Aug 30 2022 13:58:49

%S 1,11,54,179,469,1050,2100,3858,6633,10813,16874,25389,37037,52612,

%T 73032,99348,132753,174591,226366,289751,366597,458942,569020,699270,

%U 852345,1031121,1238706,1478449,1753949,2069064,2427920,2834920,3294753

%N a(n) = (n+1)(n+2)(n+3)(9n^2 + 26n + 20)/120.

%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 (pp. 167-170, Table 10.5/II/6).

%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 + 5x + 3x^2)/(1-x)^6.

%F a(n) = Sum_{i=0..n} (i+1)*Sum_{j=n-i+1..n+1} A000217(j). - _J. M. Bergot_ and _Robert Israel_, Aug 29 2022

%p a:=n->(n+1)*(n+2)*(n+3)*(9*n^2+26*n+20)/120: seq(a(n),n=0..35);

%Y Cf. A000217.

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005