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a(n) = (n+1)*(n+2)*(n+3)*(11*n^2 + 29*n + 20)/120.
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%I #20 Oct 21 2022 21:56:05

%S 1,12,61,206,546,1232,2478,4572,7887,12892,20163,30394,44408,63168,

%T 87788,119544,159885,210444,273049,349734,442750,554576,687930,845780,

%U 1031355,1248156,1499967,1790866,2125236,2507776,2943512,3437808

%N a(n) = (n+1)*(n+2)*(n+3)*(11*n^2 + 29*n + 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 (p. 168).

%H G. C. Greubel, <a href="/A114241/b114241.txt">Table of n, a(n) for n = 0..1000</a>

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

%F From _Chai Wah Wu_, Nov 11 2018: (Start)

%F a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n > 5.

%F G.f.: (4*x^2 + 6*x + 1)/(x - 1)^6. (End)

%F E.g.f.: (120 + 1320*x + 2280*x^2 + 1160*x^3 + 205*x^4 + 11*x^5)*exp(x)/5!. - _G. C. Greubel_, Nov 11 2018

%p a:=n->(n+1)*(n+2)*(n+3)*(11*n^2+29*n+20)/120: seq(a(n),n=0..38);

%t Table[Binomial[n+3,3]*(11*n^2 +29*n +20)/20, {n,0,20}] (* _G. C. Greubel_, Nov 11 2018 *)

%o (PARI) vector(35, n, n--; binomial(n+3,3)*(11*n^2 +29*n +20)/20) \\ _G. C. Greubel_, Nov 11 2018

%o (Magma) [Binomial(n+3,3)*(11*n^2 +29*n +20)/20: n in [0..35]]; // _G. C. Greubel_, Nov 11 2018

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005