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 A107970 a(n) = (n+1)*(n+2)^3*(n+3)*(2n+3)*(2n+5)/360. 0

%I

%S 1,21,168,825,3003,8918,22848,52326,109725,214291,394680,692055,

%T 1163799,1887900,2968064,4539612,6776217,9897537,14177800,19955397,

%U 27643539,37742034,50850240,67681250,89077365,116026911,149682456,191380483

%N a(n) = (n+1)*(n+2)^3*(n+3)*(2n+3)*(2n+5)/360.

%C KekulĂ© numbers for certain benzenoids.

%D S. J. Cyvin and I. Gutman, KekulĂ© structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230).

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).

%F G.f.: (x^4+13*x^3+28*x^2+13*x+1)/(x-1)^8. - _Colin Barker_, Sep 21 2012

%F From _Amiram Eldar_, May 31 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 360*zeta(3) - 3840*log(2) + 2230.

%F Sum_{n>=0} (-1)^n/a(n) = 1490 - 1680*log(2) - 270*zeta(3). (End)

%p a:=n->(1/360)*(n+1)*(n+2)^3*(n+3)*(2*n+3)*(2*n+5): seq(a(n),n=0..32);

%t Table[(n + 1)*(n + 2)^3*(n + 3)*(2 n + 3)*(2 n + 5)/360, {n, 0, 25}] (* _Amiram Eldar_, May 31 2022 *)

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Jun 12 2005

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Last modified August 9 21:08 EDT 2022. Contains 356026 sequences. (Running on oeis4.)