|
%I #17 May 31 2022 03:23:45
%S 1,25,245,1470,6468,22932,69300,185130,448305,1002001,2095093,4140500,
%T 7796880,14080080,24511824,41314284,67660425,107991345,168413245,
%U 257188162,385334180,567352500,822100500,1173831750,1653425865
%N a(n) = (n+1)*(n+2)^2*(n+3)^2*(n+4)^2*(n+5)*(2n+3)/8640.
%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. 229).
%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 G.f.: (1+15*x+40*x^2+25*x^3+3*x^4)/(1-x)^10. - _Bruno Berselli_, Apr 19 2011
%F From _Amiram Eldar_, May 31 2022: (Start)
%F Sum_{n>=0} 1/a(n) = 264*Pi^2 - 49152*log(2)/35 - 57089/35.
%F Sum_{n>=0} (-1)^n/a(n) = 12288*Pi/35 - 12*Pi^2 + 13824*log(2)/35 - 44007/35. (End)
%p a:=n->(1/8640)*(n+1)*(n+2)^2*(n+3)^2*(n+4)^2*(n+5)*(2*n+3): seq(a(n),n=0..30);
%t Table[((n+1)(n+2)^2(n+3)^2(n+4)^2(n+5)(2n+3))/8640,{n,0,30}] (* _Harvey P. Dale_, Apr 19 2011 *)
%K nonn
%O 0,2
%A _Emeric Deutsch_, Jun 12 2005
|
