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a(n) = (n+1)*(n+2)^2*(n+3)*(7n^2 + 28n + 30)/360.
2

%I #16 Mar 06 2023 08:58:03

%S 1,13,76,295,889,2254,5040,10242,19305,34243,57772,93457,145873,

%T 220780,325312,468180,659889,912969,1242220,1664971,2201353,2874586,

%U 3711280,4741750,6000345,7525791,9361548,11556181,14163745,17244184,20863744,25095400,30019297

%N a(n) = (n+1)*(n+2)^2*(n+3)*(7n^2 + 28n + 30)/360.

%C Kekulé numbers for certain benzenoids.

%C First differences of A114242. - _Peter Bala_, Sep 21 2007

%D S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 167-169, Table 10.5/II/5).

%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.: (1+x)(1 + 5x + x^2)/(1-x)^7.

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

%F Sum_{n>=0} 1/a(n) = 5*Pi*(7*sqrt(14)*coth(sqrt(2/7)*Pi) - 6*Pi) - 1295/9.

%F Sum_{n>=0} (-1)^n/a(n) = 5*Pi*(7*sqrt(14)*cosech(sqrt(2/7)*Pi) + 3*Pi) - 2755/9. (End)

%p a:=n->(n+1)*(n+2)^2*(n+3)*(7*n^2+28*n+30)/360: seq(a(n),n=0..35);

%t Table[(n + 1)*(n + 2)^2*(n + 3)*(7*n^2 + 28*n + 30)/360, {n, 0, 30}] (* _Amiram Eldar_, May 31 2022 *)

%t CoefficientList[Series[(1+x)(1+5x+x^2)/(1-x)^7,{x,0,40}],x] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,1},{1,13,76,295,889,2254,5040},40] (* _Harvey P. Dale_, Mar 06 2023 *)

%Y Cf. A114242.

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005