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%I #14 Jun 11 2021 19:01:05
%S 2,20,250,3250,42500,556250,7281250,95312500,1247656250,16332031250,
%T 213789062500,2798535156250,36633300781250,479536132812500,
%U 6277209472656250,82169738769531250,1075615844726562500
%N a(n) = (1/sqrt(5))*((sqrt(5) + 1)*((15 + 5*sqrt(5))/2)^(n-1) + (sqrt(5) - 1)*((15 - 5*sqrt(5))/2)^(n-1)).
%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. 215, K{T_m}).
%F G.f.: 2z(1-5z)/(1 - 15z + 25z^2).
%F From _Johannes W. Meijer_, Jul 01 2010: (Start)
%F a(n) = A178381(4*n+2).
%F Lim_{k->infinity} a(n+k)/a(k) = (A020876(2*n) + 5*A039717(2*n-2)*sqrt(5))/2.
%F Lim_{n->infinity} A020876(2*n)/(5*A039717(2*n-2)) = sqrt(5).
%F (End)
%F a(n) = 2*5^(n-1)*Fibonacci(2*n-1). - _Ehren Metcalfe_, Apr 21 2018
%p a:=n->(1/sqrt(5))*((sqrt(5)+1)*((15+5*sqrt(5))/2)^(n-1)+(sqrt(5)-1)*((15-5*sqrt(5))/2)^(n-1)): seq(expand(a(n)),n=1..19);
%Y Cf. A179135. - _Johannes W. Meijer_, Jul 01 2010
%K nonn
%O 1,1
%A _Emeric Deutsch_, Jun 19 2005