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a(n) = (n+1)(n+2)^3*(n+3)(n^2 + 4n + 5)/120.
3

%I #22 Apr 02 2021 08:38:14

%S 1,18,136,650,2331,6860,17472,39852,83325,162382,298584,522886,878423,

%T 1423800,2236928,3419448,5101785,7448874,10666600,15008994,20786227,

%U 28373444,38220480,50862500,66931605,87169446,112440888,143748766

%N a(n) = (n+1)(n+2)^3*(n+3)(n^2 + 4n + 5)/120.

%C Kekulé numbers for certain benzenoids.

%C First differences of A107891. Partial sums of A083200. - _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 (see p. 167, Table 10.5/I/6).

%H Seiichi Manyama, <a href="/A114239/b114239.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n-2) = (n^7-n^3)/(2^7-2^3). - _David Radcliffe_, Dec 27 2008

%F G.f.: (1+10*x+20*x^2+10*x^3+x^4)/(1-x)^8. - _Colin Barker_, Feb 09 2012

%p a:=n->(n+1)*(n+2)^3*(n+3)*(n^2+4*n+5)/120: seq(a(n),n=0..33);

%o (PARI) a(n)=n-=2;(n^7-n^3)/120 \\ _Charles R Greathouse IV_, Feb 09 2012

%Y Cf. A047819, A083200, A107891.

%K nonn,easy

%O 0,2

%A _Emeric Deutsch_, Nov 18 2005