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%I #5 Feb 27 2015 09:27:03
%S 1,6,10,20,40,50,50,110,155,175,105,245,371,455,490,196,476,756,980,
%T 1120,1176,336,840,1380,1860,2220,2436,2520,540,1380,2325,3225,3975,
%U 4515,4830,4950,825,2145,3685,5225,6600,7700,8470,8910,9075,1210,3190
%N Triangle read by rows: T(n,k) = (k+1)(k+2)(n+2)(n+3)(6n^2 - 8n*k + 18n + 3k^2 - 11k + 12)/144 for 0<=k<=n.
%C Kekulé numbers for certain benzenoids. Column 0 yields A002415. Main diagonal yields A006542.
%D S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 237, K{F(n,3,l)}).
%e Triangle begins:
%e 1;
%e 6,10;
%e 20,40,50;
%e 50,110,155,175;
%p T:=proc(n,k) if k<=n then 1/144*(k+1)*(k+2)*(n+2)*(n+3)*(6*n^2-8*n*k+18*n+3*k^2-11*k+12) else 0 fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
%Y Cf. A002415, A006542.
%K nonn,tabl
%O 0,2
%A _Emeric Deutsch_, Jun 12 2005
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