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Triangle read by rows: T(n,k) = (k+1)(k+2)(n+2)(3n-2k+3)/12 for 0<=k<=n.
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%I #6 Feb 27 2015 09:22:22

%S 1,3,6,6,14,20,10,25,40,50,15,39,66,90,105,21,56,98,140,175,196,28,76,

%T 136,200,260,308,336,36,99,180,270,360,441,504,540,45,125,230,350,475,

%U 595,700,780,825,55,154,286,440,605,770,924,1056,1155,1210,66,186,348

%N Triangle read by rows: T(n,k) = (k+1)(k+2)(n+2)(3n-2k+3)/12 for 0<=k<=n.

%C Kekulé numbers for certain benzenoids. Column 0 yields the triangular numbers (A000217). Row sums yield A006414. T(n,n) = A002415(n+2).

%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{B(n,2,l)}).

%e Triangle begins:

%e 1;

%e 3,6;

%e 6,14,20;

%e 10,25,40,50;

%p T:=proc(n,k) if k<=n then (k+1)*(k+2)*(n+2)*(3*n-2*k+3)/12 else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

%Y Cf. A000217, A006414, A002415.

%K nonn,tabl

%O 0,2

%A _Emeric Deutsch_, Jun 12 2005