%I
%S 1,6,4,20,20,10,50,60,45,20,105,140,126,84,35,196,280,280,224,140,56,
%T 336,504,540,480,360,216,84,540,840,945,900,750,540,315,120,825,1320,
%U 1540,1540,1375,1100,770,440,165,1210,1980,2376,2464,2310,1980,1540
%N Triangle read by rows: T(n,k) = (k+1)(n+2)(n+3)(nk+2)(nk+1)/12 for 0<=k<=n.
%C KekulĂ© numbers for certain benzenoids. Column 0 yields A002415. Main diagonal yields A000292. Row sums yield A006542.
%C T(n,k) = number of Dyck (n+4)paths with 4 peaks (UDs) and last descent of length k+1. For example, T(1,1)=4 counts UUDUDUDUDD, UDUUDUDUDD, UDUDUUDUDD, UDUDUDUUDD.  _David Callan_, Jun 26 2006
%D S. J. Cyvin and I. Gutman, KekulĂ© 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 6,4;
%e 20,20,10;
%e 50,60,45,20;
%p T:=proc(n,k) if k<=n then (k+1)*(n+2)*(n+3)*(nk+2)*(nk+1)/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
%t Flatten[Table[((k+1)(n+2)(n+3)(nk+2)(nk+1))/12,{n,0,10},{k,0,n}]] (* _Harvey P. Dale_, Aug 08 2013 *)
%Y Cf. A002415, A000292, A006542.
%K nonn,tabl
%O 0,2
%A _Emeric Deutsch_, Jun 12 2005
