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%I #8 Oct 06 2015 17:45:45
%S 1,0,1,1,0,0,2,5,3,0,0,0,5,21,28,12,0,0,0,0,14,84,180,165,55,0,0,0,0,
%T 0,42,330,990,1430,1001,273,0,0,0,0,0,0,132,1287,5005,10010,10920,
%U 6188,1428,0,0,0,0,0,0,0,429,5005,24024,61880,92820,81396,38760,7752,0,0,0,0
%N Triangle read by rows: T(n,k) is number of paths from (0,0) to (3n,0) that stay in the first quadrant (but may touch the horizontal axis), consisting of steps u=(2,1), U=(1,2), or d=(1,-1) and have k down steps (d).
%C Row n contains 2n+1 terms, the first n of which are equal to 0.
%C Row sums yield A027307.
%C T(n,n) = A000108(n) (the Catalan numbers).
%C T(n,2n) = A001764(n) = binomial(3n,n)/(2n+1).
%C Except for the 0's, the same as A104978.
%C Number of d steps in all paths from (0,0) to (3n,0) is given by A108430.
%H Emeric Deutsch, <a href="http://www.jstor.org/stable/2589192">Problem 10658: Another Type of Lattice Path</a>, American Math. Monthly, 107, 2000, 368-370.
%F T(n,k) = binomial(n,2n-k)*binomial(n+k, n-1)/n.
%F G.f.: G = G(t, z) satisfies G=1+tzG^2*(1+tG).
%e Example T(2,3) = 5 because we have udUdd, uUddd, Uddud, Ududd and Uuddd.
%e Triangle begins:
%e 1;
%e 0,1,1;
%e 0,0,2,5,3;
%e 0,0,0,5,21,28,12;
%e ...
%p a:=proc(n,k) if n=0 and k=0 then 1 elif n=0 then 0 elif k=0 then 0 else binomial(n,2*n-k)*binomial(n+k,n-1)/n fi end: for n from 0 to 8 do seq(a(n,k),k=0..2*n) od; # yields sequence in triangular form
%Y Cf. A027307, A000108, A001764, A104978, A108430.
%K nonn,tabf
%O 0,7
%A _Emeric Deutsch_, Jun 03 2005
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