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%I #5 Oct 06 2015 17:22:49
%S 1,1,2,5,2,1,10,28,13,11,3,1,66,196,90,89,34,18,4,1,498,1532,694,736,
%T 311,197,66,26,5,1,4066,12804,5738,6344,2800,1937,762,367,110,35,6,1,
%U 34970,111964,49758,56576,25560,18636,7953,4263,1551,615,167,45,7,1
%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 having height of the first peak equal to k.
%C Row n contains 2n terms. Row sums yield A027307. T(n,1)=A027307(n-1).
%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 G.f.=G=G(t, z)=1/(1-tzA-t^2*zA^2)-1, where A=1+zA^2+zA^3=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3 (the g.f. of A027307).
%e T(2,3)=2 because we have uUddd and Uuddd.
%e Triangle begins:
%e 1,1;
%e 2,5,2,1;
%e 10,28,13,11,3,1;
%e 66,196,90,89,34,18,4,1;
%p A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=1/(1-t*z*A-t^2*z*A^2)-1: Gserz:=simplify(series(G,z=0,10)): for n from 1 to 9 do P[n]:=sort(coeff(Gserz,z^n)) od: for n from 1 to 9 do seq(coeff(P[n],t^k),k=1..2*n) od; # yields sequence in triangular form
%Y Cf. A027307.
%K nonn,tabf
%O 1,3
%A _Emeric Deutsch_, Jun 04 2005
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