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A108438 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 abscissa of the first peak equal to k.

%I A108438 #6 Oct 06 2015 18:01:45
%S A108438 1,1,4,3,2,1,24,18,13,7,3,1,172,130,96,55,28,12,4,1,1360,1034,772,458,
%T A108438 249,119,50,18,5,1,11444,8738,6568,3982,2244,1137,526,219,80,25,6,1,
%U A108438 100520,76994,58140,35770,20624,10836,5293,2383,981,365,119,33,7,1
%N A108438 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 abscissa of the first peak equal to k.
%C A108438 Row n contains 2n terms. Row sums yield A027307. T(n,1)=A032349(n-1).
%H A108438 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 A108438 G.f.: G = G(t,z) = 1/(1-t^2zA-tzA^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 A108438 T(2,3) = 2 because we have Uuddd and uUddd.
%e A108438 Triangle begins:
%e A108438 1,1;
%e A108438 4,3,2,1;
%e A108438 24,18,13,7,3,1;
%e A108438 172,130,96,55,28,12,4,1;
%p A108438 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^2*z*A-t*z*A^2)-1: Gserz:=simplify(series(G,z=0,10)): for n from 1 to 8 do P[n]:=sort(coeff(Gserz,z^n)) od: > for n from 1 to 8 do seq(coeff(P[n],t^k),k=1..2*n) od; # yields sequence in triangular form
%Y A108438 Cf. A027307, A032349.
%K A108438 nonn,tabf
%O A108438 1,3
%A A108438 _Emeric Deutsch_, Jun 04 2005