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%I #5 Oct 06 2015 17:00:54
%S 1,0,2,4,2,4,32,18,8,8,252,146,60,24,16,2112,1186,496,176,64,32,18484,
%T 10146,4148,1488,480,160,64,166976,90162,36216,12792,4160,1248,384,
%U 128,1545548,824114,326828,113960,36720,11104,3136,896,256,14583808,7699394
%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 pyramids (a pyramid is a sequence u^pd^p or U^pd^(2p) for some positive integer p, starting at the x-axis).
%C Row sums yield A027307. Column 0 yields A108449. Number of pyramids in all paths from (0,0) to (3n,0) is given by A108450
%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. =(1-z)/[1+z-2tz-z(1-z)A(1+A)], 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(2,1)=2 because we have uudd and UUdddd.
%e Triangle begins:
%e 1;
%e 0,2;
%e 4,2,4;
%e 32,18,8,8;
%e 252,146,60,24,16;
%p A:=(2/3)*sqrt((z+3)/z)*sin((1/3)*arcsin(sqrt(z)*(z+18)/(z+3)^(3/2)))-1/3: G:=(1-z)/(1+z-2*t*z-z*(1-z)*A*(1+A)): Gser:=simplify(series(G,z=0,12)): P[0]:=1: for n from 1 to 9 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 9 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields sequence in triangular form
%Y Cf. A027307, A108449, A108450.
%K nonn,tabl
%O 0,3
%A _Emeric Deutsch_, Jun 11 2005