A109157 - OEIS

%I #5 Oct 06 2015 17:40:25

%S 1,0,1,1,4,0,2,2,2,32,8,8,4,5,5,4,252,64,84,24,28,12,14,12,8,2112,520,

%T 680,240,232,88,76,37,37,28,16,18484,4480,5804,1992,2012,776,656,264,

%U 206,106,94,64,32,166976,40008,51592,17440,17400,6776,5680,2392,1768

%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 sum of the heights of its pyramids equal to k (a pyramid is a sequence u^pd^p or U^pd^(2p) for some positive integer p, starting at the x-axis; p is the height of the pyramid).

%C Row n has 2n+1 terms. Row sums yield A027307. Column 0 yields A108449.

%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-tz)(1-t^2*z)/[1-2tz-2t^2*z+z+3t^3*z^2-t^3*z^3-z(1-z)(1-t^2*z)(1-tz)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(2,3)=2 because we have udUdd and Uddud.

%e Triangle begins:

%e 1;

%e 0,1,1;

%e 4,0,2,2,2;

%e 32,8,8,4,5,5,4;

%e 252,64,84,24,28,12,14,12,8;

%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+t*z)*(-1+t^2*z)/(z*(-1+z)*(-1+t^2*z)*(-1+t*z)*A*(1+A)+1-2*t*z-2*t^2*z+z+3*t^3*z^2-t^3*z^3): Gser:=simplify(series(G,z=0,10)): P[0]:=1: for n from 1 to 7 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 7 do seq(coeff(t*P[n],t^k),k=1..2*n+1) od; # yields sequence in triangular form

%Y Cf. A027307, A108449.

%K nonn,tabf

%O 0,5

%A _Emeric Deutsch_, Jun 20 2005