OFFSET
1,8
COMMENTS
FORMULA
G.f.: G-1, where G = G(t,z) satisfies z(2+2z+z^2-tz-tz^2)G^2+(1+2z)(1+z-tz)G+1+z-tz=0.
EXAMPLE
T(4,3) = 1 because we have U(UD)(UD)(UD)D, where U=(1,1), D=(1,-1) (the peaks at even levels are shown between parentheses).
Triangle begins:
0;
0, 1;
1, 0, 1;
1, 3, 1, 1;
3, 6, 6, 2, 1;
7, 17, 18, 11, 3, 1;
17, 48, 58, 40, 18, 4, 1;
MAPLE
G:=(1-t*z+2*z^2+3*z-2*t*z^2-sqrt(1-3*z^2-2*z*t+2*z^2*t+z^2*t^2-2*z))/2/z/(2+2*z-t*z-t*z^2+z^2)-1: Gser:=simplify(series(G, z=0, 15)): for n from 1 to 12 do P[n]:=coeff(Gser, z^n) od: for n from 1 to 12 do seq(coeff(t*P[n], t^j), j=1..n) od; # yields sequence in triangular form
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Dec 11 2005
STATUS
approved