A134423 Triangle read by rows: T(n,k) is the number of paths in the first quadrant from (0,0) to (n,0) using steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0), such that the area between the x-axis and the path is k (n>=0, 0<=k<=floor(n^2/4)).

1, 1, 2, 1, 3, 2, 1, 5, 5, 3, 2, 1, 8, 10, 8, 6, 5, 2, 1, 13, 20, 19, 17, 16, 11, 7, 3, 2, 1, 21, 38, 42, 42, 43, 36, 29, 18, 12, 8, 5, 2, 1, 34, 71, 89, 98, 108, 102, 92, 72, 55, 40, 29, 20, 13, 7, 3, 2, 1, 55, 130, 182, 218, 255, 264, 258, 228, 195, 158, 125, 96, 74, 52, 35, 22, 14, 8

Offset: 0

Author

Emeric Deutsch, Oct 25 2007

Keywords

• nonn
• tabf

Comments

Row n has 1+floor(n^2/4) terms. Row sums yield A128720. T(n,0)=fibonacci(n+1) (A000045). T(n,1)=A001629(n). Sum(k*T(n,k),k>=0)=A134424(n).

Examples

			T(4,2)=3 because we have hUhD, UhDh and UDUD.
Triangle starts:
1;
1;
2,1;
3,2,1;
5,5,3,2,1;
8,10,8,6,5,2,1;
		

Crossrefs

Cf. A128720, A000045, A001629, A134424.

Programs

Maple

P[0]:=1: P[1]:=1: for n from 2 to 9 do P[n]:=sort(expand(P[n-1]+P[n-2]+sum(P[j]*P[n-2-j]*t^(j+1),j=0..n-2))) end do: for n from 0 to 9 do seq(coeff(P[n], t, j),j=0..floor((1/4)*n^2)) end do; # yields sequence in triangular form

Formula

G.f.=G(t,z) satisfies G(t,z)=1/[1-z-z^2-(tz^2)G(t,tz)]. Rec. rel. for the row generating polynomials P[n]=P[n](t): P[n]=P[n-1]+P[n-2]+Sum(t^(j+1)P[j]P[n-2-j], j=0..n-2) for n>=2; P[0]=P[1]=1.