login
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

%I #3 Mar 30 2012 17:36:15

%S 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,

%T 38,42,42,43,36,29,18,12,8,5,2,1,34,71,89,98,108,102,92,72,55,40,29,

%U 20,13,7,3,2,1,55,130,182,218,255,264,258,228,195,158,125,96,74,52,35,22,14,8

%N 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)).

%C 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).

%F 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.

%e T(4,2)=3 because we have hUhD, UhDh and UDUD.

%e Triangle starts:

%e 1;

%e 1;

%e 2,1;

%e 3,2,1;

%e 5,5,3,2,1;

%e 8,10,8,6,5,2,1;

%p 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

%Y Cf. A128720, A000045, A001629, A134424.

%K nonn,tabf

%O 0,3

%A _Emeric Deutsch_, Oct 25 2007