A101919 Triangle read by rows: T(n,k) is the number of Schroeder paths of length 2n and having k up steps starting at even heights.

1, 1, 1, 1, 4, 1, 1, 12, 8, 1, 1, 33, 42, 13, 1, 1, 88, 183, 102, 19, 1, 1, 232, 717, 624, 205, 26, 1, 1, 609, 2622, 3275, 1650, 366, 34, 1, 1, 1596, 9134, 15473, 11020, 3716, 602, 43, 1, 1, 4180, 30691, 67684, 64553, 30520, 7483, 932, 53, 1, 1, 10945, 100284, 279106

Offset: 0

Author

Emeric Deutsch, Dec 20 2004

Keywords

• nonn
• tabl

Comments

A Schroeder path of length 2n is a lattice path starting from (0,0), ending at (2n,0), consisting only of steps U=(1,1) (up steps), D=(1,-1) (down steps) and H=(2,0) (level steps) and never going below the x-axis (Schroeder paths are counted by the large Schroeder numbers, A006318). Also number of Schroeder paths of length 2n and having k humps. A hump is an up step U followed by 0 or more level steps H followed by a down step D. The T(3,2)=8 Schroeder paths of length 6 and having 2 humps are: H(UD)(UD), (UD)H(UD), (UD)(UD)H, (UD)(UHD), (UD)(UUDD), (UHD)(UD), (UUDD)(UD) and U(UD)(UD)D, the humps being shown between parentheses. Row sums are the large Schroeder numbers (A006318). Column 1 yields the odd-indexed Fibonacci numbers minus 1 (A027941). T(n,n-1)=A034856(n)=binomial(n + 1, 2) + n - 1.

Product A085478*A090181 (Morgan-Voyce times Narayana). [From Paul Barry, Jan 29 2009]

Examples

			T(3,2)=8 because we have HU'DU'D, U'DHU'D, U'DU'DH, U'DU'HD, U'DU'UDD, U'HDU'D, U'UDDU'D and U'UU'DDD, the up steps starting at an even height being shown with a prime sign.
Triangle begins:
1;
1,1;
1,4,1;
1,12,8,1;
1,33,42,13,1;
		

Crossrefs

Cf. A006318, A027941, A034856, A101920.

Programs

Maple

G:=1/2/(-z+z^2)*(-1+z+t*z-z^2+sqrt(1-6*z-2*t*z+11*z^2+2*t*z^2-6*z^3+t^2*z^2-2*t*z^3+z^4)): Gser:=simplify(series(G,z=0,13)): P[0]:=1: for n from 1 to 11 do P[n]:=coeff(Gser,z^n) od: for n from 0 to 11 do seq(coeff(t*P[n],t^k),k=1..n+1) od; # yields the sequence in triangular form

Formula

G.f.=G=G(t, z) satisfies z(1-z)G^2-(1-z-tz+z^2)G+1-z=0.

G.f.: 1/(1-x-xy/(1-x-x/(1-x-xy/(1-x-xy/(1-x-x/(1-x-xy/(1-.... (continued fraction). [From Paul Barry, Jan 29 2009]