A108446 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 have k peaks of the form ud.

1, 1, 1, 4, 5, 1, 20, 32, 13, 1, 113, 223, 135, 26, 1, 688, 1620, 1300, 412, 45, 1, 4404, 12064, 12050, 5350, 1030, 71, 1, 29219, 91335, 109134, 62450, 17575, 2247, 105, 1, 199140, 699689, 973077, 682234, 254625, 49210, 4438, 148, 1, 1385904, 5407744

Offset: 0

Author

Emeric Deutsch, Jun 10 2005

Keywords

• nonn
• tabl

Comments

Row sums yield A027307. Column 0 yields A108447. T(n,n-1) = A008778(n-1) = n(n^2+6n-1)/6. Number of ud peaks in all paths from (0,0) to (3n,0) is given by A108448.

Examples

			T(2,1) = 5 because we have udUdd, uudd, Uddud, Ududd and Uuddd.
Triangle begins:
1;
1,1;
4,5,1;
20,32,13,1;
113,223,135,26,1;
		

Links

• Emeric Deutsch, Problem 10658: Another Type of Lattice Path, American Math. Monthly, 107, 2000, 368-370.

Crossrefs

Cf. A027307, A008778, A108447, A108448, A108425, A108426.

Programs

Maple

T:=proc(n,k) if n=0 and k=0 then 1 elif n=0 then 0 elif k=n then 1 elif k=n then 1 else (1/n)*binomial(n,k)*sum(binomial(n-k,j)*binomial(n+2*j,k+j-1),j=0..n-k) fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Mathematica

T[0, 0] = 1; T[n_, k_] := (1/n) Binomial[n, k]*Sum[Binomial[n-k, j]* Binomial[n+2j, k+j-1], {j, 0, n-k}];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 19 2018 *)

Formula

T(n,k) = (1/n) binomial(n, k)*sum(binomial(n-k,j)*binomial(n+2j,k+j-1), j=0..n-k).

G.f.: G = G(t,z) satisfies G = 1+z(G-1+t)G+zG^3.