A098978 Triangle read by rows: T(n,k) is number of Dyck n-paths with k UUDDs, 0 <= k <= n/2.

1, 1, 1, 1, 2, 3, 5, 8, 1, 13, 23, 6, 35, 69, 27, 1, 97, 212, 110, 10, 275, 662, 426, 66, 1, 794, 2091, 1602, 360, 15, 2327, 6661, 5912, 1760, 135, 1, 6905, 21359, 21534, 8022, 945, 21, 20705, 68850, 77685, 34840, 5685, 246, 1, 62642, 222892, 278192, 146092

Offset: 0

Author

David Callan, Oct 24 2004

Keywords

• nonn
• tabf

Comments

T(n,k) is the number of Łukasiewicz paths of length n having k peaks. A Łukasiewicz path of length n is a path in the first quadrant from (0,0) to (n,0) using rise steps (1,k) for any positive integer k, level steps (1,0) and fall steps (1,-1). Example: T(3,1)=3 because we have HUD, UDH and U(2)DD, where H=(1,0), U(1,1), U(2)=(1,2) and D=(1,-1). (see R. P. Stanley reference). - Emeric Deutsch, Jan 06 2005

Examples

			Table begins
\ k  0,   1,   2, ...
n
0 |  1;
1 |  1;
2 |  1,   1;
3 |  2,   3;
4 |  5,   8,   1;
5 | 13,  23,   6;
6 | 35,  69,  27,  1;
7 | 97, 212, 110, 10;
8 |275, 662, 426, 66, 1;
T(3,1) = 3 because each of UUUDDD, UDUUDD, UUDDUD has one UUDD.
		

References

• R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge Univ. Press, Cambridge, 1999, p. 223, Exercise 6.19w; the integers are the slopes of the steps. - Emeric Deutsch, Jan 06 2005

Links

• Alois P. Heinz, Rows n = 0..200, flattened
• Marilena Barnabei, Flavio Bonetti, and Niccolò Castronuovo, Motzkin and Catalan Tunnel Polynomials, J. Int. Seq., Vol. 21 (2018), Article 18.8.8.
• A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924. - From _N. J. A. Sloane_, May 05 2012
• Index entries for sequences related to Łukasiewicz

Crossrefs

Column k=0 is A025242 (apart from first term).

Cf. A243752.

Programs

Maple

b:= proc(x, y, t) option remember; `if`(y<0 or y>x, 0,
     `if`(x=0, 1, expand(b(x-1, y+1, [2, 3, 3, 2][t])
      +b(x-1, y-1, [1, 1, 4, 1][t])*`if`(t=4, z, 1))))
    end:
T:= n-> (p-> seq(coeff(p, z, i), i=0..degree(p)))(b(2*n, 0, 1)):
seq(T(n), n=0..15);  # Alois P. Heinz, Jun 10 2014

Mathematica

T[n_, k_] := Binomial[n-k, k] Binomial[2n-3k, n-k-1] HypergeometricPFQ[{k -n/2-1/2, k-n/2, k-n/2, k-n/2+1/2}, {k-2n/3, k-2n/3+1/3, k-2n/3+2/3}, 16/27]/(n-k); T[0, 0] = 1; Flatten[Table[T[n, k], {n, 0, 15}, {k, 0, n/2}]] (* Jean-François Alcover, Dec 21 2016, after 2nd formula *)

Formula

G.f.: (1 + z^2 - t*z^2 - (-4*z + (-1 - z^2 + t*z^2)^2)^(1/2))/(2*z) = Sum_{n>=0, 0<=k<=n/2} T(n, k)z^n*t^k and it satisfies G = 1 + G^2*z + G*(-z^2 + t*z^2).

T(n,k) = Sum_{j=0..floor(n/2)-k} (-1)^j * binomial(n-(j+k), j+k) * binomial(2n-3(j+k), n-(j+k)-1) * binomial(j+k, k)/(n-(j+k)). - I. Tasoulas (jtas(AT)unipi.gr), Feb 19 2006