0,5

T(n,k) is the number of Lukasiewicz paths of length n having k peaks. A Lukasiewicz 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) (see 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). 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). 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 (deutsch(AT)duke.poly.edu), Jan 06 2005

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((-1)^j * binomial(n-(j+k),j+k) * binomial(2n - 3(j+k), n-(j+k)-1) * binomial(j+k,k)/(n-(j+k)), j=0..[n/2]-k). - I. Tasoulas (jtas(AT)unipi.gr), Feb 19 2006

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.

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

Sequence in context: A093092 A031111 A089911 this_sequence A111301 A096320 A105955

Adjacent sequences: A098975 A098976 A098977 this_sequence A098979 A098980 A098981

nonn

David Callan (callan(AT)stat.wisc.edu), Oct 24 2004