A121461 Triangle read by rows: T(n,k) is the number of nondecreasing Dyck paths of semilength n, having last ascent of length k (1 <= k <= n).

1, 1, 1, 3, 1, 1, 8, 3, 1, 1, 21, 8, 3, 1, 1, 55, 21, 8, 3, 1, 1, 144, 55, 21, 8, 3, 1, 1, 377, 144, 55, 21, 8, 3, 1, 1, 987, 377, 144, 55, 21, 8, 3, 1, 1, 2584, 987, 377, 144, 55, 21, 8, 3, 1, 1, 6765, 2584, 987, 377, 144, 55, 21, 8, 3, 1, 1, 17711, 6765, 2584, 987, 377, 144, 55, 21

Offset: 1

Author

Emeric Deutsch, Jul 31 2006

Keywords

• nonn
• tabl

Comments

Also the number of directed column-convex polyominoes of area n, having k cells in the last column. Row sums are the odd-subscripted Fibonacci numbers (A001519). Sum_{k=1..n} k*T(n,k) = Fibonacci(2n) = A001906(n).

Riordan array ((1-2*x+x^2)/(1-3*x+x^2), x). - Philippe Deléham, Oct 04 2014

Antidiagonal sums are in A007598. - Philippe Deléham, May 22 2015

Examples

			T(4,2)=3 because we have UUDD(UU)DD, UUD(UU)DDD and UDUD(UU)DD, where U=(1,1) and D=(1,-1) (the last ascents are shown between parentheses).
Triangle starts:
   1;
   1,  1;
   3,  1, 1;
   8,  3, 1, 1;
  21,  8, 3, 1, 1;
  55, 21, 8, 3, 1, 1;
  ...
		

Links

• E. Barcucci, A. Del Lungo, S. Fezzi and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
• E. Barcucci, R. Pinzani and R. Sprugnoli, Directed column-convex polyominoes by recurrence relations, Lecture Notes in Computer Science, No. 668, Springer, Berlin (1993), pp. 282-298.
• A. M. Baxter, L. K. Pudwell, Ascent sequences avoiding pairs of patterns, 2014.
• E. Deutsch and H. Prodinger, A bijection between directed column-convex polyominoes and ordered trees of height at most three, Theoretical Comp. Science, 307, 2003, 319-325.

Crossrefs

Cf. A001519, A001906, A104762, A088305.

Cf. A000045, A007598.

Programs

Maple

with(combinat): T:=proc(n,k) if k
				

Formula

T(n,k) = Fibonacci(2(n-k)) if k < n; T(n,n)=1.

G.f.: G = G(t,z) = t*z*(1-z)^2/((1-3z+z^2)*(1-tz)).

From Gary W. Adamson, Jul 07 2011: (Start)

Let M be the production matrix:

1, 1, 0, 0, 0, 0, ...

2, 0, 1, 0, 0, 0, ...

3, 0, 0, 1, 0, 0, ...

4, 0, 0, 0, 1, 0, ...

5, 0, 0, 0, 0, 1, ...

...

n-th row of triangle A121461 = top row terms of (n-1)-th power of M. (End)

Let P denote Pascal's triangle. Then P^(-1)*A121461*P = A104762. - Peter Bala, Apr 11 2013