A145599 Triangular array of generalized Narayana numbers: T(n,k) = 5/(n+1)*binomial(n+1,k+4)*binomial(n+1,k-1).

1, 5, 5, 15, 35, 15, 35, 140, 140, 35, 70, 420, 720, 420, 70, 126, 1050, 2700, 2700, 1050, 126, 210, 2310, 8250, 12375, 8250, 2310, 210, 330, 4620, 21780, 45375, 45375, 21780, 4620, 330, 495, 8580, 51480, 141570, 196625, 141570, 51480, 8580, 495, 715, 15015

Offset: 4

Author

Peter Bala, Oct 15 2008

Keywords

• easy
• nonn
• tabl

Comments

T(n,k) is the number of walks of n unit steps, each step in the direction either up (U), down (D), right (R) or left (L), starting from (0,0) and finishing at lattice points on the horizontal line y = 4 and which remain in the upper half-plane y >= 0. An example is given in the Example section below.

The current array is the case r = 4 of the generalized Narayana numbers N_r(n,k) := (r + 1)/(n + 1)*binomial(n + 1,k + r)*binomial(n + 1,k - 1), which count walks of n steps from the origin to points on the horizontal line y = r that remain in the upper half-plane. Case r = 0 gives the table of Narayana numbers A001263 (but with an offset of 0 in the row numbering). For other cases see A145596 (r = 1), A145597 (r = 2) and A145598 (r = 3).

Examples

			Triangle starts
n\k|...1......2......3......4......5......6
===========================================
.4.|...1
.5.|...5......5
.6.|..15.....35.....15
.7.|..35....140....140.....35
.8.|..70....420....720....420.....70
.9.|.126...1050...2700...2700...1050....126
...
T(5,2) = 5: the 5 walks of length 5 from (0,0) to (1,4) are
UUUUR, UUURU, UURUU, URUUU and RUUUU.
		

Links

• F. Cai, Q.-H. Hou, Y. Sun, A. L. B. Yang, Combinatorial identities related to 2x2 submatrices of recursive matrices, arXiv:1808.05736 Table 2.1 for k=4.
• R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6

Crossrefs

A003519 (row sums), A001263, A145596, A145597, A145598, A145603.

Programs

Maple

with(combinat):
T:= (n,k) -> 5/(n+1)*binomial(n+1,k+4)*binomial(n+1,k-1):
for n from 4 to 13 do
seq(T(n,k),k = 1..n-3);
end do;

Mathematica

Table[5/(n+1) Binomial[n+1,k+4]Binomial[n+1,k-1],{n,4,20},{k,0,n}]/.(0-> Nothing)//Flatten (* Harvey P. Dale, Jan 25 2021 *)

Formula

T(n,k) = 5/(n+1)*binomial(n+1,k+4)*binomial(n+1,k-1) for n >=4 and 1 <= k <= n-3. In the notation of [Guy], T(n,k) equals w_n(x,y) at (x,y) = (2*k - n + 2,4). Row sums A003519.

O.g.f. for column k+2: 5/(k + 1) * y^(k+5)/(1 - y)^(k+7) * Jacobi_P(k,5,1,(1 + y)/(1 - y)).

Identities for row polynomials R_n(x) := sum {k = 1..n-3} T(n,k)*x^k:

x^4*R_(n-1)(x) = 5*(n - 1)*(n - 2)*(n - 3)*(n - 4)/((n + 1)*(n + 2)*(n + 3)*(n + 4)*(n + 5)) * sum {k = 0..n} binomial(n + 5,k) * binomial(2n - k,n) * (x - 1)^k;

sum {k = 1..n} (-1)^k*binomial(n,k)*R_k(x^2)*(1 + x)^(2*(n-k)) = R_n(1)*x^(n-2) = A003519(n)*x^(n-2).

Row generating polynomial R_(n+4)(x) = 5/(n+5)*x*(1-x)^n * Jacobi_P(n,5,5,(1+x)/(1-x)). [From Peter Bala, Oct 31 2008]