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

%I #10 Jan 25 2021 14:26:37

%S 1,5,5,15,35,15,35,140,140,35,70,420,720,420,70,126,1050,2700,2700,

%T 1050,126,210,2310,8250,12375,8250,2310,210,330,4620,21780,45375,

%U 45375,21780,4620,330,495,8580,51480,141570,196625,141570,51480,8580,495,715,15015

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

%C 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.

%C 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).

%H F. Cai, Q.-H. Hou, Y. Sun, A. L. B. Yang, <a href="http://arxiv.org/abs/1808.05736">Combinatorial identities related to 2x2 submatrices of recursive matrices</a>, arXiv:1808.05736 Table 2.1 for k=4.

%H R. K. Guy, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL3/GUY/catwalks.html">Catwalks, sandsteps and Pascal pyramids</a>, J. Integer Sequences, Vol. 3 (2000), Article #00.1.6

%F 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.

%F 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)).

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

%F 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;

%F 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).

%F 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]

%e Triangle starts

%e n\k|...1......2......3......4......5......6

%e ===========================================

%e .4.|...1

%e .5.|...5......5

%e .6.|..15.....35.....15

%e .7.|..35....140....140.....35

%e .8.|..70....420....720....420.....70

%e .9.|.126...1050...2700...2700...1050....126

%e ...

%e T(5,2) = 5: the 5 walks of length 5 from (0,0) to (1,4) are

%e UUUUR, UUURU, UURUU, URUUU and RUUUU.

%p with(combinat):

%p T:= (n,k) -> 5/(n+1)*binomial(n+1,k+4)*binomial(n+1,k-1):

%p for n from 4 to 13 do

%p seq(T(n,k),k = 1..n-3);

%p end do;

%t 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 *)

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

%K easy,nonn,tabl

%O 4,2

%A _Peter Bala_, Oct 15 2008