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Triangular array of generalized Narayana numbers: T(n, k) = 4*binomial(n+1, k+3)*binomial(n+1, k-1)/(n+1).
5

%I #15 Oct 09 2020 09:26:34

%S 1,4,4,10,24,10,20,84,84,20,35,224,392,224,35,56,504,1344,1344,504,56,

%T 84,1008,3780,5760,3780,1008,84,120,1848,9240,19800,19800,9240,1848,

%U 120,165,3168,20328,58080,81675,58080,20328,3168,165,220,5148,41184,151008

%N Triangular array of generalized Narayana numbers: T(n, k) = 4*binomial(n+1, k+3)*binomial(n+1, k-1)/(n+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 = 3 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 = 3 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 A145599 (r = 4).

%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 [math.CO], 2018; Table 2.1 for k=3.

%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) = 4/(n+1)*binomial(n+1,k+3)*binomial(n+1,k-1) for n >=3 and 1 <= k <= n-2. In the notation of [Guy], T(n,k) equals w_n(x,y) at (x,y) = (2*k - n + 1,3). Row sums A003518.

%F O.g.f. for column k+2: 4/(k + 1) * y^(k+4)/(1 - y)^(k+6) * Jacobi_P(k,4,1,(1 + y)/(1 - y)).

%F Identities for row polynomials R_n(x) = Sum_{k = 1 .. n - 2} T(n,k)*x^k:

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

%F Sum_{k = 1..n} (-1)^(k+1)*binomial(n,k)*R_k(x^2)*(1 + x)^(2*(n-k)) = R_n(1)*x^(n-1) = A003518(n)*x^(n-1).

%F Row generating polynomial R_(n+3)(x) = 4/(n+4)*x*(1-x)^n * Jacobi_P(n,4,4,(1+x)/(1-x)). - _Peter Bala_, Oct 31 2008

%F G.f.: A(x)=x*A145596(x)^2. - _Vladimir Kruchinin_, Oct 09 20

%e Triangle starts

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

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

%e .3.|..1

%e .4.|..4.....4

%e .5.|.10....24...10

%e .6.|.20....84...84....20

%e .7.|.35...224..392...224....35

%e .8.|.56...504.1344..1344...504....56

%e ...

%e Row 5:

%e T(5,3) = 10: the 10 walks of length 5 from (0,0) to (2,3) are

%e UUURR, UURUR, UURRU, URUUR, URURU, URRUU, RUUUR, RUURU, RURUU

%e and RRUUU.

%e *

%e *......*......*......y......*......*......*

%e .

%e .

%e *.....10......*.....24......*.....10......*

%e .

%e .

%e *......*......*......*......*......*......*

%e .

%e .

%e *......*......*......*......*......*......*

%e .

%e .

%e *......*......*......o......*......*......* x axis

%e .

%p T := (n,k) -> 4/(n+1)*binomial(n+1,k+3)*binomial(n+1,k-1):

%p for n from 3 to 12 do seq(T(n, k), k = 1 .. n-2) end do;

%Y A003518 (row sums), A001263, A145602, A145596, A145597, A145599.

%K easy,nonn,tabl

%O 3,2

%A _Peter Bala_, Oct 15 2008