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%I #45 Oct 06 2021 12:56:37
%S 1,1,1,1,5,1,1,14,14,1,1,30,84,30,1,1,55,330,330,55,1,1,91,1001,2145,
%T 1001,91,1,1,140,2548,10010,10010,2548,140,1,1,204,5712,37128,68068,
%U 37128,5712,204,1,1,285,11628,116280,352716,352716,116280,11628,285,1
%N Square array read by antidiagonals: S(p,q) = (p+q+1)!(2p+2q+1)!/((p+1)!(2p+1)!(q+1)!(2q+1)!) (p,q>=0).
%H Michael De Vlieger, <a href="/A111910/b111910.txt">Table of n, a(n) for n = 0..11475</a> (rows 0 <= n <= 150, flattened).
%H G. Kreweras and H. Niederhausen, <a href="https://doi.org/10.1016/S0195-6698(81)80020-0">Solution of an enumerative problem connected with lattice paths</a>, European J. Combin. 2 (1981), 55-60.
%H Anthony J. Wood, Richard A. Blythe, and Martin R. Evans, <a href="https://arxiv.org/abs/1908.00942">Combinatorial mappings of exclusion processes</a>, arXiv:1908.00942 [cond-mat.stat-mech], 2019.
%F S(n,n) = A111911(n).
%F From _Peter Bala_, Oct 13 2011: (Start)
%F Define a(n) = n!*(n+1/2)!*(n+1)!/(1/2)!.
%F S(n,k) = a(n+k)/(a(n)*a(k)) gives the sequence as a square array while T(n,k) = a(n)/(a(n-k)*a(k)) gives the sequence as a triangle.
%F S(n-1,k)*S(n,k+1)*S(n+1,k-1) = S(n-1,k+1)*S(n,k-1)*S(n+1,k). Cf. A091044.
%F (End)
%F From _G. C. Greubel_, Feb 12 2021: (Start)
%F As a number triangle:
%F T(n, k) = binomial(n+1, k)*binomial(2*n+1, 2*k)/((k+1)*(2*k+1)).
%F T(n, k) = binomial(2*n+1, 2*k)/((2*k+1)*binomial(n, k)) * A001263(n+1, k+1). (End)
%F From _Peter Bala_, Sep 19 2021: (Start)
%F As a triangle: T(n,k) = a(n)/(a(n-k)*a(k)), where a(n) = Product_{j = 1..n} s(p,j) with s(p,j) = Sum_{j = 1..n} j^p and p = 2. Note, p = 0 gives Pascal's triangle A007318, p = 1 gives the triangle of Narayana numbers A001263 and p = 3 gives the triangle A174158 whose entries are the squares of the Narayana numbers.
%F Let E(y) = Sum_{n >= 0} y^n/((n+1)!*(2*n+1)!). Then as a triangle this is the generalized Riordan array (E(y), y) as defined in Wang and Wang with respect to the sequence c_n = (n+1)!*(2*n+1)!. Cf. A001263.
%F Generating function: E(y)*E(x*y) = 1 + (1 + x)*y/(2!*3!) + (1 + 5*x + x^2)*y^2/(3!*5!) + (1 + 14*x + 14*x^2 + x^3)*y^3/(4!*7!) + ....
%F The n-th power of this array has a generating function E(y)^n*E(x*y). In particular, the matrix inverse has a generating function E(x*y)/E(y).
%F exp(y)*E(y) = 1 + 13*y/(2!*3!) + 421*y^2/(3!*5!) + 25368*y^3/(4!*7!) + ... is essentially a generating function for A081442. (End)
%e Array S(n,k) in rectangular form (n, k >= 0):
%e 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
%e 1, 5, 14, 30, 55, 91, 140, 204, 285, ...
%e 1, 14, 84, 330, 1001, 2548, 5712, 11628, 21945, ...
%e 1, 30, 330, 2145, 10010, 37128, 116280, 319770, 793155, ...
%e 1, 55, 1001, 10010, 68068, 352716, 1492260, 5393454, 17185025, ...
%e ...
%e Array T(n,k) in triangular form (n >= 0 and 0 <= k <= n):
%e 1,
%e 1, 1,
%e 1, 5, 1,
%e 1, 14, 14, 1,
%e 1, 30, 84, 30, 1,
%e 1, 55, 330, 330, 55, 1,
%e 1, 91, 1001, 2145, 1001, 91, 1,
%e ...
%p a:=(p,q)->(p+q+1)!*(2*p+2*q+1)!/(p+1)!/(2*p+1)!/(q+1)!/(2*q+1)!: for n from 0 to 10 do seq(a(j,n-j),j=0..n) od; # yields sequence in triangular form
%t Table[(# + q + 1)! (2 # + 2 q + 1)!/((# + 1)! (2 # + 1)! (q + 1)! (2 q + 1)!) &[r - q], {r, 0, 9}, {q, 0, r}] // Flatten (* _Michael De Vlieger_, Oct 21 2019 *)
%t Table[Binomial[n+1, k]*Binomial[2*n+1, 2*k]/((k+1)*(2*k+1)), {n, 0, 12}, {k, 0,
%t n}]//Flatten (* _G. C. Greubel_, Feb 12 2021 *)
%o (Sage)
%o def A111910(n,k): return binomial(n+1, k)*binomial(2*n+1, 2*k)/((k+1)*(2*k+1))
%o flatten([[A111910(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Feb 12 2021
%o (Magma)
%o T:= func< n,k | Binomial(n+1, k)*Binomial(2*n+1, 2*k)/((k+1)*(2*k+1)) >;
%o [T(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Feb 12 2021
%Y Cf. A091044, A111911 (main diagonal), A196148 (row sums of triangle).
%Y Cf. A001263, A174158.
%K nonn,tabl
%O 0,5
%A _Emeric Deutsch_, Aug 19 2005
%E Example section edited by _Petros Hadjicostas_, Sep 03 2019
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