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Square array read by ascending antidiagonals: T(n,k) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ).
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%I #21 Aug 24 2023 07:48:50

%S 1,1,2,1,6,6,1,40,90,20,1,350,5880,1680,70,1,3528,594594,1101100,

%T 34650,252,1,38808,75088728,1299170600,229265400,756756,924,1,453024,

%U 10861066216,2066315135040,3164045050530,50678855040,17153136,3432,1,5521230,1721929279200,3943172216808000

%N Square array read by ascending antidiagonals: T(n,k) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ).

%C Given two sequences of integers c = (c_1, c_2, ..., c_K) and d = (d_1, d_2, ..., d_L) where c_1 + ... + c_K = d_1 + ... + d_L we can define the factorial ratio sequence u_k(c, d) = (c_1*k)!*(c_2*k)!* ... *(c_K*k)!/ ( (d_1*k)!*(d_2*k)!* ... *(d_L*k)! ) and ask whether it is integral for all k >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1. Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.

%C Each row sequence of the present table is an integral factorial ratio sequence of height 2.

%C It is known that both row 0, the central binomial numbers, and row 1, the de Bruijn numbers, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r. We conjecture that all the row sequences of the table satisfy the same supercongruences.

%H J. W. Bober, <a href="https://arxiv.org/abs/0709.1977">Factorial ratios, hypergeometric series, and a family of step functions</a>, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.

%H K. Soundararajan, <a href="https://doi.org/10.1098/rsta.2018.0444">Integral factorial ratios: irreducible examples with height larger than 1</a>, Phil. Trans. Royal Soc., A378: 2018044, 2019.

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Dixon%27s_identity">Dixon's identity</a>

%F T(n,k) = Sum_{i = -k..k} (-1)^i * binomial(2*k, k+i) * binomial(2*n*k, n*k+i)^2 (shows that the table entries are integers).

%F For n >= 1, T(n,k) = (-1)^k * binomial(2*n*k, (n+1)*k)^2 * hypergeom([-2*k, -(n+1)*k, -(n+1)*k], [1 + (n-1)*k, 1 + (n-1)*k], 1) = (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/( (n*k)!^2 * ((n+1)*k)!^2 ) by Dixon's 3F2 summation theorem.

%F T(n,k) = (-1)^k * [x^((n + 1)*k)] ( (1 - x)^(2*(n+1)*k) * Legendre_P(2*n*k, (1 + x)/(1 - x)) ). - _Peter Bala_, Aug 15 2023

%e Square array begins:

%e n\k| 0 1 2 3 4 5

%e - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

%e 0 | 1 2 6 20 70 252 ...

%e 1 | 1 6 90 1680 34650 756756 ...

%e 2 | 1 40 5880 1101100 229265400 50678855040 ...

%e 3 | 1 350 594594 1299170600 3164045050530 8188909171581600 ...

%e 4 | 1 3528 75088728 2066315135040 63464046079757400 ...

%e 5 | 1 38808 ...

%p # display as a square array

%p T(n,k) := (2*k)!/k! * ( (2*n*k)! * ((2*n+1)*k)! )/((n*k)!^2 * ((n+1)*k)!^2):

%p seq( print(seq(T(n,k), k = 0..10)), n = 0..10);

%p # display as a sequence

%p seq( seq(T(n-k,k), k = 0..n), n = 0..10);

%Y A000984 (row 0), A006480 (row 1), A364507 (row 2), A364508 (row 3). Cf. A364303, A364509, A365025.

%K nonn,tabl,easy

%O 0,3

%A _Peter Bala_, Jul 27 2023