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%I #30 Aug 25 2023 17:21:08
%S 1,1,1,1,10,1,1,126,300,1,1,1716,79380,11440,1,1,24310,20612592,
%T 65523780,485100,1,1,352716,5318784900,328206021000,60634147860,
%U 21841260,1,1,5200300,1368494343216,1552041334596844,5876083665270000,59774707082376,1022041020,1
%N Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ) for n, k >= 0.
%C Fractional factorials are defined in terms of the gamma function; for example, ((2*n+1/2)*k)! = Gamma(1 + (2*n+1/2)*k).
%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 It is usually assumed that the c's and d's are integers but here we allow for some of the c's and d's to be half-integers. See A276098 for further examples of this type.
%C Each row sequence of the present table is an integral factorial ratio sequence of height 2.
%C Conjecture: each row sequence of the table satisfies 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.
%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_{j = 0..n*k} binomial((2*n+1)*k, n*k-j)^2 * binomial(k+j-1, j).
%F T(n,k) = binomial((2*n+1)*k,n*k)^2 * hypergeom([k, -n*k, -n*k], [1 + (n+1)*k, 1 + (n+1)*k], 1) = (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ) by Dixon's 3F2 summation theorem.
%F T(n,k) = [x^(n*k)] ( (1 - x)^(2*n*k) * Legendre_P((2*n+1)*k, (1 + x)/(1 - x)) ).
%F T(n,k) = k!!*((2*n+1)*k)!*((4*n+1)*k)!!/(k!*((n*k)!*((2*n+1)*k)!!)^2). - _Chai Wah Wu_, Aug 24 2023
%e Square array begins:
%e n\k| 0 1 2 3 4
%e - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
%e 0 | 1 1 1 1 1 ...
%e 1 | 1 10 300 11440 485100 ...
%e 2 | 1 126 79380 65523780 60634147860 ...
%e 3 | 1 1716 20612592 328206021000 5876083665270000 ...
%e 4 | 1 24310 5318784900 1552041334596844 510031828417402714500 ...
%e 5 | 1 352716 1368494343216 7108360304262169344 ...
%p # display as a square array
%p T(n, k) := (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ):
%p seq( print(seq(simplify(T(n, k)), k = 0..10)), n = 0..10);
%p # display as a sequence
%p seq( seq(simplify(T(n-k, k)), k = 0..n), n = 0..10);
%o (Python)
%o from itertools import count, islice
%o from math import factorial
%o from sympy import factorial2
%o def A365025_T(n,k): return int(factorial2(k)*factorial(r:=((m:=n<<1)+1)*k)*factorial2(((m<<1)+1)*k)//((factorial(n*k)*factorial2(r))**2*factorial(k)))
%o def A365025_gen(): # generator of terms
%o for n in count(0):
%o yield from (A365025_T(n-k,k) for k in range(n+1))
%o A365025_list = list(islice(A365025_gen(),20)) # _Chai Wah Wu_, Aug 24 2023
%Y Cf. A275652 (row 1), A365026 (row 2), A365027 (row 3).
%Y Cf. A276098, A364506, A364509, A364513, A364518.
%K nonn,tabl,easy
%O 0,5
%A _Peter Bala_, Aug 17 2023