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A364509
Square array read by ascending antidiagonals: T(n,k) = (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ) for n, k > = 0.
8
1, 1, 4, 1, 6, 36, 1, 16, 90, 400, 1, 50, 784, 1680, 4900, 1, 168, 8910, 48400, 34650, 63504, 1, 588, 113256, 2011100, 3312400, 756756, 853776, 1, 2112, 1528436, 96993024, 503909070, 240374016, 17153136, 11778624, 1, 7722, 21395520, 5056527000, 92279796840, 133954543800, 18116083216
OFFSET
0,3
COMMENTS
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.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
It is known that both row 0, the squares of 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.
LINKS
Winston de Greef, Table of n, a(n) for n = 0..3240 (80 antidiagonals)
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc. (2) 79 2009, 422-444.
K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. Royal Soc., A378: 2018044, 2019.
Wikipedia, Dixon's identity
FORMULA
T(n,k) = Sum_{i = -k..k} (-1)^i * binomial(2*k, k+i)^2 * binomial(2*n*k, n*k+i) (shows that the table entries are integers).
For n >= 1, T(n,k) = (-1)^k * binomial(2*n*k, (n+1)*k)^2 * hypergeom([-2*k, -2*k, -(n+1)*k], [1, 1 + (n-1)*k], 1) = (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ) by Dixon's 3F2 summation theorem.
T(n,k) = (-1)^(n*k) * [x^((n+1)*k)] ( (1 - x)^(2*(n+1)*k) * Legendre_P(2*k, (1 + x)/(1 - x)) ). - Peter Bala, Aug 14 2023
EXAMPLE
Square array begins:
n\k| 0 1 2 3 4 5
- + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
0 | 1 4 36 400 4900 63504 ... A002894
1 | 1 6 90 1680 34650 756756 ... A006480
2 | 1 16 784 48400 3312400 240374016 ... A364510
3 | 1 50 8910 2011100 503909070 133954543800 ... A364511
4 | 1 168 113256 96993024 92279796840 93172920645168 ...
5 | 1 588 1528436 5056527000 18592935952500 72567511917065088 ...
MAPLE
# display as a square array
T(n, k) := (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ):
seq( print(seq(T(n, k), k = 0..10)), n = 0..10):
# display as a sequence
seq( seq(T(n-k, k), k = 0..n), n = 0..10);
PROG
(PARI) T(n, k) = (2*k)!/k!^2 * ( (2*n*k)! * ((n + 2)*k)! )/( (n*k)! * ((n + 1)*k)!^2 ) \\ Winston de Greef, Oct 05 2023
CROSSREFS
A002894 (row 0), A006480 (row 1), A364510 (row 3), A364511 (row 4).
Sequence in context: A370356 A126150 A374370 * A349545 A291056 A248831
KEYWORD
nonn,tabl,easy
AUTHOR
Peter Bala, Jul 28 2023
STATUS
approved