cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

Peter Bala, Jul 28 2023

Keywords

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 [added Oct 11 2024: follows from Meštrović, Section 6, equation 39, since T(n, k) = binomial(2*k, k) * binomial(2*n*k, n*k) * binomial((n+2)*k, k)/binomial((n+1)*k, k)].

Examples

			 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 ...
		

Crossrefs

A002894 (row 0), A006480 (row 1), A364510 (row 3), A364511 (row 4).

Programs

  • 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);
  • 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

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