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.
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
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 ...
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.
- Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).
- K. Soundararajan, Integral factorial ratios: irreducible examples with height larger than 1, Phil. Trans. Royal Soc., A378: 2018044, 2019.
- Wikipedia, Dixon's identity
Programs
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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);
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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
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