A365025 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.
1, 1, 1, 1, 10, 1, 1, 126, 300, 1, 1, 1716, 79380, 11440, 1, 1, 24310, 20612592, 65523780, 485100, 1, 1, 352716, 5318784900, 328206021000, 60634147860, 21841260, 1, 1, 5200300, 1368494343216, 1552041334596844, 5876083665270000, 59774707082376, 1022041020, 1
Offset: 0
Examples
Square array begins: n\k| 0 1 2 3 4 - + - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 0 | 1 1 1 1 1 ... 1 | 1 10 300 11440 485100 ... 2 | 1 126 79380 65523780 60634147860 ... 3 | 1 1716 20612592 328206021000 5876083665270000 ... 4 | 1 24310 5318784900 1552041334596844 510031828417402714500 ... 5 | 1 352716 1368494343216 7108360304262169344 ...
Links
- J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (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
Crossrefs
Programs
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Maple
# display as a square array T(n, k) := (k/2)!/k! * ((2*n+1)*k)! * ((2*n+1/2)*k)! / ( (n*k)!^2 * ((n+1/2)*k)!^2 ): seq( print(seq(simplify(T(n, k)), k = 0..10)), n = 0..10); # display as a sequence seq( seq(simplify(T(n-k, k)), k = 0..n), n = 0..10);
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Python
from itertools import count, islice from math import factorial from sympy import factorial2 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))) def A365025_gen(): # generator of terms for n in count(0): yield from (A365025_T(n-k,k) for k in range(n+1)) A365025_list = list(islice(A365025_gen(),20)) # Chai Wah Wu, Aug 24 2023
Formula
T(n,k) = Sum_{j = 0..n*k} binomial((2*n+1)*k, n*k-j)^2 * binomial(k+j-1, j).
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.
T(n,k) = [x^(n*k)] ( (1 - x)^(2*n*k) * Legendre_P((2*n+1)*k, (1 + x)/(1 - x)) ).
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
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