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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.
4
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
0,5
COMMENTS
Fractional factorials are defined in terms of the gamma function; for example, ((2*n+1/2)*k)! = Gamma(1 + (2*n+1/2)*k).
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.
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.
Each row sequence of the present table is an integral factorial ratio sequence of height 2.
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.
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
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
EXAMPLE
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 ...
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);
PROG
(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
CROSSREFS
Cf. A275652 (row 1), A365026 (row 2), A365027 (row 3).
Sequence in context: A172378 A015124 A156767 * A174921 A010180 A109013
KEYWORD
nonn,tabl,easy
AUTHOR
Peter Bala, Aug 17 2023
STATUS
approved