A364513 -id:A364513 - cp's OEIS frontend

A365025 Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2n+1)k)! * ((2n+1/2)k)! / ( (nk)!^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

Peter Bala, Aug 17 2023

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.

			 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 ...

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

Cf. A275652 (row 1), A365026 (row 2), A365027 (row 3).

Cf. A276098, A364506, A364509, A364513, A364518.

# 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);

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

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

A364514 a(0) = 1 and a(n) = [x^n] (1 - x)^(2n) Legendre_P(n-1, (1 + x)/(1 - x)) for n >= 1.

Original entry on oeis.org

1, -2, 0, 16, 0, -252, 0, 4800, 0, -100100, 0, 2201472, 0, -50139936, 0, 1170614016, 0, -27839740500, 0, 671626956000, 0, -16388657193480, 0, 403645030064640, 0, -10018806017062752, 0, 250305475771456000, 0, -6288594802355952000, 0, 158759294846918261760

Offset: 0

Peter Bala, Aug 02 2023

Row 1 of A364513.

Paolo Xausa, Table of n, a(n) for n = 0..1000

Cf. A364513.

a := proc(n) option remember; if n = 0 then 1 elif n = 1 then -2 else -(3*n - 1)*(3*n - 5)*(3*n - 6)/(n*(n - 1)^2) * a(n-2) end if; end:
seq(a(n), n = 0..15);

A364514[n_]:=A364514[n]=Which[n==0,1,n==1,-2,True,-(3n-1)(3n-5)(3n-6)/(n(n-1)^2)A364514[n-2]];Array[A364514,40,0] (* Paolo Xausa, Oct 05 2023 *)

a(n) = Sum_{k = 0..n} (-1)^k * binomial(n-1, n-k)^2 * binomial(n+1, k).
a(2*n) = 0 for n >= 1; a(2*n+1) = (-1)^(n+1) * 2/(2*n + 1) * (3*n + 1)!/n!^3.
a(2*n+1) ~ (-1)^(n+1) * 3^(3*n) * 3*sqrt(3)/(2*Pi*n).
P-recursive: a(0) = 1, a(1) = -2 and for n >= 2, a(n) = -(3*n - 1)*(3*n - 5)*(3*n - 6)/(n*(n - 1)^2) * a(n-2).
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
a(n) = Sum_{k = 0..n} (-1)^k * binomial(n+k-1, k) * binomial(n-1, k) * binomial(2*n-k, n-k). - Peter Bala, Aug 13 2023

A364515 a(n) = (1/2) * (6n)!(2n)!^2/((3n)!(4n)!*n!^3) for n >= 1 with a(0) = 1.

Original entry on oeis.org

1, 10, 594, 44200, 3640210, 317678760, 28782923400, 2677247862432, 253909524661650, 24443479528347880, 2381221777637046344, 234220654553738055840, 23223217841360039079304, 2318164676869886054980000, 232736024465276636260692000, 23482287467017571303962819200

Offset: 0

Peter Bala, Aug 01 2023

Row 4 of A364513.

			Examples of supercongruences:
a(7) - a(1) = 2677247862432 - 10 = 2*(7^3)*1999*1952323 == 0 (mod 7^3).
a(2*7) - a(2) = 232736024465276636260692000 - 594 = 2*(3^3)*(7^3)*1097* 11454314520615738059 == 0 (mod 7^3).

Paolo Xausa, Table of n, a(n) for n = 0..400

Cf. A364513.

seq( (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3), n = 1..15);

A364515[n_]:=If[n==0,1,(1/2)(6n)!(2n)!^2/((3n)!(4n)!n!^3)];Array[A364515,15,0] (* Paolo Xausa, Oct 05 2023 *)

a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(4*n-1, (1 + x)/(1 - x)) for n >= 1.
a(n) = Sum_{k = 0..n} binomial(4*n-1, n-k)^2 * binomial(2*n+k-2, k).
a(n) = (1/2) *  binomial(6*n, 2*n)*binomial(2*n, n)^2 / binomial(3*n, n).
a(n) =  (4*n - 1)!*(3*n - 1/2)!*(n - 1/2)!/((2*n - 1)! * (2*n - 1/2)!^2 * n!^2) for n >= 1 (fractional factorials are defined in terms of the gamma function, for example, (3*n - 1/2)! = gamma(3*n + 1/2)).
a(n) ~ 108^n * sqrt(2)/(4*Pi*n).
P-recursive: a(0) = 1; for n >= 1, a(n) = 12*(6*n-1)*(6*n-5)*(2*n-1)^2 / ((4*n-1)*(4*n-3)*n^2) * a(n-1) with a(1) = 10.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

A364516 a(n) = (2/3) * (8n)!(3n)!^2/((6n)!(4n)!(2n)!*n!^2) for n >= 1, with a(0) = 1.

Original entry on oeis.org

1, 28, 3900, 685216, 133501500, 27583083528, 5919115212192, 1304298034300800, 293086491979934268, 66857471357130883000, 15434267149448839091400, 3597756971630997935635200, 845406187463509505329860000, 200002748013094535687584437696

Offset: 0

Peter Bala, Aug 02 2023

Row 6 of A364513.

			Examples of supercongruences:
a(7) - a(1) = 1304298034300800 - 28 = (2^2)*(7^4)*103553*1311481 == 0 (mod 7^4).
a(11) - a(1) = 3597756971630997935635200 - 28 = (2^2)*(11^3)*22567*7702811* 3887502719 == 0 (mod 11^3).

Paolo Xausa, Table of n, a(n) for n = 0..400

Cf. A364513.

seq( (2/3) * (8*n)!*(3*n)!^2/((6*n)!*(4*n)!*(2*n)!*n!^2), n = 0..15);

A364516[n_]:=If[n==0,1,(2/3)(8n)!(3n)!^2/((6n)!(4n)!(2n)!n!^2)];Array[A364516,15,0] (* Paolo Xausa, Oct 05 2023 *)

a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(6*n-1, (1 + x)/(1 - x)) for n >= 1.
a(n) = Sum_{k = 0..n} binomial(6*n - 1, n - k)^2 * binomial(4*n + k - 2, k).
a(n) = (6*n-1)!*(4*n-1/2)!*(2*n-1/2)!/((4*n-1)! * (3*n-1/2)!^2 * n!^2) for n >= 1 (fractional factorials are defined in terms of the gamma function, for example, (4*n - 1/2)! = gamma(4*n + 1/2)).
a(n) ~ 2^(8*n) * sqrt(6)/(6*Pi*n).
P-recursive: a(0) = 1; for n >= 1, a(n) = (8*n-1)*(8*n-3)*(8*n-5)*(8*n-7)*(3*n-1)*(3*n-2)/((6*n-1)*(6*n-5)*(2*n-1)^2*n^2) * a(n-1) with a(1) = 28.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.

A364517 a(n) = (5/7) * (9n)!(7n/2)!^2/((9n/2)!(7n)!(5n/2)!*n!^2) for n >= 1, with a(0) = 1.

Original entry on oeis.org

1, 40, 7650, 1847560, 494944450, 140625140040, 41500392000480, 12576565436409000, 3886690320522202050, 1219380045859742166400, 387154587452271772676400, 124120231850529022319265600, 40113527971798583517288018400, 13052024252899352166622940568000

Offset: 0

Peter Bala, Aug 03 2023

Fractional factorials are defined in terms of the gamma function; for example, (7*n/2)! = gamma(7*n/2 + 1).

Row 7 of A364513.

Paolo Xausa, Table of n, a(n) for n = 0..300

Cf. A364513.

seq( (5/7) * (9*n)!*(7*n/2)!^2/((9*n/2)!*(7*n)!*(5*n/2)!*n!^2), n = 1..15);

A364517[n_]:=If[n==0,1,(5/7)(9n)!(7n/2)!^2/((9n/2)!(7n)!(5n/2)!n!^2)];Array[A364517,15,0] (* Paolo Xausa, Oct 05 2023 *)

from math import factorial
from sympy import factorial2
def A364517(n): return int(5*factorial(9*n)*factorial2(7*n)**2//factorial2(9*n)//factorial(7*n)//factorial2(5*n)//factorial(n)**2//7) if n else 1 # Chai Wah Wu, Aug 08 2023

a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(7*n-1, (1 + x)/(1 - x)) for n >= 1.
a(n) = Sum_{k = 0..n} binomial(7*n - 1, n - k)^2 * binomial(5*n + k - 2, k).
a(n) = (5/7) * binomial(9*n,2*n)*binomial(9*n/2,2*n)*binomial(2*n,n)^2 / binomial(9*n/2,n)^2 for n >= 1.
a(n) = (7*n-1)! * ((9*n-1)/2)! * ((5*n-1)/2)!/( (5*n-1)! * ((7*n-1)/2)!^2 * n!^2 ) for n >= 1.
a(n) ~ c^n * sqrt(35)/(14*Pi*n), where c = (3^9)/(5^3) * sqrt(5) = 352.1002080....
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r.
For n > 0, a(n) = 5*(9*n)!*(7*n)!!^2/(7*(9*n)!!*(7*n)!*(5*n)!!*n!^2). - Chai Wah Wu, Aug 08 2023

A364518 Square array read by ascending antidiagonals: T(n,k) = [x^(2*k)] ( (1 + x)^(n+2)/(1 - x)^(n-2) )^k for n, k >= 0.

Original entry on oeis.org

1, 1, -2, 1, 0, 6, 1, 6, -10, -20, 1, 16, 70, 0, 70, 1, 30, 630, 924, 198, -252, 1, 48, 2310, 28672, 12870, 0, 924, 1, 70, 6006, 204204, 1385670, 184756, -4420, -3432, 1, 96, 12870, 860160, 19122246, 69206016, 2704156, 0, 12870, 1, 126, 24310, 2704156, 130378950, 1848483780, 3528923580, 40116600, 104006, -48620

Offset: 0

Peter Bala, Aug 07 2023

Compare with A364303 and A364519.
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_n(c, d) = (c_1*n)!*(c_2*n)!* ... *(c_K*n)!/ ( (d_1*n)!*(d_2*n)!* ... *(d_L*n)! ) and ask whether it is integral for all n >= 0. The integer L - K is called the height of the sequence. Bober completed the classification of integral factorial ratio sequences of height 1 (see A295431). Soundararajan gives many examples of two-parameter families of integral factorial ratio sequences of height 2.
Each row of the present table is an integral factorial ratio sequence of height 1. It is usually assumed that the c's and d's are integers but here some of the c's and d's are half-integers. See A276098 and the cross references there for further examples of this type.
It is known that the unsigned version of row 0 (the central binomial numbers A000984) and row 2 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.

			 Square array begins:
 n\k|  0   1      2        3           4             5
  - + - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |  1  -2      6      -20          70          -252   ... see A000984
  1 |  1   0    -10        0         198             0   ... see A211419
  2 |  1   6     70      924       12870        184756   ... A001448
  3 |  1  16    630    28672     1385670      69206016   ... A091496
  4 |  1  30   2310   204204    19122246    1848483780   ... A061162
  5 |  1  48   6006   860160   130378950   20392706048   ... A276098
  6 |  1  70  12870  2704156   601080390  137846528820   ... A001448 bisected
  7 |  1  96  24310  7028736  2149374150  678057476096   ... A276099

Peter Bala, Some integer ratios of factorials
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
Wikipedia, Hypergeometric function

Cf. A000984 (row 0 unsigned), A211419 (row 1 unsigned without 0's), A001448 (row 2), A091496 (row 3), A061162 (row 4), A276098 (row 5), A001448 bisected (row 6), A276099 (row 7).

Cf. A330843, A364303, A364506, A364509, A364513, A364519.

T(n,k) = add( binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j), j = 0..2*k):
# display as a square array
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);

T(n,k) = sum(j = 0, 2*k, binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j));
lista(nn) = for( n=0, nn, for (k=0, n, print1(T(n-k, k), ", "))); \\ Michel Marcus, Aug 13 2023

T(n,k) = Sum_{j = 0..2*k} binomial((n+2)*k, j)*binomial(n*k-j-1, 2*k-j).
T(2,k) = binomial(4*k,2*k).
For n >= 3, T(n,k) = binomial(n*k-1,2*k) * hypergeom([-(n+2)*k, -2*k], [1 - n*k], -1) except when (n,k) = (3,1).
For n >= 2, T(n,k) = ((n+2)*k)!*((n-2)*k/2)!/(((n+2)*k/2)!*((n-2)*k)!*(2*k)!) by Kummer's Theorem.
T(n,k) = [x^k] (1 - x)^(2*k) * Chebyshev_T(n*k, (1 + x)/(1 - x)).
T(n,k) = Sum_{j = 0..k} binomial(2*n*k, 2*j)*binomial((n-1)*k-j-1, k-j).
For n >= 3, T(n,k) = binomial((n-1)*k-1,k) * hypergeom([-n*k, -k, -n*k + 1/2], [1 - (n-1)*k, 1/2], 1).
The row generating functions are algebraic functions over the field of rational functions Q(x).

cp's OEIS Frontend

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.

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A364514 a(0) = 1 and a(n) = [x^n] (1 - x)^(2*n) * Legendre_P(n-1, (1 + x)/(1 - x)) for n >= 1.

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A364515 a(n) = (1/2) * (6*n)!*(2*n)!^2/((3*n)!*(4*n)!*n!^3) for n >= 1 with a(0) = 1.

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A364516 a(n) = (2/3) * (8*n)!*(3*n)!^2/((6*n)!*(4*n)!*(2*n)!*n!^2) for n >= 1, with a(0) = 1.

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A364517 a(n) = (5/7) * (9*n)!*(7*n/2)!^2/((9*n/2)!*(7*n)!*(5*n/2)!*n!^2) for n >= 1, with a(0) = 1.

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A364518 Square array read by ascending antidiagonals: T(n,k) = [x^(2*k)] ( (1 + x)^(n+2)/(1 - x)^(n-2) )^k for n, k >= 0.

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A365025 Square array read by antidiagonals: T(n, k) := (k/2)!/k! * ((2n+1)k)! * ((2n+1/2)k)! / ( (nk)!^2 ((n+1/2)*k)!^2 ) for n, k >= 0.

A364514 a(0) = 1 and a(n) = [x^n] (1 - x)^(2n) Legendre_P(n-1, (1 + x)/(1 - x)) for n >= 1.

A364515 a(n) = (1/2) * (6n)!(2n)!^2/((3n)!(4n)!*n!^3) for n >= 1 with a(0) = 1.

A364516 a(n) = (2/3) * (8n)!(3n)!^2/((6n)!(4n)!(2n)!*n!^2) for n >= 1, with a(0) = 1.

A364517 a(n) = (5/7) * (9n)!(7n/2)!^2/((9n/2)!(7n)!(5n/2)!*n!^2) for n >= 1, with a(0) = 1.