A262732 -id:A262732 - cp's OEIS frontend

A352651 a(n) = ( binomial(5n,2n)binomial(5n/2,2n)binomial(2n,n)^2 ) / binomial(5n/2,n)^2.

1, 12, 378, 14700, 629850, 28540512, 1341310320, 64676424384, 3178603964250, 158529793422000, 7999466594747628, 407514796591710600, 20924507330066816112, 1081581197431986720000, 56225684939117297889600, 2937292879652230377427200, 154108110471294720105987930

Offset: 0

Peter Bala, Mar 25 2022

We write x! as shorthand for Gamma(x+1) and binomial(x,y) as shorthand for x!/(y!*(x-y)!) = Gamma(x+1)/(Gamma(y+1)*Gamma(x-y+1)). Given two sequences of numbers 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. 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 rational numbers. See A276098 and the cross references for further examples of this type.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for all primes p >= 5 and positive integers n and k. The case n = k = 1 is easily proved.
More generally, for an integer N not equal to 0 or 1, the height 2 factorial ratio sequence whose n-th term is given by ( binomial(N*n,2*n)* binomial(N*n/2,2*n)* binomial(2*n,n)^2 )/binomial(N*n/2,n)^2 is conjectured to be integral and satisfy the same supercongruences. This is the case N = 5. See A352652 (N = 7)

			Examples of supercongruences:
a(2*7) - a(2) = 56225684939117297889600 - 378 = 2*(3^3)*(7^4)*6553*411473* 160830097 == 0 (mod 7^4).
a(13) - a(1) = 1081581197431986720000 - 12 = (2^2)*3*(13^3)* 41024927834622467 == 0 (mod 13^3)

Peter Bala, Some integer ratios of factorials
J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, 2007, arXiv:0709.1977v1 [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. R. Soc.A378: 2018044, 2019.

Cf. A008978, A262732, A275652, A276098, A276100, A276101, A276102, A352652.

a := n -> if n = 0 then 1 elif n = 1 then 12 else
5*(3*n - 2)*(3*n - 4)*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)/(n^2*(n - 1)^2*(3*n - 1)*(3*n - 5))*a(n-2) end if:
seq(a(n), n = 0..20);

from math import factorial
from sympy import factorial2
def A352651(n): return int(factorial(5*n)*factorial2(3*n)**2//factorial(3*n)//factorial2(5*n)//factorial(n)**2//factorial2(n)) # Chai Wah Wu, Aug 08 2023

a(n) = (5*n)!*(3*n/2)!^2/( (3*n)!*(5*n/2)!*n!^2*(n/2)! ).
a(n) = 3*Sum_{k = 0..n} (-1)^(n+k)*binomial(5*n,n-k)*binomial(3*n+k-1,k)^2 for n >= 1 (this formula shows the sequence is integral).
a(n) = 3*Sum_{k = 0..n} binomial(2*n-k-2,n-k)*binomial(3*n-1,k)^2 for n >= 1.
a(n) = 3 * [x^n] ( (1 - x)^(2*n) * P(3*n-1,(1 + x)/(1 - x)) ) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
a(n) ~ (sqrt(3)/Pi)*(5^n)^(5/2)*( 1/(2*n) - 2/(15*n^2) + 4/(225*n^3) + O(1/n^4) ).
a(n) = A008978(n)/A275652(n).
a(n) = binomial(3*n/2,n)*A262732(n).
a(n) = 3*(-1)^n*binomial(5*n,n)*hypergeom([-n, 3*n, 3*n], [1, 4*n+1], 1) for n >= 1.
a(n) = 5*(3*n-2)*(3*n-4)*(5*n-1)*(5*n-3)*(5*n-7)*(5*n-9)/(n^2*(n-1)^2*(3*n- 1)*(3*n-5)) * a(n-2) with a(0) = 1 and a(1) = 12.
a(p) == 12 (mod p^3) for prime p >= 5.
O.g.f.: A(x) = hypergeom([1/10, 3/10, 7/10, 9/10, 1/3, 2/3], [1/6, 5/6, 1/2, 1/2, 1], (5^5)*x^2) + 12*x*hypergeom([3/5, 4/5, 6/5, 7/5, 5/6, 7/6], [2/3, 4/3, 3/2, 3/2, 1], (5^5)*x^2).

A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 2, -2, 1, 4, 6, 0, 1, 6, 30, 20, 6, 1, 8, 70, 256, 70, 0, 1, 10, 126, 924, 2310, 252, -20, 1, 12, 198, 2240, 12870, 21504, 924, 0, 1, 14, 286, 4420, 41990, 184756, 204204, 3432, 70, 1, 16, 390, 7680, 104006, 811008, 2704156, 1966080, 12870, 0

Offset: 0

Seiichi Manyama, Feb 07 2020

			Square array begins:
    1,   1,     1,      1,      1,       1, ...
    0,   2,     4,      6,      8,      10, ...
   -2,   6,    30,     70,    126,     198, ...
    0,  20,   256,    924,   2240,    4420, ...
    6,  70,  2310,  12870,  41990,  104006, ...
    0, 252, 21504, 184756, 811008, 2521260, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=1..7 give A000984, A091527, A001448, A262732, A211419, A262733, A211421.

Main diagonal is A332231.

T[n_, k_] := Sum[Binomial[(k + 1)*n, j] * Binomial[k*n - j - 1, n - j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 05 2021 *)

T(n,k) = Sum_{j=0..n} binomial((k+1)*n,j) * binomial(k*n-j-1,n-j).

T(n,k) = 1/n! * ((k+1)*n)!/Gamma(1 + (k+1)*n/2) * Gamma(1 + (k-1)*n/2)/((k-1)*n)!.

A115293 Row sums of correlation triangle for (1+x)^3/(1-x).

Original entry on oeis.org

1, 8, 31, 80, 160, 272, 416, 592, 800, 1040, 1312, 1616, 1952, 2320, 2720, 3152, 3616, 4112, 4640, 5200, 5792, 6416, 7072, 7760, 8480, 9232, 10016, 10832, 11680, 12560, 13472, 14416, 15392, 16400, 17440, 18512, 19616, 20752, 21920, 23120, 24352

Offset: 0

Paul Barry, Jan 19 2006

Row sums of number triangle A115292.

If Y_i (i=1,2,3,4,5) are 2-blocks of a (n+5)-set X then a(n-2) is the number of 7-subsets of X intersecting each Y_i (i=1,2,3,4,5). - Milan Janjic, Oct 28 2007

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Milan Janjic, Two Enumerative Functions
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A115291, A115292, A262732.

seq(add(binomial(5,n-k)*binomial(k+2,k), k = 0..n), n = 0..40); # Peter Bala, Sep 26 2021

LinearRecurrence[{3,-3,1},{1,8,31,80,160,272},50] (* Harvey P. Dale, Dec 03 2018 *)

a(n) = sum(k = 0, n, binomial(5,n-k)*binomial(k+2,k)); \\ Michel Marcus, Oct 01 2021

G.f.: A(x) = (1+x)^5/(1-x)^3.
a(n) = Sum_{k = 0..n} Sum_{j = 0..n} [j<=k]*A115291(k-j)*[j<=n-k]*A115291(n-k-j).
From Peter Bala, Sep 26 2021: (Start)
a(n) = Sum_{k = 0..n} binomial(5,n-k)*binomial(k+2,k).
A262732(n) = [x^n] A(x)^n. (End)

A364402 a(n) = (3n)!(10n)!/((2n)!(5n)!(6n)!).

Original entry on oeis.org

1, 126, 41990, 15967980, 6421422150, 2663825039876, 1127155102890908, 483537022180231320, 209536624110664757830, 91505601042318156186900, 40205863224219682380130740, 17753412284992688334256754280, 7871411119532225034145860092700, 3502017467737750755575471520717480

Offset: 0

Neven Sajko, Jul 22 2023

nonn

Member of Bober's second infinite family of integral factorial ratio sequences with a=5 and b=3 (see equation 11 at p. 16 in Bober).

Jonathan W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, Journal of the London Mathematical Society, Vol. 79, No. 2 (2009), pp. 422-444; arXiv preprint, arXiv:0709.1977 [math.NT], 2007.

Bisection of A262732. Cf. A182400, A211419.

seq( (3*n)!*(10*n)!/((2*n)!*(5*n)!*(6*n)!), n = 0..20); # Peter Bala, Sep 24 2023

a(n) = (3*n)!*(10*n)!/((2*n)!*(5*n)!*(6*n)!); \\ Michel Marcus, Sep 20 2023

a(n) = 10*(10*n - 1)*(10*n - 3)*(10*n - 7)*(10*n - 9)/(3*n*(2*n - 1)*(6*n - 1)*(6*n - 5))*a(n-1).
a(n) ~ 2^(2*n-1) * 5^(5*n) / (sqrt(Pi*n) * 3^(3*n)). - Vaclav Kotesovec, Sep 21 2023
From Peter Bala, Sep 24 2023: (Start)
a(n) = A262732(2*n).
a(n) = [x^(2*n)] (1 + 4*x)^((10*n-1)/2) = 16^n * binomial((10*n-1)/2, 2*n).
O.g.f. A(x) = hypergeom([9/10, 7/10, 3/10, 1/10], [5/6, 1/2, 1/6], (12500/27)*x).
(End)

cp's OEIS Frontend

A352651 a(n) = ( binomial(5*n,2*n)*binomial(5*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(5*n/2,n)^2.

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A330843 Square array T(n,k) = [x^n] ((1+x)^(k+1) / (1-x)^(k-1))^n, n>=0, k>=0, read by descending antidiagonals.

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A115293 Row sums of correlation triangle for (1+x)^3/(1-x).

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Author

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Crossrefs

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Maple

Mathematica

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Formula

A364402 a(n) = (3*n)!*(10*n)!/((2*n)!*(5*n)!*(6*n)!).

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Author

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Crossrefs

Programs

Maple

PARI

Formula

A352651 a(n) = ( binomial(5n,2n)binomial(5n/2,2n)binomial(2n,n)^2 ) / binomial(5n/2,n)^2.

A364402 a(n) = (3n)!(10n)!/((2n)!(5n)!(6n)!).