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.
1, 12, 378, 14700, 629850, 28540512, 1341310320, 64676424384, 3178603964250, 158529793422000, 7999466594747628, 407514796591710600, 20924507330066816112, 1081581197431986720000, 56225684939117297889600, 2937292879652230377427200, 154108110471294720105987930
Offset: 0
Examples
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)
Links
- 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.
Programs
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Maple
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);
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Python
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
Formula
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) = 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).
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