%I #36 Dec 26 2024 04:22:57
%S 1,12,378,14700,629850,28540512,1341310320,64676424384,3178603964250,
%T 158529793422000,7999466594747628,407514796591710600,
%U 20924507330066816112,1081581197431986720000,56225684939117297889600,2937292879652230377427200,154108110471294720105987930
%N a(n) = ( binomial(5*n,2*n)*binomial(5*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(5*n/2,n)^2.
%C 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.
%C 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.
%C 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.
%C 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)
%H Peter Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>
%H J. W. Bober, <a href="http://arxiv.org/abs/0709.1977">Factorial ratios, hypergeometric series, and a family of step functions</a>, 2007, arXiv:0709.1977v1 [math.NT], 2007; J. London Math. Soc., 79, Issue 2, (2009), 422-444.
%H K. Soundararajan, <a href="http://doi.org/10.1098/rsta.2018.0444">Integral factorial ratios: irreducible examples with height larger than 1</a>, Phil. Trans. R. Soc.A378: 2018044, 2019.
%F a(n) = (5*n)!*(3*n/2)!^2/( (3*n)!*(5*n/2)!*n!^2*(n/2)! ).
%F 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).
%F a(n) = 3*Sum_{k = 0..n} binomial(2*n-k-2,n-k)*binomial(3*n-1,k)^2 for n >= 1.
%F 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.
%F a(n) ~ (sqrt(3)/Pi)*(5^n)^(5/2)*( 1/(2*n) - 2/(15*n^2) + 4/(225*n^3) + O(1/n^4) ).
%F a(n) = A008978(n)/A275652(n).
%F a(n) = binomial(3*n/2,n)*A262732(n).
%F a(n) = 3*(-1)^n*binomial(5*n,n)*hypergeom([-n, 3*n, 3*n], [1, 4*n+1], 1) for n >= 1.
%F 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.
%F a(p) == 12 (mod p^3) for prime p >= 5.
%F 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).
%e Examples of supercongruences:
%e a(2*7) - a(2) = 56225684939117297889600 - 378 = 2*(3^3)*(7^4)*6553*411473* 160830097 == 0 (mod 7^4).
%e a(13) - a(1) = 1081581197431986720000 - 12 = (2^2)*3*(13^3)* 41024927834622467 == 0 (mod 13^3)
%p a := n -> if n = 0 then 1 elif n = 1 then 12 else
%p 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:
%p seq(a(n), n = 0..20);
%o (Python)
%o from math import factorial
%o from sympy import factorial2
%o 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
%Y Cf. A008978, A262732, A275652, A276098, A276100, A276101, A276102, A352652.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Mar 25 2022