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A352652 a(n) = ( binomial(7*n,2*n)*binomial(7*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(7*n/2,n)^2.

%I A352652 #24 Dec 26 2024 04:22:14
%S A352652 1,30,2860,343200,45643500,6435891280,942422020540,141696569678400,
%T A352652 21724714133822700,3381208130986900500,532553441617598475360,
%U A352652 84695057996350934903680,13578009523892192555221500,2191530567314796197691108600,355765014009052303028935320000
%N A352652 a(n) = ( binomial(7*n,2*n)*binomial(7*n/2,2*n)*binomial(2*n,n)^2 ) / binomial(7*n/2,n)^2.
%C A352652 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)).
%C A352652 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. Usually, it is 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 factorial ratios of this type.
%C A352652 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.
%H A352652 Peter Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>
%H A352652 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 A352652 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 A352652 a(n) = (5/3)*Sum_{k = 0..n} (-1)^(n+k)*binomial(7*n,n-k)*binomial(5*n+k-1,k)^2 for n >= 1 (this formula shows 3*a(n) is integral; how to show a(n) is integral?).
%F A352652 a(n) = (5/3)*Sum_{k = 0..n} binomial(4*n-k-2,n-k)*binomial(5*n-1,k)^2 for n >= 1.
%F A352652 a(n) = (7*n)!*(5*n/2)!^2/((5*n)!*(7*n/2)!*(3*n/2)!*n!^2!).
%F A352652 a(n) = (5/3) * [x^n] ( (1 - x)^(2*n) * P(5*n-1,(1 + x)/(1 - x)) ) for n >= 1, where P(n,x) denotes the n-th Legendre polynomial.
%F A352652 a(n) = (5/3)*(-1)^n*binomial(7*n,n)*hypergeom([-n, 5*n, 5*n], [1, 6*n+1], 1) for n >= 1.
%F A352652 a(n) ~ sqrt(15)/Pi * 7^(7*n/2)/3^(3*n/2) * ( 1/(6*n) - 29/(945*n^2) + 841/(297675*n^3) + O(1/n^4) ).
%F A352652 a(n) = 7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) * a(n-2) with a(0) = 1 and a(1) = 30.
%F A352652 a(n)*A275654(n) = (7*n)!/(n!^4*(3*n)!) = A071549(n)/A006480(n).
%F A352652 a(p) == 30 (mod p^3) for all primes p >= 5.
%e A352652 Examples of supercongruences:
%e A352652 a(11) - a(1) = 84695057996350934903680 - 30 = 2*(5^2)*(11^3)*23*593* 3671*5693*4464799 == 0 (mod 11^3)
%e A352652 a(2*7) - a(2) = 355765014009052303028935320000 - 2860 = (2^2)*5*(7^3)*11* 269*3307*375101*14129010228023 == 0 (mod 7^3)
%p A352652 a := n -> if n = 0 then 1 elif n = 1 then 30 else
%p A352652 7*(5*n-2)*(5*n-4)*(5*n-6)*(5*n-8)*(7*n-1)*(7*n-3)*(7*n-5)*(7*n-9)*(7*n-11)*(7*n-13)/(3*n^2*(n-1)^2*(3*n-2)*(3*n-4)*(5*n-1)*(5*n- 3)*(5*n -7)*(5*n-9)) *a(n-2) end if:
%p A352652 seq(a(n), n = 0..20);
%o A352652 (Python)
%o A352652 from math import factorial
%o A352652 from sympy import factorial2
%o A352652 def A352652(n): return int(factorial(7*n)*factorial2(5*n)**2//factorial(5*n)//factorial2(7*n)//factorial2(3*n)//factorial(n)**2) # _Chai Wah Wu_, Aug 08 2023
%Y A352652 Cf. A275654, A276098, A276100, A276101, A276102, A352651.
%K A352652 nonn,easy
%O A352652 0,2
%A A352652 _Peter Bala_, Apr 03 2022