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%I #61 Aug 11 2023 10:00:02
%S 1,8,126,2240,41990,811008,15967980,318636032,6421422150,130395668480,
%T 2663825039876,54684895150080,1127155102890908,23311847679590400,
%U 483537022180231320,10054732930602762240,209536624110664757830,4375058594685417160704,91505601042318156186900
%N a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!.
%C Sequence terms are given by the coefficient of x^n in the expansion of ( (1 + x)^(k+2)/(1 - x)^k )^n when k = 3. See the cross references for related sequences obtained from other values of k.
%C Let a > b be nonnegative integers. Then the ratio of factorials ((2*a + 1)*n)!*((b + 1/2)*n)!/(((a + 1/2)*n)!*((2*b + 1)*n)!*((a - b)*n)!) is an integer for n >= 0. This is the case a = 2, b = 1. - _Peter Bala_, Aug 28 2016
%D R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
%H Michael De Vlieger, <a href="/A262732/b262732.txt">Table of n, a(n) for n = 0..751</a>
%H Peter Bala, <a href="/A100100/a100100_1.pdf">Notes on logarithmic differentiation, the binomial transform and series reversion</a>
%H Peter Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>
%H Peter Bala, <a href="/A262732/a262732.pdf">A supercongruence for A262732</a>
%F a(n) = Sum_{i = 0..n} binomial(5*n,i)*binomial(4*n-i-1,n-i).
%F a(n) = [x^n] ( (1 + x)^5/(1 - x)^3 )^n.
%F D-finite with recurrence a(n) = 20*(5*n - 1)*(5*n - 3)*(5*n - 7)*(5*n - 9)/( n*(3*n - 1)*(3*n - 3)*(3*n - 5) ) * a(n-2).
%F The o.g.f. exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 8*x + 95*x^2 + 1336*x^4 + ... has integer coefficients and equals (1/x) * (series reversion of x*(1 - x)^3/(1 + x)^5). See A262737.
%F a(n) ~ 2^n*3^(-3*n/2)*5^(5*n/2)/sqrt(2*Pi*n). - _Ilya Gutkovskiy_, Jul 31 2016
%F From _Peter Bala_, Aug 22 2016: (Start)
%F a(n) = Sum_{k = 0..floor(n/2)} binomial(8*n,n - 2*k) * binomial(3*n + k - 1,k).
%F O.g.f.: A(x) = Hypergeom([9/10, 7/10, 3/10, 1/10], [5/6, 1/2, 1/6], (12500/27)*x^2) + 8*x*Hypergeom([7/5, 6/5, 4/5, 3/5], [4/3, 3/2, 2/3], (12500/27)*x^2).
%F The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^5/(1 - x)^3) and hence is algebraic by Stanley 1999, Theorem 6.33, p. 197. (End)
%F From _Karol A. Penson_, Apr 26 2018: (Start)
%F Integral representation of a(n) as the n-th moment of a positive function w(x) on the support (0, sqrt(12500/27)):
%F a(n) = Integral_{x=0..sqrt(12500/27)} x^n*w(x) dx,
%F where w(x) = sqrt(5)*2^(3/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*hypergeom([1/10, 4/15, 3/5, 14/15], [1/5, 2/5, 4/5], 27*x^2*(1/12500))/(10*Pi*x^(4/5)) + sqrt(5)*2^(4/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*hypergeom([3/10, 7/15, 4/5, 17/15], [2/5, 3/5, 6/5], 27*x^2*(1/12500))/(50*Pi*x^(2/5)) + sqrt(5)*2^(1/5)*csc(2*Pi*(1/5))*sin(3*Pi*(1/10))*x^(2/5)*hypergeom([7/10, 13/15, 6/5, 23/15], [4/5, 7/5, 8/5], 27*x^2*(1/12500))/(625*Pi) + 11*sqrt(5)*2^(2/5)*csc((1/5)*Pi)*sin((1/10)*Pi)*x^(4/5)*hypergeom([9/10, 16/15, 7/5, 26/15], [6/5, 8/5, 9/5], 27*x^2*(1/12500))/(50000*Pi). The function w(x) involves four different hypergeometric functions of type 4F3. The function w(x) is singular at both ends of the support. It is the solution of the Hausdorff moment problem and as such it is unique. (End)
%F From _Peter Bala_, Sep 15 2021: (Start)
%F a(n) = [x^n] (1 + 4*x)^((5*n-1)/2) = 4^n*binomial((5*n-1)/2,n).
%F a(p) == a(1) (mod p^3) for prime p >= 5.
%F More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) for prime p >= 5 and positive integers n and k. (End)
%p a := n -> 1/n! * (5*n)!/GAMMA(1 + 5*n/2) * GAMMA(1 + 3*n/2)/(3*n)!:
%p seq(a(n), n = 0..18);
%t Table[1/n!*(5 n)!/(5 n/2)!*(3 n/2)!/(3 n)!, {n, 0, 18}] (* or *)
%t Table[Sum[Binomial[8 n, n - 2 k] Binomial[3 n + k - 1, k], {k, 0, Floor[n/2]}], {n, 0, 18}] (* _Michael De Vlieger_, Aug 28 2016 *)
%o (PARI) a(n) = sum(k=0, n, binomial(5*n,k)*binomial(4*n-k-1,n-k));
%o vector(30, n, a(n-1)) \\ _Altug Alkan_, Oct 03 2015
%o (Python)
%o from math import factorial
%o from sympy import factorial2
%o def A262732(n): return int((factorial(5*n)*factorial2(3*n)<<n)//(factorial2(5*n)*factorial(3*n)*factorial(n))) # _Chai Wah Wu_, Aug 10 2023
%Y Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A262737, A276098, A276099.
%Y Cf. A115293.
%K nonn,easy
%O 0,2
%A _Peter Bala_, Sep 29 2015
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