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A262732 a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!.

%I A262732 #67 Aug 09 2025 10:05:12
%S A262732 1,8,126,2240,41990,811008,15967980,318636032,6421422150,130395668480,
%T A262732 2663825039876,54684895150080,1127155102890908,23311847679590400,
%U A262732 483537022180231320,10054732930602762240,209536624110664757830,4375058594685417160704,91505601042318156186900
%N A262732 a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!.
%C A262732 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 A262732 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 A262732 R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
%H A262732 Michael De Vlieger, <a href="/A262732/b262732.txt">Table of n, a(n) for n = 0..751</a>
%H A262732 Peter Bala, <a href="/A100100/a100100_1.pdf">Notes on logarithmic differentiation, the binomial transform and series reversion</a>
%H A262732 Peter Bala, <a href="/A276098/a276098.pdf">Some integer ratios of factorials</a>
%H A262732 Peter Bala, <a href="/A262732/a262732.pdf">A supercongruence for A262732</a>
%F A262732 a(n) = Sum_{i = 0..n} binomial(5*n,i) * binomial(4*n-i-1,n-i).
%F A262732 a(n) = [x^n] ( (1 + x)^5/(1 - x)^3 )^n.
%F A262732 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 A262732 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 A262732 a(n) ~ 2^n*3^(-3*n/2)*5^(5*n/2)/sqrt(2*Pi*n). - _Ilya Gutkovskiy_, Jul 31 2016
%F A262732 From _Peter Bala_, Aug 22 2016: (Start)
%F A262732 a(n) = Sum_{k = 0..floor(n/2)} binomial(8*n,n - 2*k) * binomial(3*n + k - 1,k).
%F A262732 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 A262732 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 A262732 From _Karol A. Penson_, Apr 26 2018: (Start)
%F A262732 Integral representation of a(n) as the n-th moment of a positive function w(x) on the support (0, sqrt(12500/27)):
%F A262732 a(n) = Integral_{x=0..sqrt(12500/27)} x^n*w(x) dx,
%F A262732 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 A262732 From _Peter Bala_, Sep 15 2021: (Start)
%F A262732 a(n) = [x^n] (1 + 4*x)^((5*n-1)/2) = 4^n*binomial((5*n-1)/2,n).
%F A262732 a(p) == a(1) (mod p^3) for prime p >= 5.
%F A262732 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)
%F A262732 From _Seiichi Manyama_, Aug 09 2025: (Start)
%F A262732 a(n) = [x^n] 1/((1-x)^(n+1) * (1-2*x)^(3*n)).
%F A262732 a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(5*n,k) * binomial(2*n-k,n-k).
%F A262732 a(n) = Sum_{k=0..n} 2^k * binomial(3*n+k-1,k) * binomial(2*n-k,n-k).
%F A262732 a(n) = [x^n] 1/(1-4*x)^((3*n+1)/2). (End)
%p A262732 a := n -> 1/n! * (5*n)!/GAMMA(1 + 5*n/2) * GAMMA(1 + 3*n/2)/(3*n)!:
%p A262732 seq(a(n), n = 0..18);
%t A262732 Table[1/n!*(5 n)!/(5 n/2)!*(3 n/2)!/(3 n)!, {n, 0, 18}] (* or *)
%t A262732 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 A262732 (PARI) a(n) = sum(k=0, n, binomial(5*n,k)*binomial(4*n-k-1,n-k));
%o A262732 vector(30, n, a(n-1)) \\ _Altug Alkan_, Oct 03 2015
%o A262732 (Python)
%o A262732 from math import factorial
%o A262732 from sympy import factorial2
%o A262732 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 A262732 Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A262737, A276098, A276099.
%Y A262732 Cf. A115293.
%K A262732 nonn,easy
%O A262732 0,2
%A A262732 _Peter Bala_, Sep 29 2015