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 A262732 a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!. 14
 1, 8, 126, 2240, 41990, 811008, 15967980, 318636032, 6421422150, 130395668480, 2663825039876, 54684895150080, 1127155102890908, 23311847679590400, 483537022180231320, 10054732930602762240, 209536624110664757830, 4375058594685417160704, 91505601042318156186900 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. 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 REFERENCES R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197. LINKS Michael De Vlieger, Table of n, a(n) for n = 0..751 Peter Bala, Some integer ratios of factorials Peter Bala, A supercongruence for A262732 FORMULA a(n) = Sum_{i = 0..n} binomial(5*n,i)*binomial(4*n-i-1,n-i). a(n) = [x^n] ( (1 + x)^5/(1 - x)^3 )^n. 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). 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. a(n) ~ 2^n*3^(-3*n/2)*5^(5*n/2)/sqrt(2*Pi*n). - Ilya Gutkovskiy, Jul 31 2016 From Peter Bala, Aug 22 2016: (Start) a(n) = Sum_{k = 0..floor(n/2)} binomial(8*n,n - 2*k) * binomial(3*n + k - 1,k). 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). 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) From Karol A. Penson, Apr 26 2018: (Start) Integral representation of a(n) as the n-th moment of a positive function w(x) on the support (0, sqrt(12500/27)): a(n) = Integral_{x=0..sqrt(12500/27)} x^n*w(x), 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) From Peter Bala, Sep 15 2021: (Start) a(n) = [x^n] (1 + 4*x)^((5*n-1)/2) = 4^n*binomial((5*n-1)/2,n). a(p) == a(1) (mod p^3) for prime p >= 5. 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) MAPLE a := n -> 1/n! * (5*n)!/GAMMA(1 + 5*n/2) * GAMMA(1 + 3*n/2)/(3*n)!: seq(a(n), n = 0..18); MATHEMATICA Table[1/n!*(5 n)!/(5 n/2)!*(3 n/2)!/(3 n)!, {n, 0, 18}] (* or *) 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 *) PROG (PARI) a(n) = sum(k=0, n, binomial(5*n, k)*binomial(4*n-k-1, n-k)); vector(30, n, a(n-1)) \\ Altug Alkan, Oct 03 2015 CROSSREFS Cf. A000984 (k = 0), A091527 (k = 1), A001448 (k = 2), A211419 (k = 4), A262733 (k = 5), A211421 (k = 6), A262737, A276098, A276099. Cf. A115293. Sequence in context: A055762 A281714 A236549 * A220728 A029472 A001081 Adjacent sequences:  A262729 A262730 A262731 * A262733 A262734 A262735 KEYWORD nonn,easy AUTHOR Peter Bala, Sep 29 2015 STATUS approved

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Last modified May 22 09:12 EDT 2022. Contains 353941 sequences. (Running on oeis4.)