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A262732 a(n) = (1/n!) * (5*n)!/(5*n/2)! * (3*n/2)!/(3*n)!. 11
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

Table of n, a(n) for n=0..18.

Peter Bala, Notes on logarithmic differentiation, the binomial transform and series reversion

Peter Bala, Some integer ratios of factorials

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)), in Maple notation:

a(n) = int(x^n*w(x), x=0..sqrt(12500/27)), n = 0,1...,

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)

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.

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 April 1 05:04 EDT 2020. Contains 333155 sequences. (Running on oeis4.)