A001460 a(n) = (5*n)!/((2*n)!*(n!)^3).

1, 60, 18900, 8408400, 4364860500, 2473653742560, 1483630051503600, 925833064837824000, 594927307937311420500, 391004487919622186610000, 261614105944603801295306400, 177601637048592673099585584000, 122027661025630720013771117910000

Offset

0,2

Links

James Spahlinger, Table of n, a(n) for n = 0..400

R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Formula

a(n) = A008978(n)/A000984(n). - Zerinvary Lajos, Jun 28 2007

From Gheorghe Coserea, Jul 18 2016, (Start):

a(n) = [(xyzw)^(3n)] 1/(1-(w*x*y+w*z+x*z+y*z)).

a(n) ~ sqrt(5)/(4*Pi^(3/2)) * n^(-3/2) * (3125/4)^n.

0 = (-4*x^3+3125*x^6)*y'''' + (-18*x^2+37500*x^5)*y''' + (-10*x+117500*x^4)*y'' + (2+95000*x^3)*y' + (9720*x^2)*y, where y(x) = A(x^3).

From Peter Bala, Dec 30 2019: (Start)

a(n) = binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n).

a(n) = ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ).

a(n) = [x^n]( F(x)^(60*n) ), where [x^n] is the coefficient extraction operator and where F(x) = 1 + x + 98*x^2 + 23861*x^3 + 7987534*x^4 + 3169655645*x^5 + 1398711076599*x^6 + ... appears to have integer coefficients. Cf. A008978. (End)

From Peter Bala, Feb 16 2020: (Start)

Congruences: a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k - apply Mestrovic, equation 39, p. 12.

a(n) = [(x*y*z)^n] (1 + x + y + z)^(5*n). (End)

a(n) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/(2*n^3*(2*n - 1)). - Neven Sajko, Jul 22 2023

Maple

f := n->(5*n)!/((2*n)!*(n!)^3);
seq((5*n)!/(n!)^5/binomial(2*n, n), n=0..15); # Zerinvary Lajos, Jun 28 2007

Mathematica

Table[(5 n)!/((2 n)! (n!)^3), {n, 0, 15}] (* or *)
Table[(5 n)!/(n!)^5/Binomial[2 n, n], {n, 0, 15}] (* Michael De Vlieger, Jul 18 2016 *)

Prog

(PARI) a(n) = (5*n)!/((2*n)!*n!^3);  \\ Gheorghe Coserea, Jul 18 2016
(Magma) [Factorial(5*n)/(Factorial(2*n)*Factorial(n)^3):n in [0..15]]; // Marius A. Burtea, Feb 17 2020
(SageMath) f=factorial; [f(5*n)/(f(2*n)*f(n)^3) for n in range(16)] # G. C. Greubel, Sep 03 2023

Crossrefs

Cf. A000984, A008978.

Cf. A268545 - A268555.

Sequence in context: A009564 A269762 A291912 * A003794 A275051 A068295

Adjacent sequences: A001457 A001458 A001459 * A001461 A001462 A001463

Keyword

nonn

Author

N. J. A. Sloane

Status

approved