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a(n) = (5*n)!/((2*n)!*(n!)^3).
4

%I #52 Sep 04 2023 02:01:16

%S 1,60,18900,8408400,4364860500,2473653742560,1483630051503600,

%T 925833064837824000,594927307937311420500,391004487919622186610000,

%U 261614105944603801295306400,177601637048592673099585584000,122027661025630720013771117910000

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

%H James Spahlinger, <a href="/A001460/b001460.txt">Table of n, a(n) for n = 0..400</a>

%H R. Mestrovic, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.

%F a(n) = A008978(n)/A000984(n). - _Zerinvary Lajos_, Jun 28 2007

%F From _Gheorghe Coserea_, Jul 18 2016, (Start):

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

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

%F 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).

%F From _Peter Bala_, Dec 30 2019: (Start)

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

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

%F 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)

%F From _Peter Bala_, Feb 16 2020: (Start)

%F 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.

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

%F 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

%p f := n->(5*n)!/((2*n)!*(n!)^3);

%p seq((5*n)!/(n!)^5/binomial(2*n,n), n=0..15); # _Zerinvary Lajos_, Jun 28 2007

%t Table[(5 n)!/((2 n)! (n!)^3), {n, 0, 15}] (* or *)

%t Table[(5 n)!/(n!)^5/Binomial[2 n, n], {n, 0, 15}] (* _Michael De Vlieger_, Jul 18 2016 *)

%o (PARI) a(n) = (5*n)!/((2*n)!*n!^3); \\ _Gheorghe Coserea_, Jul 18 2016

%o (Magma) [Factorial(5*n)/(Factorial(2*n)*Factorial(n)^3):n in [0..15]]; // _Marius A. Burtea_, Feb 17 2020

%o (SageMath) f=factorial; [f(5*n)/(f(2*n)*f(n)^3) for n in range(16)] # _G. C. Greubel_, Sep 03 2023

%Y Cf. A000984, A008978.

%Y Cf. A268545 - A268555.

%K nonn

%O 0,2

%A _N. J. A. Sloane_