A008978 a(n) = (5*n)!/(n!)^5.
1, 120, 113400, 168168000, 305540235000, 623360743125120, 1370874167589326400, 3177459078523411968000, 7656714453153197981835000, 19010638202652030712978200000, 48334775757901219912115629238400, 125285878026462826569986857692288000
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..100
- V. Batyrev, Review of "Mirror Symmetry and Algebraic Geometry", by D. A. Cox and S. Katz, Bull. Amer. Math. Soc., 37 (No. 4, 2000), 473-476.
- R. M. Dickau, 5-D shortest path diagrams
- R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.
Crossrefs
Programs
-
Magma
[Factorial(5*n)/Factorial(n)^5: n in [0..10]]; // Vincenzo Librandi, Mar 08 2014
-
Mathematica
Table[(5 n)!/(n)!^5, {n, 0, 20}] (* Vincenzo Librandi, Mar 08 2014 *) -
PARI
a(n) = (5*n)!/(n!)^5; \\ Michel Marcus, Mar 08 2014
Formula
a(n) ~ 5^(5*n+1/2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Mar 07 2014
From Peter Bala, Jul 12 2016: (Start)
a(n) = binomial(2*n,n)*binomial(3*n,n)*binomial(4*n,n)*binomial(5*n,n) = ( [x^n](1 + x)^(2*n) ) * ( [x^n](1 + x)^(3*n) ) * ( [x^n](1 + x)^(4*n) ) * ( [x^n](1 + x)^(5*n) ) = [x^n]( F(x)^(120*n) ), where F(x) = 1 + x + 353*x^2 + 318986*x^3 + 408941594*x^4 + 633438203535*x^5 + 1105336091531052*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977, A186420 and A188662. (End)
From Peter Bala, Jul 17 2016: (Start)
a(n) = Sum_{k = 0..4*n} (-1)^k*binomial(5*n,n + k)*binomial(n + k,k)^5.
a(n) = Sum_{k = 0..5*n} (-1)^(n+k)*binomial(5*n,k)*binomial(n + k,k)^5. (End)
From Ilya Gutkovskiy, Nov 23 2017: (Start)
O.g.f.: 4F3(1/5,2/5,3/5,4/5; 1,1,1; 3125*x).
E.g.f.: 4F4(1/5,2/5,3/5,4/5; 1,1,1,1; 3125*x). (End)
From Peter Bala, Feb 16 2020: (Start)
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*u)^n] (1 + x + y + z + u )^(5*n). (End)
a(n) = 120*A322252(n). - R. J. Mathar, Jun 21 2023
a(n) = a(n-1)*5*(5*n - 1)*(5*n - 2)*(5*n - 3)*(5*n - 4)/n^4. - Neven Sajko, Jul 21 2023
Comments