cp's OEIS Frontend

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

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

a(n) = 2F1[-3n,-2n,1,1] (see Mathematica code below). - John M. Campbell, Jul 15 2011

G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/3, 1/2, 2/3], (3125/108)*x). - Robert Israel, Aug 07 2014

From Peter Bala, Oct 05 2015: (Start)

a(n) = [x^n] ( (1 + x)*C(x) )^(5*n), where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.

a(n) = 5*A259550(n) for n >= 1.

exp( (1/5) * Sum_{n >= 1} a(n)*x^n/n ) = 1 + 2*x + 23*x^2 + 377*x^3 + ... is the o.g.f. for the sequence of Duchon numbers A060941. (End)

a(n) = [x^(2*n)] 1/(1 - x)^(3*n+1). - Ilya Gutkovskiy, Oct 10 2017

D-finite with recurrence 6*n*(3*n-1)*(2*n-1)*(3*n-2)*a(n) -5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1)=0. - R. J. Mathar, Feb 08 2021

a(n) = Sum_{k = 0..2*n} binomial(3*n+k-1, k). Cf. A066802. - Peter Bala, Jun 04 2024

Right-hand side of the identity Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)* binomial(4*n-k, 2*n-k) = binomial(5*n, 2*n). Compare with the identity Sum_{k = 0..n} (-1)^k*binomial(n, k)*binomial(4*n-k, 2*n-k) = binomial(3*n, n). - Peter Bala, Jun 05 2024

From Karol A. Penson, May 07 2025: (Start)

G.f. denoted by h(x) satisfies the following algebraic equation of order 10:

8 - 3125*x + 20*(-13 + 3125*x)*h(x) - 45*(-74 + 9375*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^2 + 5*(-4023 + 175000*x)*h(x)^3 + 25*(1809 + 53125*x)*h(x)^4 + (34375*x - 738)*(3125*x - 108)*h(x)^5 + 15*(3125*x + 297)*(3125*x - 108)*h(x)^6 + 5*(3125*x - 108)^2*h(x)^7 + 135*(3125*x - 108)^2*h(x)^8 + (3125*x - 108)^3*h(x)^10=0.

a(n) = Integral_{x=0..3125/108} x^n*W(x)*dx, n>=0, where W(x) = W1(x)+W2(x)+W3(x)+W4(x) can be expressed with four generalized hypergeometric functions of type 4F3:

W1(x) = sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([1/5, 8/15, 7/10, 13/15], [2/5, 3/5, 4/5], (108*x)/3125)/(10*Pi*x^(4/5)),

W2(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([2/5, 11/15, 9/10, 16/15], [3/5, 4/5, 6/5], (108*x)/3125)/(50*Pi*x^(3/5)),

W3(x) = sqrt(5)*sec((3*Pi)/10)*sin(Pi/10)*hypergeom([3/5, 14/15, 11/10, 19/15], [4/5, 6/5, 7/5], (108*x)/3125)/(125*Pi*x^(2/5)), and

W4(x) = (7*sqrt(5)*csc((2*Pi)/5)*sin((3*Pi)/10)*hypergeom([4/5, 17/15, 13/10, 22/15], [6/5, 7/5, 8/5], (108*x)/3125))/(1250*Pi*x^(1/5)).

Using the formula for a(n) only, W(x) can be shown to be a positive function. It is singular at x=0 and at x=3125/108. This integral representation is unique since W(x) is the solution of the Hausdorff moment problem. (End)

From Peter Bala, Jun 21 2025: (Start)

a(n) = [x^(3*n)] 1/(1 - x)^(2*n+1).

a(n) = Sum_{k = 0..3*n} binomial(2*n+k-1, k). (End)

a(n) = n! * Sum_{k=0..floor(n/5)} 1/(k!^3 * (2*k)! * (n-5*k)!) = Sum_{k=0..floor(n/5)} binomial(n,5*k) * A001460(k).

From Vaclav Kotesovec, Mar 22 2023: (Start)

Recurrence: 2*n^3*(2*n - 5)*a(n) = 2*(10*n^4 - 40*n^3 + 50*n^2 - 30*n + 7)*a(n-1) - 10*(n-1)*(4*n^3 - 18*n^2 + 26*n - 13)*a(n-2) + 40*(n-2)^3*(n-1)*a(n-3) - 10*(n-3)*(n-2)*(n-1)*(2*n - 5)*a(n-4) + 3129*(n-4)*(n-3)*(n-2)*(n-1)*a(n-5).

a(n) ~ sqrt(c) * (1 + 5/2^(2/5))^n / (Pi^(3/2) * n^(3/2)), where c = 3.154712586460560795509193778252140601572145506226776094640234924884123818... is the real root of the equation -30634915689 + 95407210000*c - 127160000000*c^2 + 79846400000*c^3 - 25600000000*c^4 + 3276800000*c^5 = 0. (End)