A186420 a(n) = binomial(2n,n)^4.

1, 16, 1296, 160000, 24010000, 4032758016, 728933458176, 138735983333376, 27435582641610000, 5588044012339360000, 1165183173971324375296, 247639903129149250277376, 53472066459540320483696896, 11701285507234585729600000000, 2589980371199606611713600000000

Offset

0,2

Links

Seiichi Manyama, Table of n, a(n) for n = 0..417

Formula

a(n) = A000984(n)^4 = A002894(n)^2.

a(n) = binomial(2*n,n)^4 = ( [x^n](1 + x)^(2*n) )^4 = [x^n](F(x)^(16*n)), where F(x) = 1 + x + 25*x^2 + 1798*x^3 + 183442*x^4 + 22623769*x^5 + 3142959012*x^6 + ... appears to have integer coefficients. For similar results see A000897, A002894, A002897, A006480, A008977 and A188662. - Peter Bala, Jul 14 2016

a(n) ~ 256^n/(Pi*n)^2. - Ilya Gutkovskiy, Jul 13 2016

Example

G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,1},256x) where 4F3 is a hypergeometric series.

Mathematica

Table[Binomial[2n, n]^4, {n, 0, 20}]
Table[Coefficient[Series[HypergeometricPFQ[{1/2, 1/2, 1/2, 1/2}, {1, 1, 1}, 256 x], {x, 0, n}], x, n], {n, 0, 14}] (* Michael De Vlieger, Jul 13 2016 *)

Prog

(Maxima)
makelist(binomial(2*n, n)^4, n, 0, 40);

Crossrefs

Cf. binomial(2n,n)^k: A000984 (k=1), A002894 (k=2), A002897 (k=3), this sequence (k=4).

Cf. A000108, A000888, A186414, A186415, A186416, A186418, A186419, A000897, A006480, A008977, A188662.

Sequence in context: A007480 A369169 A307814 * A163395 A134375 A321583

Adjacent sequences: A186417 A186418 A186419 * A186421 A186422 A186423

Keyword

easy,nonn

Author

Emanuele Munarini, Feb 21 2011

Status

approved