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%I #29 Nov 01 2016 11:26:56
%S 1,16,1296,160000,24010000,4032758016,728933458176,138735983333376,
%T 27435582641610000,5588044012339360000,1165183173971324375296,
%U 247639903129149250277376,53472066459540320483696896,11701285507234585729600000000,2589980371199606611713600000000
%N a(n) = binomial(2n,n)^4.
%H Seiichi Manyama, <a href="/A186420/b186420.txt">Table of n, a(n) for n = 0..417</a>
%F a(n) = A000984(n)^4 = A002894(n)^2.
%F 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
%F a(n) ~ 256^n/(Pi*n)^2. - _Ilya Gutkovskiy_, Jul 13 2016
%e G.f.: 4F3({1/2,1/2,1/2,1/2},{1,1,1},256x) where 4F3 is a hypergeometric series.
%t Table[Binomial[2n,n]^4,{n,0,20}]
%t 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 *)
%o (Maxima)
%o makelist(binomial(2*n,n)^4,n,0,40);
%Y Cf. binomial(2n,n)^k: A000984 (k=1), A002894 (k=2), A002897 (k=3), this sequence (k=4).
%Y Cf. A000108, A000888, A186414, A186415, A186416, A186418, A186419, A000897, A006480, A008977, A188662.
%K easy,nonn
%O 0,2
%A _Emanuele Munarini_, Feb 21 2011
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