%I
%S 2,8,27,64,125,216,343,512,729,1000,1331,1728,2197,2744,3375,4096,
%T 4913,5832,6859,8000,9261,10648,12167,13824,15625,17576,19683,21952,
%U 24389,27000,29791,32768,35937,39304,42875,46656,50653,54872,59319,64000,68921,74088,79507,85184,91125,97336,103823,110592,117649,125000,132651,140608,148877,157464,166375
%N Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.
%C Is a(n) = A066023(n) for n>=2?
%H Ray Chandler, <a href="/A184628/b184628.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a> signature (4, 6, 4, 1).
%F a(n) = floor(1/{(4+n^4)^(1/4)}), where {}=fractional part.
%F It appears that a(n)=4a(n1)6a(n2)+4a(n3)a(n4) for n>=6 and that a(n)=n^3 for n>=2.
%F Empirical g.f.: x*(x^44*x^3+7*x^2+2) / (x1)^4.  _Colin Barker_, Sep 06 2014
%t p[n_]:=FractionalPart[(n^4+4)^(1/4)];
%t q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,80}]
%t FindLinearRecurrence[Table[q[n],{n,1,1000}]]
%t Join[{2}, LinearRecurrence[{4, 6, 4, 1}, {8, 27, 64, 125}, 54]] (* _Ray Chandler_, Aug 01 2015 *)
%Y Cf. A000578, A066023, A184536.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jan 18 2011
