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 A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x. 1

%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(n-1)-6a(n-2)+4a(n-3)-a(n-4) for n>=6 and that a(n)=n^3 for n>=2.

%F Empirical g.f.: x*(x^4-4*x^3+7*x^2+2) / (x-1)^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

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