cp's OEIS Frontend

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

%I A184628 #29 Aug 28 2025 05:56:35
%S A184628 2,8,27,64,125,216,343,512,729,1000,1331,1728,2197,2744,3375,4096,
%T A184628 4913,5832,6859,8000,9261,10648,12167,13824,15625,17576,19683,21952,
%U A184628 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 A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.
%C A184628 Is a(n) = A066023(n) for n>=2? _R. J. Mathar_, Jan 28 2011
%H A184628 Ray Chandler, <a href="/A184628/b184628.txt">Table of n, a(n) for n = 1..10000</a>
%H A184628 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -6, 4, -1).
%F A184628 a(n) = floor(1/{(4+n^4)^(1/4)}), where {}=fractional part.
%F A184628 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 A184628 Empirical g.f.: x*(x^4-4*x^3+7*x^2+2) / (x-1)^4. - _Colin Barker_, Sep 06 2014
%t A184628 p[n_]:=FractionalPart[(n^4+4)^(1/4)];
%t A184628   q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,80}]
%t A184628   FindLinearRecurrence[Table[q[n],{n,1,1000}]]
%t A184628 Join[{2}, LinearRecurrence[{4, -6, 4, -1}, {8, 27, 64, 125}, 54]] (* _Ray Chandler_, Aug 01 2015 *)
%Y A184628 Cf. A000578, A066023, A184536.
%K A184628 nonn,changed
%O A184628 1,1
%A A184628 _Clark Kimberling_, Jan 18 2011