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

2, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 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

Offset: 1

Author

Clark Kimberling, Jan 18 2011

Keywords

• nonn
• changed

Comments

Is a(n) = A066023(n) for n>=2? R. J. Mathar, Jan 28 2011

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Crossrefs

Cf. A000578, A066023, A184536.

Programs

Mathematica

p[n_]:=FractionalPart[(n^4+4)^(1/4)];
  q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,80}]
  FindLinearRecurrence[Table[q[n],{n,1,1000}]]
Join[{2}, LinearRecurrence[{4, -6, 4, -1}, {8, 27, 64, 125}, 54]] (* Ray Chandler, Aug 01 2015 *)

Formula

a(n) = floor(1/{(4+n^4)^(1/4)}), where {}=fractional part.

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.

Empirical g.f.: x*(x^4-4*x^3+7*x^2+2) / (x-1)^4. - Colin Barker, Sep 06 2014