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

1,1

Comments

Is a(n) = A066023(n) for n>=2?

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).

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

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 *)

Crossrefs

Cf. A000578, A066023, A184536.

Sequence in context: A365806 A087241 A347595 * A092071 A289864 A290056

Adjacent sequences: A184625 A184626 A184627 * A184629 A184630 A184631

Keyword

nonn

Author

Clark Kimberling, Jan 18 2011

Status

approved