A184629 Floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

1, 7, 22, 51, 100, 173, 274, 409, 583, 800, 1064, 1382, 1757, 2195, 2700, 3276, 3930, 4665, 5487, 6400, 7408, 8518, 9733, 11059, 12500, 14060, 15746, 17561, 19511, 21600, 23832, 26214, 28749, 31443, 34300, 37324, 40522, 43897, 47455, 51200, 55136, 59270, 63605, 68147, 72900, 77868, 83058, 88473, 94119, 100000

Offset: 1

Author

Clark Kimberling, Jan 18 2011

Keywords

• nonn

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (3, -3, 1, 0, 1, -3, 3, -1).

Crossrefs

Cf. A184536.

Programs

Mathematica

p[n_]:=FractionalPart[(n^4+5)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 7, 22, 51, 100, 173}, LinearRecurrence[{3, -3, 1, 0, 1, -3, 3, -1}, {274, 409, 583, 800, 1064, 1382, 1757, 2195}, 44]] (* Ray Chandler, Aug 01 2015 *)

Formula

a(n)=floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-5)-3a(n-6)+3a(n-7)-a(n-8) for n>=15.