A184633 Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

1, 4, 12, 28, 55, 96, 152, 227, 324, 444, 591, 768, 976, 1219, 1500, 1820, 2183, 2592, 3048, 3555, 4116, 4732, 5407, 6144, 6944, 7811, 8748, 9756, 10839, 12000, 13240, 14563, 15972, 17468, 19055, 20736, 22512, 24387, 26364, 28444, 30631, 32928, 35336, 37859, 40500, 43260, 46143, 49152, 52288, 55555, 58956, 62492, 66167, 69984, 73944, 78051, 82308, 86716, 91279, 96000, 100880, 105923, 111132, 116508, 122055, 127776, 133672, 139747, 146004, 152444, 159071, 165888, 172896, 180099, 187500

Offset

1,2

Links

Edward Jiang and Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Edward Jiang)

Index entries for linear recurrences with constant coefficients, signature (3, -3, 2, -3, 3, -1).

Formula

a(n) = floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

It appears that a(n) = 3a(n-1)-3a(n-2)+2a(n-3)-3a(n-4)+3a(n-5)-a(n-6) for n>=9.

Empirical g.f.: x*(x+1)*(x^6-3*x^5+3*x^4-x^3+3*x^2+1) / ((x-1)^4*(x^2+x+1)). - Colin Barker, Jun 13 2015

Mathematica

p[n_]:=FractionalPart[(n^4+9)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 4}, LinearRecurrence[{3, -3, 2, -3, 3, -1}, {12, 28, 55, 96, 152, 227}, 73]] (* Ray Chandler, Aug 02 2015 *)

Prog

(PARI) a(n)=my(t=(9+n^4)^(1/4)); 1\(t-t\1) \\ Charles R Greathouse IV, Sep 12 2014

Crossrefs

Cf. A184536.

Sequence in context: A109629 A112087 A166019 * A006000 A161216 A085622

Adjacent sequences: A184630 A184631 A184632 * A184634 A184635 A184636

Keyword

nonn

Author

Clark Kimberling, Jan 18 2011

Status

approved