A184632 Floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.

1, 4, 13, 32, 62, 108, 171, 256, 364, 500, 665, 864, 1098, 1372, 1687, 2048, 2456, 2916, 3429, 4000, 4630, 5324, 6083, 6912, 7812, 8788, 9841, 10976, 12194, 13500, 14895, 16384, 17968, 19652, 21437, 23328, 25326, 27436, 29659, 32000, 34460, 37044, 39753, 42592, 45562, 48668, 51911, 55296, 58824, 62500, 66325, 70304, 74438, 78732, 83187, 87808, 92596, 97556, 102689, 108000

Offset: 1

Author

Clark Kimberling, Jan 18 2011

Keywords

• nonn

Links

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

Crossrefs

Cf. A184536, A036487.

Programs

Mathematica

p[n_]:=FractionalPart[(n^4+8)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1},LinearRecurrence[{3,-2,-2,3,-1},{4,13,32,62,108},59]] (* Ray Chandler, Aug 02 2015 *)

Formula

a(n)=floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-2a(n-2)-2a(n-3)+3a(n-4)-a(n-5) for n>=7.