A184630 Floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.

1, 6, 18, 43, 83, 144, 228, 341, 486, 666, 887, 1152, 1464, 1829, 2250, 2730, 3275, 3888, 4572, 5333, 6174, 7098, 8111, 9216, 10416, 11717, 13122, 14634, 16259, 18000, 19860, 21845, 23958, 26202, 28583, 31104, 33768, 36581, 39546, 42666, 45947, 49392, 53004, 56789, 60750, 64890, 69215, 73728, 78432, 83333, 88434

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, 2, -3, 3, -1).

Crossrefs

Cf. A184536.

Programs

Mathematica

p[n_]:=FractionalPart[(n^4+6)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,6,18,43},LinearRecurrence[{3,-3,2,-3,3,-1},{83,144,228,341,486,666},47]] (* Ray Chandler, Aug 02 2015 *)

Formula

a(n)=floor(1/{(6+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>=11, which implies a(n) = (2*n^3-1+A049347(n))/3 for n>=5.