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

1,2

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).

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.

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 *)

Crossrefs

Cf. A184536.

Sequence in context: A272250 A272700 A191101 * A009957 A344992 A011929

Adjacent sequences: A184627 A184628 A184629 * A184631 A184632 A184633

Keyword

nonn

Author

Clark Kimberling, Jan 18 2011

Status

approved