A184636 a(n) = floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.

3, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418, 4608, 4802, 5000, 5202, 5408, 5618, 5832, 6050, 6272, 6498, 6728, 6962, 7200, 7442, 7688, 7938, 8192, 8450, 8712, 8978, 9248, 9522, 9800

Offset

1,1

Comments

Is a(n) = A001105(n) for n>=2 ?

Links

Table of n, a(n) for n=1..70.

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

Formula

a(n)=floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3) for n>=5, and that a(n)=2*n^2 for n>=2.

Mathematica

p[n_]:=FractionalPart[(n^4+2*n)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{3}, LinearRecurrence[{3, -3, 1}, {8, 18, 32}, 69]] (* Ray Chandler, Aug 02 2015 *)

Crossrefs

Cf. A184536, A184635, A184637.

Sequence in context: A004210 A247022 A119881 * A075342 A083726 A319006

Adjacent sequences: A184633 A184634 A184635 * A184637 A184638 A184639

Keyword

nonn

Author

Clark Kimberling, Jan 18 2011

Status

approved