login
A184636
a(n) = floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.
1
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 ?
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
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 18 2011
STATUS
approved