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