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%I #13 Feb 17 2023 14:23:11
%S 3,8,18,32,50,72,98,128,162,200,242,288,338,392,450,512,578,648,722,
%T 800,882,968,1058,1152,1250,1352,1458,1568,1682,1800,1922,2048,2178,
%U 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
%N a(n) = floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.
%C Is a(n) = A001105(n) for n>=2 ?
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1).
%F a(n)=floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.
%F 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.
%t p[n_]:=FractionalPart[(n^4+2*n)^(1/4)];
%t q[n_]:=Floor[1/p[n]];
%t Table[q[n], {n, 1, 80}]
%t FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
%t Join[{3},LinearRecurrence[{3,-3,1},{8,18,32},69]] (* _Ray Chandler_, Aug 02 2015 *)
%Y Cf. A184536, A184635, A184637.
%K nonn
%O 1,1
%A _Clark Kimberling_, Jan 18 2011
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