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Floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.
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%I #13 Mar 21 2017 10:48:11

%S 1,4,13,32,62,108,171,256,364,500,665,864,1098,1372,1687,2048,2456,

%T 2916,3429,4000,4630,5324,6083,6912,7812,8788,9841,10976,12194,13500,

%U 14895,16384,17968,19652,21437,23328,25326,27436,29659,32000,34460,37044,39753,42592,45562,48668,51911,55296,58824,62500,66325,70304,74438,78732,83187,87808,92596,97556,102689,108000

%N Floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (3, -2, -2, 3, -1).

%F a(n)=floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.

%F It appears that a(n)=3a(n-1)-2a(n-2)-2a(n-3)+3a(n-4)-a(n-5) for n>=7.

%t p[n_]:=FractionalPart[(n^4+8)^(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[{1},LinearRecurrence[{3,-2,-2,3,-1},{4,13,32,62,108},59]] (* _Ray Chandler_, Aug 02 2015 *)

%Y Cf. A184536, A036487.

%K nonn

%O 1,2

%A _Clark Kimberling_, Jan 18 2011