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A184632
Floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.
0
1, 4, 13, 32, 62, 108, 171, 256, 364, 500, 665, 864, 1098, 1372, 1687, 2048, 2456, 2916, 3429, 4000, 4630, 5324, 6083, 6912, 7812, 8788, 9841, 10976, 12194, 13500, 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
OFFSET
1,2
FORMULA
a(n)=floor(1/{(8+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=3a(n-1)-2a(n-2)-2a(n-3)+3a(n-4)-a(n-5) for n>=7.
MATHEMATICA
p[n_]:=FractionalPart[(n^4+8)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1}, LinearRecurrence[{3, -2, -2, 3, -1}, {4, 13, 32, 62, 108}, 59]] (* Ray Chandler, Aug 02 2015 *)
CROSSREFS
Sequence in context: A270976 A272510 A036487 * A212747 A011936 A037235
KEYWORD
nonn
AUTHOR
Clark Kimberling, Jan 18 2011
STATUS
approved