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A184631
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Floor(1/{(7+n^4)^(1/4)}), where {}=fractional part.
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1
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1, 5, 15, 36, 71, 123, 196, 292, 416, 571, 760, 987, 1255, 1568, 1928, 2340, 2807, 3332, 3919, 4571, 5292, 6084, 6952, 7899, 8928, 10043, 11247, 12544, 13936, 15428, 17023, 18724, 20535, 22459, 24500, 26660, 28944, 31355, 33896, 36571, 39383, 42336, 45432, 48676, 52071, 55620, 59327, 63195, 67228, 71428, 75800
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OFFSET
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1,2
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (3, -3, 1, 0, 0, 0, 1, -3, 3, -1).
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FORMULA
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a(n)=floor(1/{(7+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-7)-3a(n-8)+3a(n-9)-a(n-10) for n>=13.
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MATHEMATICA
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p[n_]:=FractionalPart[(n^4+7)^(1/4)]; q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 5}, LinearRecurrence[{3, -3, 1, 0, 0, 0, 1, -3, 3, -1}, {15, 36, 71, 123, 196, 292, 416, 571, 760, 987}, 49]] (* Ray Chandler, Aug 02 2015 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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