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A184629
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Floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.
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1
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1, 7, 22, 51, 100, 173, 274, 409, 583, 800, 1064, 1382, 1757, 2195, 2700, 3276, 3930, 4665, 5487, 6400, 7408, 8518, 9733, 11059, 12500, 14060, 15746, 17561, 19511, 21600, 23832, 26214, 28749, 31443, 34300, 37324, 40522, 43897, 47455, 51200, 55136, 59270, 63605, 68147, 72900, 77868, 83058, 88473, 94119, 100000
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OFFSET
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1,2
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LINKS
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FORMULA
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a(n)=floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-5)-3a(n-6)+3a(n-7)-a(n-8) for n>=15.
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MATHEMATICA
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p[n_]:=FractionalPart[(n^4+5)^(1/4)]; q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 7, 22, 51, 100, 173}, LinearRecurrence[{3, -3, 1, 0, 1, -3, 3, -1}, {274, 409, 583, 800, 1064, 1382, 1757, 2195}, 44]] (* Ray Chandler, Aug 01 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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