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 A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x. 1
 2, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507, 85184, 91125, 97336, 103823, 110592, 117649, 125000, 132651, 140608, 148877, 157464, 166375 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Is a(n) = A066023(n) for n>=2? LINKS Ray Chandler, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1). FORMULA a(n) = floor(1/{(4+n^4)^(1/4)}), where {}=fractional part. It appears that a(n)=4a(n-1)-6a(n-2)+4a(n-3)-a(n-4) for n>=6 and that a(n)=n^3 for n>=2. Empirical g.f.: x*(x^4-4*x^3+7*x^2+2) / (x-1)^4. - Colin Barker, Sep 06 2014 MATHEMATICA p[n_]:=FractionalPart[(n^4+4)^(1/4)];   q[n_]:=Floor[1/p[n]]; Table[q[n], {n, 1, 80}]   FindLinearRecurrence[Table[q[n], {n, 1, 1000}]] Join[{2}, LinearRecurrence[{4, -6, 4, -1}, {8, 27, 64, 125}, 54]] (* Ray Chandler, Aug 01 2015 *) CROSSREFS Cf. A000578, A066023, A184536. Sequence in context: A027267 A060414 A087241 * A092071 A289864 A290056 Adjacent sequences:  A184625 A184626 A184627 * A184629 A184630 A184631 KEYWORD nonn AUTHOR Clark Kimberling, Jan 18 2011 STATUS approved

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Last modified July 28 13:15 EDT 2021. Contains 346332 sequences. (Running on oeis4.)