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A117896
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Number of perfect powers between consecutive squares n^2 and (n+1)^2.
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4
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0, 1, 0, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0
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OFFSET
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1,5
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COMMENTS
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a(n)=2 only 14 times for n^2 < 2^63. What is the least n such that a(n)=3? Is a(n) bounded?
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LINKS
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FORMULA
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Trivially, a(n) << log n/log log n. Turk gives a(n) << sqrt(log n) and Loxton improves this to a(n) <= exp(40 sqrt(log log n log log log n)). Stewart improves the constant from 40 to 30 and conjectures that a(n) < 3 for all but finitely many n. - Charles R Greathouse IV, Dec 11 2012
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EXAMPLE
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a(5)=2 because powers 27 and 32 are between 25 and 36.
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MATHEMATICA
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nn=151^2; powers=Join[{1}, Union[Flatten[Table[n^i, {i, Prime[Range[PrimePi[Log[2, nn]]]]}, {n, 2, nn^(1/i)}]]]]; t=Table[0, {Sqrt[nn]-1}]; Do[n=Floor[Sqrt[i]]; If[i>n^2, t[[n]]++], {i, powers}]; t (* revised, T. D. Noe, Apr 19 2011 *)
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PROG
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(PARI) a(n)=my(k); -sum(e=3, 2*log(n+1)\log(2), k=round((n+1/2)^(2/e))^e; if(n^2<k&&k<(n+1)^2, moebius(e))) \\ Charles R Greathouse IV, Dec 19 2011
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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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