

A117896


Number of perfect powers between consecutive squares n^2 and (n+1)^2.


4



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

1,5


COMMENTS

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?


REFERENCES

J. Turk, Multiplicative properties of integers in short intervals, Proc. Kon. Ned. Akad. Wet. (A) 83 (1980), pp. 429436.


LINKS

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
J. H. Loxton, Some problems involving powers of integers, Acta Arith., 46 (1986), pp. 113123.
C. L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta Arith. 133 (2008), pp. 97108.


FORMULA

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


EXAMPLE

a(5)=2 because powers 27 and 32 are between 25 and 36.


MATHEMATICA

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 *)


PROG

(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


CROSSREFS

Cf. A001597 (perfect powers), A014085 (primes between squares), A097055, A097056, A117934.
Sequence in context: A045837 A126825 A045833 * A275962 A132976 A143840
Adjacent sequences: A117893 A117894 A117895 * A117897 A117898 A117899


KEYWORD

nonn


AUTHOR

T. D. Noe, Mar 31 2006, Feb 15 2010


STATUS

approved



