A117896 Number of perfect powers between consecutive squares n^2 and (n+1)^2.

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

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?

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. 113-123.

C. L. Stewart, On heights of multiplicatively dependent algebraic numbers, Acta Arith. 133 (2008), pp. 97-108.

J. Turk, Multiplicative properties of integers in short intervals, Proc. Kon. Ned. Akad. Wet. (A) 83 (1980), pp. 429-436.

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: A319687 A126825 A045833 * A356735 A275962 A132976

Adjacent sequences: A117893 A117894 A117895 * A117897 A117898 A117899

Keyword

nonn

Author

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

Status

approved