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

Author

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

Keywords

• nonn

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?

Examples

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

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.

Crossrefs

Cf. A001597 (perfect powers), A014085 (primes between squares), A097055, A097056, A117934.

Programs

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 *)

PARI

a(n)=my(k);-sum(e=3,2*log(n+1)\log(2),k=round((n+1/2)^(2/e))^e;if(n^2Charles R Greathouse IV, Dec 19 2011

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