# Prime Power Gaps

In 2000, Baker, Harman and Pintz proved that there is always a prime in the interval $(n-n^{0.525}, n)$. There are also conditional results implying smaller intervals. Nevertheless, I could not find any information about prime power gaps. So, what I'm asking is:

What is the asymptotically largest function $f(n)$ s.t. there is always a prime power in the interval $(f(n), n)$?

For example, Bertrand’s postulate is almost trivial in this case, since there is always a power of $2$ in the interval $(n, 2n]$. On the other hand, the distribution of prime powers with exponent $e>1$ is much smaller than the distribution of primes, so adding them might not change the answer.

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I expect this is the same as for the primes, between $n-\log^{2+\varepsilon} x$ and $n-\log^2 x$. –  Charles Oct 29 '12 at 21:56
Just looking at powers of integers, usually most integers are near a square. I expect that the 0.525 exponent is not going to change much if you throw in prime powers. Gerhard "Not A Paid Professional Opinion" Paseman, 2012.10.29 –  Gerhard Paseman Oct 29 '12 at 22:00
@Charles, I think you meant $x$ and $n$ to be the same letter? Also, I think you are referring to conjectural results, while it appears OP refers to unconditional results. –  Gerry Myerson Oct 29 '12 at 23:12

The paper shows that for $x \gt x_0$ there are primes in the interval $[x-x^{0.525},x]$ where the value of $x_0$ could be found effectively "with enough effort." The result fails for $x=126$ but I can't immediately rule out that $x_0=127$ suffices. The paper seems to say that for large enough $n$ the number of primes in $[n,n+n^{0.525}]$ eventually exceeds $\frac{9n^{0.525}}{100\log{n}}.$
Call an integer power proper if it is at least a square and respectable if it is at least a cube, To expand on Gerhard's observation: If we looked for intervals which contain either a prime or a proper integer power then we can say that there is always a square in $(n-2\sqrt{n},n)$ so this eventually improves on $x-x^{.525}.$
There is always a respectable power in $(n-3n^{2/3},n)$ but that is pretty near best possible, so I would guess that if we said "a prime or a respectable power" then we would not be able to get any improvement over just "prime". Likewise, there seems no reason to think that "prime or prime power" is essentially better than "prime or prime square" nor that "prime or prime square" is essentially better than "prime."
Ami, it seems likely that for any $\epsilon \gt 0$, $[x-x^{\epsilon},x]$ contains a prime whne $x$ is large enough. At the moment $\epsilon=0.525$ is perhaps the best value with a proof. Can we get to $\epsilon=0.5$ by including powers? (forget prime). Imagine that there actually are infinitely many prime free intervals $[n-n^{0.5},n].$ Maybe each starts near $m^2-3$ for some $m$, but one would imagine that about half contain a square and half don't. The chance that any particular one contains a cube is under $\frac{1}{n^{1/6}}$ and under $\frac{1}{n^{3/10}}$ for a fifth power. (cont) –  Aaron Meyerowitz Nov 4 '12 at 4:20