If f(k) is the least number such that none of [f(k), f(k)+1,...f(k)+k-1] are k power free, how fast does f(k) grow?

Question

By a k-powerfree number, I mean a positive integer that is not divisible by the kth power of any prime number. For example, 32 is 6 powerfree but not 5 powerfree. The interval of integers [f(k),...f(k)+k-1] might be called a "k gap" in the k powerfree numbers, as I recall David Bernier did with gaps in the squarefree numbers. f(1)=2, f(2)=8. I have found that none of 1375, 1376, 1377 are cubefree, so f(3) is probably equal to 1375, because I couldn't find a smaller number than 1375 with the same property, and furthermore, a mathematician named Peter has stated here on stack exchange that 1375 was indeed the smallest such number, along with providing examples of larger gaps in the cubefrees and squarefrees. His larger examples make me think he has put in the time to do a computer search, so I'm reasonably sure he's correct. Anyway, I'm wondering if f(k) increases extremely fast. Thanks.

Answer 1

I have an estimate, based on some logic but no proof.
The $k$-powers would be coprime. I expect the roots are most often all prime.
Suppose the powers are $p_1^k,p_2^k,...,p_k^k$. By the Chinese Remainder Theorem, there are $k!$ different values $n\in[1,\prod_i p_i^k]$ for which $n+1,...,n+k$ have these $k$-powers, in some order.
The average number of successes in $[1,N]$ for these particular primes is then $\frac{Nk!}{\prod p_i^k}$.
For all primes, it is $$Nk!\sum_{p_1..p_k\text{ prime}}\prod p_i^{-k}$$ The sum would be dominated by the special case where the primes are the first $k$ primes.
I would expect your $f(k)$ to be $N$ that makes the expected number of successes near $1$; so somewhere around the following, where in this case the $p_i$ are the first $k$ primes: $$f(k)=O(\prod_i p_i^k/k!)$$ The first few numbers would be $$\frac{6^2}{2!}=18,\frac{30^3}{3!}=4500,\frac{210^4}{4!}=81033750$$ which you can compare with $8,1375$ and ??