How many $k$-full numbers are there?

Question

Recall that $n\in\mathbb{N}$ is $k$-full if, for a prime number $p$ : $p\mid n$ implies $p^{k}\mid n$. In the paper I am currently reading it is said that there are $O(B^{1/k})$ $k$-full positive integers in the dyadic interval $(B/2,B]$ for any $B\geq 1$. How to obtain this bound?

Obviously, any $k^{\text{th}}$ power is a $k$-full number, so there are at least $\approx B^{1/k}$ $k$-full numbers in $(0,B]$, and there are $O(B^{1/k})$ powers in $(B/2,B]$. Note that if number $n$ is $k$-full, then $n = x^{k}y,\;y\mid x$, so $B/2<x^{k}y\leq B$ and $y\leq x/2$, so $B<x^{k+1}$, so $x>B^{\frac{1}{k+1}}$ and clearly $x\leq B^{\frac{1}{k}}$. Thus, we are left with $B^{1/k} - B^{1/(k+1)}$ choices for $x$, which is still bad, because it is hard to find bound for choices of $y$. Trivial bound would be something like $O(B^{1/k+1})$, where $+1$ comes from the bound for choices of $y$. Maybe on average there are constant number of choices for $y$, in which case we get the desired $O(B^{1/k})$, but I don't know how to make the estimate. Any help is appreciated.

Answer 1

Claim: There are at most $O(B^{1/2})$ square-full integers in $[1,B]$.

Proof (due to answers to this post): Every square-full number $x$ can be written in the form $x=y^2z^3$, for some integers $y$ and $z$. For a fixed $z$, and $y$ varying, the number of square-full numbers in $[1,B]$ of the form $y^2z^3$ is at most $(B/z^3)^{1/2}$. Summing over all $z$, we get $$ \begin{align} \text{# square-full integers in $[1,B]$} &\le \sum_{z^{3}\le B}\left(\frac{B}{z^3}\right)^{1/2} \\&\le B^{1/2}\sum_{z=1}^\infty \frac1{z^{3/2}} \\&=\zeta(3/2)B^{1/2}\in O( B^{1/2} ). \end{align} $$

Claim: There are at most $O(B^{1/3})$ 3-full integers in $[1,B]$.

Proof: Every $3$-full number $x$ can be written in the form $$ x = y^3\cdot z^4\cdot w^5. $$ for some integers $y,z$ and $w$ (prove this). For a fixed $z$ and $w$, with $y$ varying, the number of 3-full numbers in $[1,B]$ of the form $y^3z^4w^5$ is at most $(B/(z^4w^5))^{1/3}$. Therefore, the number of 3-full integers in $[1,B]$ is at most $$ \begin{align} B^{1/3}\cdot \sum_{z=1}^\infty\sum_{w=1}^\infty \frac{1}{z^{4/3}w^{5/3}} &=B^{1/3} \left(\sum_{z=1}^\infty\frac1{z^{4/3}}\right) \left(\sum_{w=1}^\infty \frac1{w^{5/3}}\right) \\&=B^{1/3}\cdot \zeta(4/3)\cdot\zeta(5/3) \in O(B^{1/3}) \end{align} $$

I got the idea to represent $3$-full numbers as $y^3z^4w^5$ from https://oeis.org/A036966, the OEIS entry for 3-full numbers.

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If you have understood what is going on so far, you should be able to generalize this strategy in order to show that, for all $k\ge 1$, that there are at most $O(B^{1/k})$ integers in $[1,B]$ which are $k$-full.

For example, for $k=4$, you will be using the fact that every $4$-full integer can be written in the form $y^4\cdot z^5\cdot w^6\cdot u^7$ for some integers $y,z,w,u$.