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# On the distribution of powered numbers

Jörg Brüdern, Olivier Robert

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Jörg Brüdern, Olivier Robert. On the distribution of powered numbers. Open Mathematics Journal, 2024, 22 (1), 10.1515/math-2024-0007. hal-04954775

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Submitted on 18 Feb 2025

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# ON THE DISTRIBUTION OF POWERED NUMBERS

J. BRÜDERN AND O. ROBERT

ABSTRACT. Asymptotic formulae are established for the number of natural numbers $m$ with largest square-free divisor not exceeding $m^\vartheta$, for any fixed positive parameter $\vartheta$. Related counting functions are also considered.

## 1. INTRODUCTION

Motivated by questions about diophantine equations and the abc-conjecture, Mazur [1] proposed to smooth out the set of positive $l$-th powers in a multiplicative way, by what he named powered numbers. To introduce the latter, let $k(m)$ denote the largest square-free divisor of the natural number $m$. Mazur's powered numbers (relative to the real number $l > 0$) are the numbers $m \in \mathbb{N}$ with $k(m)^l \le m$. Note here that $l$ need not be integral in this definition, but for $l \in \mathbb{N}$ the powered numbers (relative to $l$) contain the $l$-th powers. It is proposed in [1] to replace, within a given diophantine equation, an $l$-th power by the corresponding powered numbers, and to consider the resulting equation between powered numbers as the associated “rounded” diophantine equation.

In this note we analyse the distribution of powered numbers. We find it more appropriate to work with the real number $\vartheta = 1/l$. The condition $k(m)^l \le m$ is then expressed equivalently as $k(m) \le m^\vartheta$, and for any $\vartheta > 0$ we define the set

$$ \mathcal{A}(\vartheta) = \{m \in \mathbb{N}: k(m) \le m^\vartheta\}. $$

Thus, Mazur's powered numbers (relative to $l$) are exactly the elements of $\mathcal{A}(1/l)$. For approaches to rounded diophantine problems it is relevant to determine the density of the set $\mathcal{A}(\vartheta)$. We therefore consider the number $S_\vartheta(x)$ of elements in $\mathcal{A}(\vartheta)$ that do not exceed $x$. It is not difficult to see that for any $0 < \vartheta < 1$ the number $S_\vartheta(x)$ obeys the inequalities

$$ (1.1) \qquad x^{\vartheta} \ll S_{\vartheta}(x) \ll x^{\vartheta+\varepsilon} $$

whenever $\varepsilon$ is a given positive real number and $x$ is large in terms of $\varepsilon$. It now transpires that for $\vartheta = 1/l$ the powered numbers are not much denser than the $l$-th powers. A weaker version of (1.1) occurs in Mazur [1] who refers to Granville, showing him an “easy” argument supposedly confirming the inequalities $x^{\vartheta-\varepsilon} \ll S_\vartheta(x) \ll x^{\vartheta+\varepsilon}$. In Mazur's article there is no indication how this would go but the simplest argument we know allows one to take $\varepsilon = 0$ in the lower bound. To substantiate this claim, fix a number $\vartheta \in (0, 1)$. Define $l \in \mathbb{N}$ by $1/l < \vartheta \le 1/(l-1)$, and $t \in \mathbb{R}$ by $l - t = 1/\vartheta$. Then $t \in (0, 1]$ and

$$ (1.2) \qquad 1/\vartheta = t(l-1) + (1-t)l. $$

Keywords: powered numbers, nuclear numbers

MSC(2020): 11N25, 11N37
Let $W(x)$ be the number of natural numbers $w$ that have a representation

$$ (1.3) \qquad w = 8^l n^{l-1} m^l $$

with $n, m$ square-free and constrained to the intervals

$$ (1.4) \qquad \frac{1}{2} \left(\frac{x}{8^l}\right)^{\vartheta t} < n \le \left(\frac{x}{8^l}\right)^{\vartheta t}, \quad \frac{1}{2} \left(\frac{x}{8^l}\right)^{\vartheta(1-t)} < m \le \left(\frac{x}{8^l}\right)^{\vartheta(1-t)}. $$

The numbers $w$ counted by $W(x)$ uniquely determine the square-free numbers $n, m$ with (1.3) and (1.4). This follows from unique factorisation. Hence $W(x)$ equals the number of pairs $n, m$ of square-free numbers satisfying (1.4). In particular, we see that $W(x) \gg x^{\vartheta}$.

Next, we observe that (1.3) gives $k(w) \le 2nm$, and therefore $k(w) \le 2(8^{-l}x)^{\vartheta}$. By (1.2) and (1.4), we also have $w \ge 2^{-(2l-1)}x$. It follows that $k(w) \le w^{\vartheta}$. Hence, numbers counted by $W(x)$ are also counted by $S_{\vartheta}(x)$. This yields $S_{\vartheta}(x) \ge W(x)$, and the left inequality in (1.1) follows at once.

For the upper bound in (1.1) one may refer to authorities like Tenenbaum [2, Theorem II.1.15]. However, the shortest argument available to us uses Rankin's trick within the following chain of obvious inequalities:

$$ S_{\vartheta}(x) \le \sum_{\substack{m \le x \\ k(m) \le x^{\vartheta}}} 1 \le \sum_{m=1}^{\infty} \left( \frac{x^{\vartheta}}{k(m)} \right) \left( \frac{x}{m} \right)^{\varepsilon} \ll x^{\vartheta+\varepsilon}. $$

In this argument, one needs to realise that the series $\sum k(m)^{-1}m^{-\varepsilon}$ converges, but this is immediately seen after converting this series to an Euler product.

For some of the finer questions on rounded diophantine equations, such as the asymptotic distribution of their solutions, the rough bounds displayed in (1.1) do not suffice. Instead one desires an asymptotic formula for $S_{\vartheta}(x)$, and our primary goal in this note is to supply such a formula. Let $1 \le y \le x$. Following Robert and Tenenbaum [3] we write $N(x, y)$ for the number of $m \in \mathbb{N}$ with $m \le x$ and $k(m) \le y$. The numbers counted here are their nuclear numbers. It seems natural to expect that the trivial upper bound $S_{\vartheta}(x) \le N(x, x^{\vartheta})$ should not be very wasteful. However, it was conjectured by Erdős<fnref n="1" /> (1962) and proved instantly by de Bruijn and van Lint [4] (1963) that

$$ (1.5) \qquad \sum_{m \le x} \frac{m}{k(m)} = o\left(\sum_{m \le x} \frac{x}{k(m)}\right) \quad (x \to \infty). $$

For a quantitative version of this estimate, see [3, Théorème 4.4]. These results suggest that the order of magnitude of $S_{\vartheta}(x)$ is somewhat smaller than that of $N(x, x^{\vartheta})$, and this is indeed the case.

Before we make this precise, we recall an estimate for $N(x, y)$. We are primarily interested in the case $y = x^{\vartheta}$ with $\vartheta > 0$, but we work in the wider range $y > \exp((\log x)^{2/3})$. The multiplicative function

$$ (1.6) \qquad \psi(m) = \prod_{p|m} (p+1) \quad (m \ge 1) $$

<footnote n="1">¹The authors of [4] acknowledge a letter by Erdős where the conjecture is made.</footnote>
and the function $F: [0, \infty) \to [0, \infty)$ defined by

$$F(v) = \frac{6}{\pi^2} \sum_{m \ge 1} \frac{1}{\psi(m)} \min \left( 1, \frac{e^v}{m} \right)$$

are featured in the uniform asymptotic formula

$$(1.7) \qquad N(x, y) = (1 + o(1)) yF(\log(x/y)) \quad (y > \exp((\log x)^{2/3}), x \to \infty)$$

that is contained in [3, Proposition 10.1]. As we shall see momentarily, for each pair $\vartheta, x$ with $0 < \vartheta < 1$ and $x \ge 2$, there is exactly one real number $\alpha = \alpha_{\vartheta}(x) > 0$ with

$$(1.8) \qquad \sum_p \frac{p^\alpha \log p}{(p^\alpha - 1)(1 + (p+1)(p^\alpha - 1))} = (1 - \vartheta) \log x.$$

We are now in a position to state our first result.

**Theorem 1.** Let $0 < \vartheta < 1$ fixed. Then, for $x \ge 9$, one has

$$(1.9) \qquad S_{\vartheta}(x) = x^{\vartheta} F((1-\vartheta) \log x) \frac{\alpha_{\vartheta}(x)}{\vartheta} \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} \right) \right).$$

For $x \ge 27$, one also has

$$(1.10) \qquad S_{\vartheta}(x) = \frac{\sqrt{2} x^{\vartheta} F((1-\vartheta) \log x)}{\vartheta \sqrt{(1-\vartheta)(\log x) \log \log x}} \left( 1 + O\left( \frac{\log \log \log x}{\log \log x} \right) \right).$$

This result calls for several comments. First we take $y = x^{\vartheta}$ in (1.7) and substitute the resulting equation

$$(1.11) \qquad N(x, x^{\vartheta}) = (1 + o(1)) x^{\vartheta} F((1 - \vartheta) \log x) \quad (x \to \infty)$$

within (1.10) to infer that

$$S_{\vartheta}(x) = (1 + o(1)) N(x, x^{\vartheta}) \frac{1}{\vartheta} \left( \frac{2}{1 - \vartheta} \right)^{1/2} \left( (\log x) \log \log x \right)^{-1/2} \quad (x \to \infty).$$

This is an analogue of (1.5), in quantitative form that is of strength comparable to [3, Théorème 4.4].

Our second comment concerns the implicitly defined function $\alpha_{\vartheta}(x)$. It originates in the Dirichlet series

$$(1.12) \qquad \mathcal{G}(s) = \frac{6}{\pi^2} \sum_{m \ge 1} \frac{1}{\psi(m)m^s} = \frac{6}{\pi^2} \prod_p \left( 1 + \frac{1}{(p+1)(p^s - 1)} \right)$$

that converges absolutely in $\text{Re } s > 0$, and therefore has no zeros in this half plane. For real numbers $\sigma > 0$ we have $\mathcal{G}(\sigma) > 0$. We may then define

$$g(\sigma) = \log \mathcal{G}(\sigma).$$

Note that $g$ extends to a holomorphic function on the right half plane, and one computes the logarithmic derivative of $\mathcal{G}(s)$ from the Euler product representation (1.12) to

$$g'(s) = - \sum_p \frac{p^s \log p}{(p^s - 1)(1 + (p+1)(p^s - 1))}.$$

Note that the sum on the right hand side here coincides with the sum in (1.8). On differentiating again, it transpires that the real function $g' : (0, \infty) \to \mathbb{R}$ is
increasing. Considering $\sigma \to 0$ and $\sigma \to \infty$ one finds that its range is the open interval $(-\infty, 0)$. We conclude that for a given $v > 0$ there is a unique positive number $\sigma_v$ with

$$g'(\sigma_v) + v = 0.$$

From [3, Lemme 6.6] we infer that for $v \ge 3$ one has

$$(1.13) \qquad \sigma_v = \sqrt{\frac{2}{v \log v}} \left( 1 + O\left( \frac{\log \log v}{\log v} \right) \right).$$

Here we choose $v = (1 - \vartheta) \log x$ and then have $\alpha_{\vartheta}(x) = \sigma_v$. In particular, we see that (1.9) and (1.13) imply (1.10). Thus, it only remains to prove (1.9).

Our last comment concerns the actual size of $S_{\vartheta}(x)$. This requires some more information on the function $F$. From [3, (2.12)] we have the asymptotic relation

$$(1.14) \qquad \log F(v) = (1 + o(1)) \sqrt{\frac{8v}{\log v}} \quad (v \to \infty).$$

By inserting (1.14) into (1.10), we deduce that there exists some positive number $\beta(x; \vartheta) > 0$ with

$$(1.15) \qquad S_{\vartheta}(x) = x^{\vartheta+\beta(x;\vartheta)} \quad (0 < \vartheta < 1, x \ge x_0(\vartheta)).$$

and the property that for any fixed $\vartheta \in (0, 1)$ one has

$$\beta(x; \vartheta) = (1 + o(1)) \sqrt{\frac{8(1 - \vartheta)}{(\log x)(\log \log x)}} \quad (x \to \infty).$$

Note that (1.15) yields another proof of (1.1).

We now turn to local estimates for $S_{\vartheta}(x)$. In our case, this amounts to comparing the respective behaviour of $S_{\vartheta}(zx)$ and $S_{\vartheta}(x)$ uniformly for large $x$, when $z$ is in some sense sufficiently close to 1. Such estimates are often obtained with the saddle-point method, and we follow this route here, too. In a suitable range for $z$, the fraction $S_{\vartheta}(zx)/S_{\vartheta}(x)$ may be approximated by a simple function of $z$.

**Theorem 2.** Let $0 < \vartheta < 1$. Then for $x$ large, we have

$$S_{\vartheta}(zx) = z^{\vartheta} S_{\vartheta}(x) \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} \right) \right)$$

uniformly for $z > 0$ with $|\log z| \ll \log \log x$.

Finally, we consider the counting function for a variation of the powered numbers. For given $\vartheta \in (0, 1)$ and $\Theta \in \mathbb{R}$, we consider

$$S_{\vartheta, \Theta}(x) = \#\{n \le x : k(n) \le n^{\vartheta} (\log n)^{\Theta}\}.$$

Note that $S_{\vartheta}(x) = S_{\vartheta,0}(x)$. The set of integers such that $k(n) \le n^{\vartheta}(\log n)^{\Theta}$ plays a prominent role in our companion paper [5], and therefore, we provide an estimate for $S_{\vartheta, \Theta}(x)$. It turns out that the conditions $k(n) \le n^{\vartheta}$ and $k(n) \le n^{\vartheta}(\log n)^{\Theta}$ are relatively close, and that the ratio $S_{\vartheta, \Theta}(x)/S_{\vartheta}(x)$ is roughly of size $(\log x)^{\Theta}$.

**Theorem 3.** Let $0 < \vartheta < 1$ and $\Theta \in \mathbb{R}$ be fixed. Then for $x$ large, one has

$$S_{\vartheta, \Theta}(x) = (\log x)^{\Theta} S_{\vartheta}(x) \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} \right) \right).$$
## 2. PROOF OF THEOREM 1

In this section we derive Theorem 1. Before we embark on the main argument, we fix some notation and recall a pivotal result concerned with the distribution of square-free numbers. This involves the function $\psi(m)$ as defined in (1.6), the Möbius function $\mu(m)$, and for a parameter $0 \le \gamma \le \frac{1}{2}$ at our disposal, the product

$$r_{\gamma}(m) = \prod_{p|m} (1 + 4\gamma p^{-1/2}).$$

One then has the estimate ([3, (10.1)])

$$(2.1) \qquad \sum_{l \le z} \mu^2(lk) = \frac{6kz}{\pi^2 \psi(k)} + O(r_{\gamma}(k)z^{1-\gamma})$$

that holds uniformly relative to the square-free number $k$ and the real parameters $z, \gamma$ in the ranges $z \ge 1, 0 \le \gamma \le \frac{1}{2}$.

The first steps of our argument follow the pattern laid out in [3, Sect. 10]. Unique factorisation shows that for all natural numbers $n$ there exists exactly one pair of coprime natural numbers $l, m$ with $\mu(l)^2 = 1$ and $n = lmk(m)$. Note that the two conditions $(l, m) = 1$ and $\mu(l)^2 = 1$ are equivalent to the single condition $\mu(lk(m))^2 = 1$. Further, one has $k(n) = lk(m)$. With $\vartheta \in (0, 1)$ now fixed, it follows that $S_{\vartheta}(x)$ equals the number of $(l, m) \in \mathbb{N}^2$ satisfying the conditions

$$\mu(lk(m))^2 = 1, \quad lmk(m) \le x, \quad lk(m) \le (lmk(m))^{\vartheta}.$$

These last three conditions we recast more compactly as

$$(2.2) \qquad \mu(lk(m))^2 = 1, \quad lk(m) \le \min(x/m, m^{\vartheta/(1-\vartheta)}).$$

From now on, the number $\kappa = \vartheta/(1 - \vartheta)$ features prominently, and we also put $y = x^{\vartheta}$. Note that

$$(2.3) \qquad \min(x/m, m^{\kappa}) = m^{\kappa} \quad \text{if and only if} \quad m \le x/y.$$

Hence, we consider the ranges $m \le x/y$ and $x/y < m \le x$ separately. By (2.2) and (2.3), this leads to the decomposition

$$(2.4) \qquad S_{\vartheta}(x) = S_{\vartheta}^{(1)}(x) + S_{\vartheta}^{(2)}(x)$$

in which

$$S_{\vartheta}^{(1)}(x) = \sum_{\substack{m \le x/y \\ k(m) \le m^{\kappa}}} \sum_{l \le m^{\kappa}/k(m)} \mu(lk(m))^2, \quad S_{\vartheta}^{(2)}(x) = \sum_{\substack{x/y < m \le x \\ mk(m) \le x}} \sum_{l \le x/mk(m)} \mu(lk(m))^2.$$

We apply (2.1) with $k = k(m)$ to both inner sums and obtain

$$(2.5) \qquad S_{\vartheta}^{(1)}(x) = \frac{6}{\pi^2} \sum_{\substack{m \le x/y \\ k(m) \le m^{\kappa}}} \frac{m^{\kappa}}{\psi(m)} + O(R_1),$$

$$(2.6) \qquad S_{\vartheta}^{(2)}(x) = \frac{6}{\pi^2} \sum_{\substack{x/y < m \le x \\ mk(m) \le x}} \frac{x}{m\psi(m)} + O(R_2).$$
where

$$R_1 = \sum_{\substack{m \le x/y \\ k(m) \le m^\kappa}} r_\gamma(m) \left(\frac{m^\kappa}{k(m)}\right)^{1-\gamma}, \quad R_2 = \sum_{\substack{x/y < m \le x \\ mk(m) \le x}} r_\gamma(m) \left(\frac{x}{mk(m)}\right)^{1-\gamma}.$$

It turns out that $R_1$ and $R_2$ are small. In order to couch their estimation, as well as the analysis of other error terms that arise later, under the umbrella of a single treatment, we choose parameters $\gamma$ and $\sigma$ with $0 < \gamma < \sigma \le \frac{1}{2}$ and introduce the series

$$(2.7) \qquad E = \sum_{m=1}^{\infty} r_{\gamma}(m) \left(\frac{y}{k(m)}\right)^{1-\gamma} \left(\frac{x/y}{m}\right)^{\sigma}.$$

If one turns the sum over $m$ into an Euler product, the factor at the prime $p$ becomes $1 + O_\sigma(p^{\gamma-1-\sigma})$, so the conditions $\sigma > \gamma > 0$ ensure convergence. It is routine to show that

$$R_1 + R_2 \ll E.$$

In fact, by Rankin's trick,

$$R_1 \le \sum_{\substack{m \le x/y \\ k(m) \le m^\kappa}} r_\gamma(m) \left(\frac{m^\kappa}{k(m)}\right)^{1-\gamma} \left(\frac{x/y}{m}\right)^{\sigma}.$$

For $m \le x/y$ one has $m^\kappa \le y$, and it follows that $R_1 \le E$.

Likewise, one confirms $R_2 \le E$ by observing that $1 - \gamma > \frac{1}{2} \ge \sigma$ so that

$$(2.8) \qquad \left(\frac{x}{mk(m)}\right)^{1-\gamma} = \left(\frac{y}{k(m)}\right)^{1-\gamma} \left(\frac{x/y}{m}\right)^{1-\gamma} \le \left(\frac{y}{k(m)}\right)^{1-\gamma} \left(\frac{x/y}{m}\right)^{\sigma}.$$

The appearance of $k(m)$ in the summation conditions on the right hand sides of (2.5) and (2.6) is a nuisance, and we proceed by removing these. If the condition $k(m) \le m^\kappa$ is removed from the sum in (2.5) one imports an error no larger than

$$\sum_{\substack{m \le x/y \\ k(m) > m^\kappa}} \frac{m^\kappa}{\psi(m)} \le \sum_{\substack{m \le x/y \\ k(m) > m^\kappa}} \frac{m^\kappa}{k(m)} \le \sum_{m \le x/y} \left(\frac{m^\kappa}{k(m)}\right)^{1-\gamma} \left(\frac{x/y}{m}\right)^{\sigma} \le E.$$

Here, the penultimate inequality is obtained by the argument that completed the estimation of $R_1$. Similarly, if the condition $mk(m) \le x$ is removed from the summation condition in (2.6), then the resulting error does not exceed

$$\sum_{\substack{x/y < m \le x \\ mk(m) > x}} \frac{x}{m\psi(m)} \le \sum_{\substack{x/y < m \le x \\ mk(m) > x}} \frac{x}{mk(m)} \le \sum_{\substack{m > x/y \\ mk(m) > x}} \frac{x}{mk(m)} = R,$$

say. By Rankin's trick and (2.8),

$$R \le \sum_{m > x/y} \left(\frac{x}{mk(m)}\right)^{1-\gamma} \le \sum_{m > x/y} \left(\frac{y}{k(m)}\right)^{1-\gamma} \left(\frac{x/y}{m}\right)^{\sigma} \le E.$$

Collecting together, we deduce from (2.5) and (2.6) the asymptotic relations

$$S_{\vartheta}^{(1)}(x) = \frac{6}{\pi^2} \sum_{m \le x/y} \frac{m^\kappa}{\psi(m)} + O(E), \qquad S_{\vartheta}^{(2)}(x) = \frac{6}{\pi^2} \sum_{x/y < m \le x} \frac{x}{m\psi(m)} + O(E),$$
and by (2.4), we infer that

$$S_{\vartheta}(x) = \frac{6}{\pi^2} \sum_{m \le x} \frac{m^{\kappa}}{\psi(m)} \min(1, x m^{-\kappa-1}) + O(E).$$

Note that the sum on the right is a partial sum of a convergent series. If one completes the sum, then it is immediate that the error thus imported is bounded by $R$, and hence by $E$. We have now reached the provisional expansion

$$(2.9) \qquad S_{\vartheta}(x) = \frac{6}{\pi^2} \sum_{m=1}^{\infty} \frac{m^{\kappa}}{\psi(m)} \min(1, x m^{-\kappa-1}) + O(E).$$

It remains to estimate $E$. In its definition (2.7), we encounter a sum over a multiplicative function, and so

$$\begin{aligned} E &= y^{1-\gamma} (x/y)^{\sigma} \prod_{p} \left(1 + \frac{p^{\gamma} r_{\gamma}(p)}{p(p^{\sigma} - 1)}\right) \\ &\le y^{1-\gamma} (x/y)^{\sigma} \prod_{p} \left(1 + \frac{p^{\gamma}}{p(p^{\sigma} - 1)}\right) \prod_{p} \left(1 + \frac{4\gamma p^{\gamma-\frac{1}{2}}}{p(p^{\sigma} - 1)}\right). \end{aligned}$$

Now, since $p^{\gamma-\frac{1}{2}} \le 1$, $\gamma < \sigma$, $p^{\sigma}-1 \ge \sigma \log p$, and since the sum $\sum_p \frac{1}{p \log p}$ converges,

$$\prod_{p} \left(1 + \frac{4\gamma p^{\gamma-\frac{1}{2}}}{p(p^{\sigma}-1)}\right) \le \prod_{p} \left(1 + \frac{4}{p \log p}\right) \ll 1.$$

As in the discussion following (1.13), we write $v = (1 - \vartheta) \log x = \log \frac{x}{y}$, and then choose $\sigma = \sigma_v$ and $\gamma = \sigma_v - \frac{1}{\log y}$. Then $\gamma < \sigma$, and for large $x$, we have $\sigma < \frac{1}{2}$. By (1.12) one has

$$\prod_{p} \left(1 + \frac{p^{\gamma}}{p(p^{\sigma} - 1)}\right) = \frac{\pi^2}{6} \mathcal{G}(\sigma) \prod_{p} \left(1 + \frac{(p^{\gamma} - 1)(p + 1) + 1}{p + p(p + 1)(p^{\sigma} - 1)}\right).$$

Now, on recalling (1.13),

$$\prod_{p \le e^{1/\sigma}} \left(1 + \frac{(p^{\gamma} - 1)(p + 1) + 1}{p + p(p + 1)(p^{\sigma} - 1)}\right) \le \prod_{p \le e^{1/\sigma}} \left(1 + \frac{1}{p}\right) \ll \frac{1}{\sigma} \ll (v \log v)^{1/2}$$

while one also has

$$\prod_{p > e^{1/\sigma}} \left(1 + \frac{(p^{\gamma} - 1)(p + 1) + 1}{p + p(p + 1)(p^{\sigma} - 1)}\right) \le \prod_{p > e^{1/\sigma}} \left(1 + \frac{1}{p^{1+\sigma-\gamma}}\right) \ll \zeta \left(1 + \frac{1}{\log y}\right) \ll \log y.$$

On collecting together, this shows that

$$E \ll y^{1-\sigma_v} e^{v\sigma_v} \mathcal{G}(\sigma_v) (v \log v)^{1/2} \log y.$$

From [3, (2.11)] we deduce that

$$F(v) \asymp \left(\frac{\log v}{v}\right)^{1/4} e^{v\sigma_v} \mathcal{G}(\sigma_v) \quad (v \ge 2),$$

and hence

$$E \ll y^{1-\sigma_v} F(v) v^{3/4} (\log v)^{1/4} \log y.$$

With the choice of $y$ and $v$, one has $\sigma_v = \alpha_{\vartheta}(x)$. Moreover, $\log y$ and $v$ have the order of magnitude $\log x$ so that the last inequality now reads

$$(2.10) \qquad E \ll x^{\vartheta} F\left((1-\vartheta) \log x\right) x^{-\vartheta \alpha_{\vartheta}(x)} (\log x)^{7/4} (\log \log x)^{1/4}.$$
Our final task is to compare our estimate for $E$ with the size of the sum on the right of (2.9).

Recall that in view of (1.1) and (1.11), $S_\vartheta(x)$ and $N(x, x^\vartheta)$ are of comparable size. In order to mimick the estimate (1.11), we introduce the function

$$H_{\vartheta}(x) = \frac{6}{\pi^2 x^{\vartheta}} \sum_{m=1}^{\infty} \frac{m^{\kappa}}{\psi(m)} \min(1, x m^{-\kappa-1})$$

so that (2.9) now reads

$$S_{\vartheta}(x) = x^{\vartheta} H_{\vartheta}(x) + O(E).$$

Our aim is to give an estimate of $H_\vartheta(x)$ by using the saddle-point method, and to describe more precisely $H_\vartheta(x)/F((1-\vartheta)\log x)$ as $x \to \infty$.

**Lemma 1.** Fix $\vartheta$ with $0 < \vartheta < 1$. Then for $x \ge 27$ one has

$$H_{\vartheta}(x) = F((1 - \vartheta) \log x) \frac{\alpha_{\vartheta}(x)}{\vartheta} \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} \right) \right).$$

*Proof.* The argument is modelled on [3, Section 8]. Recall the identity

$$\frac{1}{2\pi i} \int_{\sigma+i\mathbb{R}} \frac{y^s}{s(1-s)} ds = \min(1, y), \quad (0 < \sigma < 1, y > 0).$$

We then have the integral representation

$$\begin{align*}
H_{\vartheta}(x) &= \frac{1}{x^{\vartheta}} \frac{6}{\pi^2} \sum_{m \ge 1} \frac{m^{\kappa}}{\psi(m)} \frac{1}{2\pi i} \int_{\sigma+i\mathbb{R}} \frac{1}{s(1-s)} \left( \frac{x}{m^{1/(1-\vartheta)}} \right)^s ds \\
&= \frac{1}{2\pi i} \int_{\sigma+i\mathbb{R}} \mathcal{G} \left( \frac{s-\vartheta}{1-\vartheta} \right) x^{s-\vartheta} \frac{ds}{s(1-s)} \\
&= \frac{1}{2\pi} \int_{\mathbb{R}} \mathcal{G} \left( \frac{\sigma+it-\vartheta}{1-\vartheta} \right) x^{\sigma+it-\vartheta} \frac{dt}{(\sigma+it)(1-\sigma-it)}.
\end{align*}$$

After a linear change of variable in $t$, we arrive at

$$H_{\vartheta}(x) = \frac{1}{2\pi} \int_{\mathbb{R}} \mathcal{G} \left( \frac{\sigma - \vartheta}{1 - \vartheta} + it \right) x^{\sigma + i(1-\vartheta)t - \vartheta} \frac{(1-\vartheta)dt}{(\sigma + i(1-\vartheta)t)(1 - \sigma - i(1-\vartheta)t)}.$$

Recall again that $v = (1 - \vartheta) \log x$, and that $\alpha_\vartheta(x) = \sigma_v$. We take $\sigma = \vartheta + (1 - \vartheta)\sigma_v$. For large $x$ one then has $0 < \sigma < 1$, and the previous formula for $H_\vartheta(x)$ becomes

$$H_{\vartheta}(x) = \frac{1}{2\pi} \int_{\mathbb{R}} \frac{\mathcal{G}(\sigma_v + it)e^{(\sigma_v + it)v}}{(\vartheta + (1-\vartheta)(\sigma_v + it))(1-\sigma_v - it)} dt.$$

After truncation, we have

$$H_{\vartheta}(x) = \frac{1}{2\pi} \int_{-v^2}^{v^2} \frac{\mathcal{G}(\sigma_v + it)e^{(\sigma_v + it)v}}{(\vartheta + (1-\vartheta)(\sigma_v + it))(1-\sigma_v - it)} dt + O\left(\frac{\mathcal{G}(\sigma_v)e^{v\sigma_v}}{v^2}\right).$$

Moreover, following the proof of [3, Théorème 8.6], we set

$$\eta_v = (\log v) / \sqrt{g''(\sigma_v)} \asymp (\log v)^{3/4} / v^{3/4}$$

and recall [3, Lemme 8.5], asserting that for some $c > 0$ we have

$$|\mathcal{G}(\sigma_v + it)| \ll \mathcal{G}(\sigma_v) e^{-c(\log v)^2} \quad (\eta_v \le |t| \le \exp((\log v)^{38/37}).)$$
It now follows that

$$H_{\vartheta}(x) = \frac{1}{2\pi} \int_{-\eta_v}^{\eta_v} \frac{\mathcal{G}(\sigma_v + it)e^{(\sigma_v + it)v}}{(\vartheta + (1-\vartheta)(\sigma_v + it))(1-\sigma_v - it)} dt + O\left(\frac{\mathcal{G}(\sigma_v)e^{v\sigma_v}}{v^2}\right).$$

Setting

$$D_m = v^{(m+1)/2} (\log v)^{(m-1)/2} \quad (m \ge 0),$$

we have

$$H_{\vartheta}(x) = \frac{\mathcal{G}(\sigma_v)e^{v\sigma_v}}{2\pi} \int_{-\eta_v}^{\eta_v} \Upsilon(t)e^{-g''(\sigma_v)t^2/2}dt + O\left(\frac{\mathcal{G}(\sigma_v)e^{v\sigma_v}}{v^2}\right)$$

where

$$\Upsilon(t) = \frac{e^{-ig'''(\sigma_v)t^3/6 + O(t^4D_4)}}{(\vartheta + (1-\vartheta)(\sigma_v + it))(1-\sigma_v - it)}.$$

By Taylor expansion, we infer

$$\Upsilon(t) = \frac{Z(t)}{(\vartheta + (1 - \vartheta)\sigma_v)(1 - \sigma_v)}$$

where

$$Z(t) = 1 + it \left( \frac{1}{1 - \sigma_v} - \frac{1 - \vartheta}{\vartheta + (1 - \vartheta)\sigma_v} \right) - ig'''(\sigma_v) \frac{t^3}{6} + O(t^2 + D_4t^4 + D_3^2t^6).$$

Now, still following the pattern of the proof of [3, Théorème 8.6], one is lead to

$$H_{\vartheta}(x) = \frac{x^{(1-\vartheta)\alpha_{\vartheta}(x)} \mathcal{G}(\alpha_{\vartheta}(x))}{\vartheta \sqrt{2\pi g''(\alpha_{\vartheta}(x))}} \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} \right) \right).$$

We omit the details. From [3, Théorème 8.6] we import the relation

$$F(v) = \frac{e^{v\sigma_v} \mathcal{G}(\sigma_v)}{\sigma_v \sqrt{2\pi g''(\sigma_v)}} \left( 1 + O\left( \sqrt{\frac{\log v}{v}} \right) \right) \quad (v \ge 2),$$

and the lemma follows.

□

We may now complete the proof of Theorem 1. Using the lemma and (1.13), we obtain

$$H_{\vartheta}(x) \asymp F((1 - \vartheta) \log x) ((\log x) \log \log x)^{-1/2},$$

so that the estimate (2.10) implies

$$E \ll x^{\vartheta} H_{\vartheta}(x) x^{-\vartheta \alpha_{\vartheta}(x)} (\log x)^{9/4} (\log \log x)^{3/4}.$$

We then have

$$S_{\vartheta}(x) = x^{\vartheta} H_{\vartheta}(x) \left(1 + O\left(x^{-\vartheta \alpha_{\vartheta}(x)} (\log x)^{9/4} (\log \log x)^{3/4}\right)\right).$$

We may now replace $H_{\vartheta}(x)$ by the estimate from Lemma 1. Since

$$x^{-\vartheta \alpha_{\vartheta}(x)} (\log x)^{9/4} (\log \log x)^{3/4} \ll \sqrt{\frac{\log \log x}{\log x}},$$

this yields (1.9). As remarked earlier, (1.10) follows from (1.13) and (1.9). The proof of Theorem 1 is complete.
### 3. PROOF OF THEOREM 2

Subject to the hypotheses of Theorem 2, when $x$ is large, one has $\log zx \asymp \log x$. Hence, Theorem 1 implies that

$$ (3.1) \qquad S_{\vartheta}(zx) = (zx)^{\vartheta} F\left((1-\vartheta)\log zx\right) \frac{\alpha_{\vartheta}(zx)}{\vartheta} \left(1 + O\left(\sqrt{\frac{\log \log x}{\log x}}\right)\right) $$

holds uniformly for $|\log z| \ll \log \log x$. We recall [3, Proposition 8.7]. This asserts that uniformly for $|t| \le v^{3/4}(\log v)^{1/4}$, one has

$$ F(v+t) = e^{t\sigma_v} F(v) \left(1 + O\left(\frac{\log v + t^2/v}{\sqrt{v \log v}}\right)\right). $$

Using this estimate with $v = (1 - \vartheta) \log x$ and $t = (1 - \vartheta) \log z$, one finds

$$ F((1 - \vartheta) \log zx) = F((1 - \vartheta) \log x) z^{(1-\vartheta)\alpha_{\vartheta}(x)} \left(1 + O\left(\sqrt{\frac{\log \log x}{\log x}}\right)\right). $$

Moreover, in the ranges for $x$ and $z$ considered here, one has

$$ z^{(1-\vartheta)\alpha_{\vartheta}(x)} = 1 + O(\alpha_{\vartheta}(x) \log z), $$

which yields an admissible error term.

Finally, [3, (8.9)] implies that uniformly for $|t| \le v/2$ one has

$$ \sigma_{v+t} = \sigma_v \left(1 + O\left(\frac{|t|}{v}\right)\right). $$

Hence

$$ \alpha_{\vartheta}(zx) = \alpha_{\vartheta}(x) \left(1 + O\left(\frac{\log z}{\log x}\right)\right) $$

which again leads to an admissible error term. Inserting these estimates in (3.1) completes the proof of Theorem 2.

It may be worth pointing out that the above argument actually proves a little more. A close inspection of the proof of Theorem 2 shows that the estimate

$$ S_{\vartheta}(zx) = z^{\vartheta+(1-\vartheta)\alpha_0(x)} S_{\vartheta}(x) \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} + \frac{(\log z)^2}{(\log x)^{3/2}(\log \log x)^{1/2}} \right) \right) $$

holds uniformly in the range $z > 0, x > 27, |\log z| \le (\log x)^{3/4}(\log \log x)^{1/4}$.

### 4. PROOF OF THEOREM 3

Before proving Theorem 3, we briefly sketch the main steps. We first choose a suitable real number $U = U(x)$ such that $\log U = (\log x)(1 + o(1))$, and count the integers $n$ not exceeding $x$ such that $k(n) \le n^{\vartheta}(\log U)^{\Theta}$. The first step is to show that the number of these integers is essentially $S_{\vartheta}(x)$ multiplied by $(\log x)^{\Theta}$. The second step is to prove that the number of these integers is close to $S_{\vartheta, \Theta}(x)$.

In light of this description, for any $x \ge 1$ and any $z > 0$, we set

$$ B(x, z) = \#\{n \le x : k(n) \le zn^{\vartheta}\}. $$

**Theorem 4.** Let $0 < \vartheta < 1$ be fixed. Then for $x$ large, one has

$$ B(x, z) = z S_{\vartheta}(x) \left( 1 + O\left( \sqrt{\frac{\log \log x}{\log x}} \right) \right) $$
uniformly for $z > 0$ with $|\log z| \ll \log \log x$.

*Proof.* We follow very closely the proof of Theorem 1 which corresponds to the case $z = 1$. We redefine the meaning of $y$, now set to $y = x^\vartheta z$, and keep the notation $\kappa$ and $E$. Note that in the sequel of this proof, the error term $E$ is to be interpreted with the current specific choices for $x$ and $y$.

For any $n \ge 1$, recall that there is a unique pair $l, m$ with $n = lmk(m)$ and $\mu^2(lk(m)) = 1$. The conditions $n \le x$ and $k(n) \le z^\vartheta$ become

$$ lmk(n) \le x \quad \text{and} \quad lk(m) \le m^\kappa z^{1+\kappa}. $$

By keeping the same dichotomy $m \le x/y$ and $m > x/y$ it is straightforward that

$$ B(x, z) = \frac{6}{\pi^2} \sum_{m=1}^{\infty} \frac{1}{\psi(m)} \min(m^{\kappa} z^{1+\kappa}, x/m) + O(E). $$

Moreover, (2.9) with the parameter $xz^{-\kappa-1}$ reads

$$ S_{\vartheta}(xz^{-\kappa-1}) = \frac{6}{\pi^2} \sum_{m=1}^{\infty} \frac{1}{\psi(m)} \min(m^{\kappa}, xz^{-\kappa-1}/m) + O(z^{-(1+\kappa)(1-\gamma)}E). $$

Hence, one has

$$ B(x, z) = z^{1+\kappa} S_{\vartheta}(xz^{-\kappa-1}) + O(z^{(1+\kappa)\gamma}E). $$

Still choosing $\gamma = \sigma_v - \frac{1}{\log y}$ and $\sigma = \sigma_v$, we have $z^{(1+\kappa)\gamma} \ll 1$. Moreover, (2.10) implies that for some $c > 0$ one has

$$ E \ll x^{\vartheta} F((1-\vartheta) \log x) \exp \left( -c \sqrt{\frac{\log x}{\log \log x}} \right). $$

This is sufficient to ensure that

$$ E \ll z S_{\vartheta}(x) \sqrt{\frac{\log \log x}{\log x}}. $$

Finally we estimate the term $S_{\vartheta}(xz^{-\kappa-1})$ by Theorem 2. Inserting this in the estimate for $B(x, z)$ and noticing that in the main term the exponent $1+\kappa-(1+\kappa)\vartheta$ of $z$ is equal to 1 gives the expected result. $\square$

We may now complete the proof of Theorem 3. It is sufficient to prove the result for $\Theta \ne 0$, since in the case $\Theta = 0$ one has $S_{\vartheta,\Theta}(x) = S_{\vartheta}(x)$. We put

$$ C = 1 + (1 + |\Theta|)/\vartheta \quad \text{and} \quad U = U(x) = x(\log x)^{-C}. $$

Note that we have $C > 0$, $C\vartheta \ge 1$ and $\Theta + C\vartheta \ge 1$.

First consider the case $\Theta > 0$. Any integer $n$ counted by $S_{\vartheta,\Theta}(x)$ satisfies $k(n) \le n^\vartheta (\log x)^\Theta$, whence $S_{\vartheta,\Theta}(x) \le B(x, (\log x)^\Theta)$. Now, a lower bound is obtained by noticing that the set of integers $U < n \le x$ counted in $S_{\vartheta,\Theta}(x)$ contains the integers such that $k(n) \le n^\vartheta (\log U)^\Theta$. These deliberations yield the inequalities

$$ B(x, (\log U)^\Theta) - B(U, (\log U)^\Theta) \le S_{\vartheta,\Theta}(x) \le B(x, (\log x)^\Theta). $$

Now, using Theorem 4 to estimate $B(x, (\log U)^\Theta)$ and $B(x, (\log x)^\Theta)$, and then replacing $(\log U)^\Theta$ by $(\log x)^\Theta$ at the price of an admissible error term, one obtains the main term to estimate $S_{\vartheta,\Theta}(x)$. For the remaining term, Theorem 4 and the definition of $U$ imply that

$$ B(U, (\log U)^\Theta) \ll (\log U)^\Theta S_{\vartheta}(U) \ll (\log x)^\Theta S_{\vartheta}(U). $$
Finally, Theorem 2 with the choice $z = (\log x)^{-C}$ implies that

$$S_{\vartheta}(U) \ll z^{\vartheta} S_{\vartheta}(x) \ll (\log x)^{-\vartheta C} S_{\vartheta}(x) \ll (\log x)^{-1} S_{\vartheta}(x).$$

Gathering these estimates, we obtain an upper bound for $B(U, (\log U)^{\Theta})$ that provides an admissible error term, and therefore proves the result for $\Theta > 0$.

The case $\Theta < 0$ is very similar. Any integer counted by $S_{\vartheta,\Theta}(x)$ either satisfies $n \le U$ with $k(n) \le n^{\vartheta}$, or $U < n \le x$ with $k(n) \le n^{\vartheta}(\log U)^{\Theta}$. Moreover, any integer $n \le x$ such that $k(n) \le n^{\vartheta}(\log x)^{\Theta}$ is counted in $S_{\vartheta,\Theta}(x)$. We conclude that

$$B(x, (\log x)^{\Theta}) \le S_{\vartheta,\Theta}(x) \le B(x, (\log U)^{\Theta}) + S_{\vartheta}(U).$$

As before, one uses Theorem 4 to estimate $B(x, (\log U)^{\Theta})$ and $B(x, (\log x)^{\Theta})$. This provides the main term. Finally, Theorem 2 with $z = (\log x)^{-C}$ provides

$$S_{\vartheta}(U) \ll z^{\vartheta} S_{\vartheta}(x) \ll (\log x)^{-\vartheta C} S_{\vartheta}(x) \ll (\log x)^{\Theta-1} S_{\vartheta}(x),$$

which yields an admissible error term.

**Funding information.** First author supported by Deutsche Forschungsgemeinschaft project number 462335009. Second author supported by the joint FWF-ANR project Arithrand: FWF: I 4945-N and ANR-20-CE91-0006.

**Conflict of interest.** Authors state no conflict of interest.

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[1] B. Mazur. *Questions about powers of numbers*. Notices Amer. Math. Soc. **47** (2000), no. 2, 195–202.

[2] G. Tenenbaum. *Introduction to analytic and probabilistic number theory*, Graduate Studies in Mathematics, vol 163. American Mathematical Society, Providence, RI, 3rd edn., 2015.

[3] O. Robert and G. Tenenbaum. *Sur la répartition du noyau d'un entier*. Indag. Math. (N.S.) **24** (2013), no. 4, 802–914.

[4] N. G. de Bruijn, J. H. van Lint, *On the number of integers $\le x$ whose prime factors divide $n$*. Acta Arith. **8** (1962), 349–356.

[5] J. Brüdern and O. Robert, *On the sum of two powered numbers*. Acta Arith. (2024), doi: 0.4064/aa230609-26-9

Jörg Brüdern
Universität Göttingen
Mathematisches Institut
Bunsenstrasse 3–5
D 37073 Göttingen
Germany
jbruede@gwdg.de

Olivier Robert
Université Jean Monnet,
ICJ UMR5208, CNRS, Ecole Centrale de Lyon, INSA Lyon,
Université Claude Bernard Lyon 1,
42023 Saint-Etienne, France
olivier.robert@univ-st-etienne.fr