What is the natural density of the set of natural numbers whose prime signature is the multi-set $\underbrace{\lbrace{k, k, \ldots, k \rbrace} }_m$ for $k,m\in \mathbb{N}$. That is, $n$ is in the set iff its prime factorisation $n = \prod_{i}{p_i}^{k_i}$ satisfies $k_i = k$ for all $i$. Alternatively, the set can be defined as the set of powers of squarefree numbers.

The square-free numbers are a subset of the set, so the natural density is lower-bounded by $6/\pi^2$. It is not immediately clear to me whether the natural density is larger than this, since the powers die out rather quickly. Are there in general arguments one can make about the relation of the natural density of a subset, and the natural density of all powers of the subset, besides the above trivial argument? It would seem not, since the natural density of a subset $A$ and a subset $B = \lbrace n + l\mid n\in A\rbrace$ for some fixed positive integer $l$ will have the same density. At the same time, the set of powers of elements of $B$ can be made to have arbitrarily small contribution to the natural density.


1 Answer 1


Let, for convenience, $A\subseteq\Bbb N\setminus\{0,1\}$. Call $A_k=\{x^k\,:\, x\in A\}$ and $A_*=\bigcup_{k=1}^\infty A_k$. Also call $a(n)=\lvert A\cap\{1,\cdots, n\}\rvert$ and $a_k(n)=\lvert A_k\cap\{1,\cdots, n\}\rvert$ with $k\in\Bbb N\cup\{*\}$. Now, notice that for $k\in\Bbb N\setminus\{0\}$, $a_k(n)=a(\lfloor n^{1/k}\rfloor)\le n^{1/k}$ and also that $a_k(n)=0$ whenever $2^k>n$. Therefore \begin{align}\frac{a(n)}{n}\le \frac{a_*(n)}{n}&\le \frac1n\sum_{k=1}^{\lfloor\log_2(n)\rfloor}a(\lfloor n^{1/k}\rfloor)\le\frac{a(n)}{n}+\sum_{k=2}^{\lfloor\log_2n\rfloor} n^{(1-k)/k}\le\\&\le \frac{a(n)}{n}+\sum_{k=2}^{\lfloor \log_2n\rfloor} n^{-1/2}\le\frac{a(n)}n+\frac{\log_2n}{\sqrt n}\end{align}

Since $n^{-1/2}\log_2n\to 0$ and the bound holds for all $n$, $A_*$ has the same upper and lower densities as $A$.

Added: The discussion above could have been stated in one sentence as the fact that the set $\{x^k\,:\, x\in\Bbb N\land k\ge2\}$ has density $0$ (with essentially the same proof).


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