# Natural density of integers such that $\forall 1\leq k\leq n-1$ satisfy $\operatorname{rad}(k)+\operatorname{rad}(n-k)\geq \operatorname{rad}(n)$

For an integer $n>1$ we defined its square-free kernel $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p \text{ prime}}}p$$ as the product of distinct prime factors dividing it, with the definition $\operatorname{rad}(1)=1$. You can see the Wikpedia Radical of an integer to see the properties of such arithmetic function. Then I was inspired in Problem 1.43 of [1] to encourage myself to study next definition (it is a variation of the mentioned problem).

Definition. I call that an integer $n>1$ is $\operatorname{rad}$-superadditive if $$\operatorname{rad}(k)+\operatorname{rad}(n-k)\leq \operatorname{rad}(n)$$ $\forall k$ such that $1\leq k\leq n-1$, and $\operatorname{rad}$-subadditive when for all integer $k$ of the segment $1\leq k\leq n-1$ $$\operatorname{rad}(k)+\operatorname{rad}(n-k)\geq \operatorname{rad}(n)$$ holds.

Claim. A) It is easy to prove that there exist infinitely many $\operatorname{rad}$-superadditive numbers. B) It is easy to prove that there exist infinitely many $\operatorname{rad}$-subadditive numbers.

Sketch of proof. For A) consider the set of square-free integers, these are the integers $m\geq 1$ such that $\operatorname{rad}(m)=m$, and use the property $\operatorname{rad}(L)\leq L$ of the square-free kernel. For B) consider the powers of two. $\square$

I would like to propose the following problem.

Question. We denote the set of $\operatorname{rad}$-subadditive numbers as $A$. What work can be done about the calculation (if it exists) of the asymptotic density $d(A)$ (I mean the definition of $d(A)$ from this section of the Wikipedia's article dedicated to Natural density)? I am asking if we can deduce a statement about such limits $\underline{d}(A)$ and/or $\overline{d}(A)$. Thanks you in advance.

Upto $1000$, the $\operatorname{rad}$-subadditive numbers are

$$1,2,3,4,8,9,16,25,27,32, 49, 54, 64, 81, 108, 121, 125, 128, 162, 216, 243,$$ $$250, 256, 324, 343, 432, 486, 500, 512, 625, 648, 686, 729, 864, 972$$ and $1000.$ We see that some of these are prime powers.

## References:

[1] Valentin Boju and Louis Funar, The Math Problems, $\mathcal{Notebook }$ , Birkhäuser (2007).

• +1 You have a knack for finding great NT questions! – Isky Mathews Jun 1 '18 at 10:10
• Zero, I guess. Most numbers have their radical comparable to them and every sufficiently large number is a sum of a multiple of $3^{100}$ and a multiple of $2^{100}$. – fedja Jun 9 '18 at 3:07
• Feel free to add an answer @fedja My purpose with this kind of questions is learn, but your answer will be interesting for many people. Thus this is an invitation if you want add an answer with your details, heuristics and reasonings. Many thanks. – user243301 Jun 9 '18 at 8:20
• @user243301 Yes, learning is the main objective. However you can learn from hints too, not only from full expanded answers ;-). Anyway, I shared what I knew :-) – fedja Jun 9 '18 at 21:54

The density is certainly $0$ (that is easy no matter how you look at it) but the interesting thing is to estimate the order of magnitude of the number $S(x)$ of rad-subadditive numbers up to $x$.
The upper bound is not hard. Note that if $p>3$ is a prime that occurs in the factorization of $n=pm$ in the first power, then $n=4m+(p-4)m$ is a bad decomposition. Thus every prime except, perhaps, $2,3$, should be at least squared in the prime factorization of a rad-subadditive number $n$. In other words, we should have $\operatorname{rad}(n)\le 3\sqrt n$. The numbers up to $x$ with radical at most $3\sqrt x$ are rather few: for every $\delta>0$ their number can be estimated by $O(x^{\frac 12+\delta})$. The easiest way to see it is to consider for every squarefree $q=p_1p_2\dots p_k$ the sum $$\sum_{n:\operatorname{rad}(n)=q} n^{-s}=q^{-s}\prod_{j=1}^k(1-p_j^{-s})^{-1}$$ For fixed $s>0$, the RHS is $O(q^{\delta/2})$ because $k=o(\log q)$, so the total number of $n\le x$ with $\operatorname{rad}(n)=q$ is at most $x^sO(q^{\delta/2})$. Taking $s=\delta/2$, we get the claim.
The interesting part is the lower bound. So I wonder if anyone can get $S(x)=\Omega(x^{\frac 12-\delta})$ (or, at least, some positive power of $x$).
• Feel free to continue studying these sequences $\operatorname{rad}$-superadditive and $\operatorname{rad}$-subadditive, and your own questions about these: the order of magnitude of the number $S(x)$ or other questions that arise. – user243301 Jun 12 '18 at 7:54