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).
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 $x$).