The $abc$ conjecture states that, for every positive real $\varepsilon$, there exist only finitely many triples $(a, b, c)$ of coprime positive integers such that $a + b = c$ and $$c > \operatorname{rad}(abc)^{1+\varepsilon }$$ where the operator "$\operatorname{rad}$" stands for the radical, defined by $$\operatorname{rad}({p_1}^{n_1} \ldots {p_k}^{n_k}) = p_1 \ldots p_k$$ for distinct primes $p_1, \ldots, p_k$.
The condition $\varepsilon > 0$ is mandatory because the set of "$abc$ triples" for which the above inequality holds with $\varepsilon=0$ is known to be infinite.
Now, define the function $$F({p_1}^{n_1} \ldots {p_k}^{n_k}) = n_1 \ldots n_k p_1 \ldots p_k$$ that takes into account the exponents, unlike the radical. Then, let us consider the triples $(a, b, c)$ of coprime positive integers such that $a + b = c$ and $$c > F(abc).$$
We get a subset of the $abc$ triples since the function $F$ is lower bounded by the radical. It is clearly not empty as it contains $(1, 239^2, 2 \cdot 13^4)$.
Question: Should we expect this subset to be infinite as well?
I am not aware of any reference to this variant of the $abc$ conjecture.
