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Define $\text{rad}_{23}(2^m3^nr)=2^{\text{sign}(m)}3^{\text{sign}(n)}r$, where $m,n\ge0$ and $2,3\nmid r\in\mathbb{N}$.

For a triple $a+b=c$ define the quality $q_{23}(a,b,c)=\frac{\log(c)}{\log(\text{rad}_{23}(abc))}$.

Has anyone attempted to specifically prove that only finitely many primitive triples have quality $q_{23}>1+\epsilon$ for any given $\epsilon>0$? Is there reason to hope this may be within reach of well established methods?

Based on this table of $abc$-triples, only 2 mini-$abc$-triples: are known with quality $\ge1.4$:

$37 + 2^{15} = 3^8 \times 5,\ \ \ \ \ \ \ q_{23}=1.48291$

$5 + 3^{11} = 2^{10} \times 173,\ \ \ \ \ q_{23}=1.41268$

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    $\begingroup$ We are dealing with triples of the form $p+q2^m=r3^n$ or similar, with $p,q,r$ "small", which in particular gives $q2^m\approx r3^n$. Using Baker's theorem it should be possible to give some nontrivial bounds on the quality. $\endgroup$ – Wojowu May 5 at 13:13
  • $\begingroup$ what is $\mathrm{sign}(m)$? $\endgroup$ – Fedor Petrov May 5 at 13:26
  • $\begingroup$ @FedorPetrov: sign(0)=0, sign(a)=1 if a>0. $\endgroup$ – Yaakov Baruch May 5 at 13:27
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    $\begingroup$ It's worth noting that the $2^{sign(m)}3^{sign(n)}$ is immaterial, since its log is obviously $o(\log c)$, so you might as well get rid of it. $\endgroup$ – Wojowu May 5 at 15:24

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