How balanced can abc triples be?

Question

I was looking at the $241$ known "good" abc triples (i.e. with quality $\geqslant1.4$), wondering how frequently $a$ and $b$ would have more or less the same order of magnitude. The outcome is not very surprising. With rank numbers (rk) corresponding to the quality ranking, there are only 15 such triples with $1<b/a<10$, which are the following:

rk  quality size    merit   b/a
95  1.4316  13.28   12.18   1.1951826
240 1.4003  16.79   14.68   1.5557841
66  1.4420  15.51   15.53   1.5715695
105 1.4290  10.44   8.74    1.6514252
151 1.4158  23.92   24.63   1.6673082
43  1.4526  9.43    8.28    2.8166179
173 1.4121  29.38   31.48   3.3264647
206 1.4061  8.95    6.49    3.6854690
225 1.4022  10.67   8.12    4.6701482
160 1.4145  8.81    6.56    4.8162473
226 1.4020  13.49   11.09   7.4162550
108 1.4284  11.77   10.25   7.7411486
9   1.5270  9.78    11.02   8.7781887
199 1.4071  16.28   14.49   9.3202338
72  1.4403  16.98   17.38   9.4437408

Let's define the balance of an abc triple as the proportion $a/b$. (The convention being of course always $a<b$.) As this criterion has nothing to do with prime factors, I am not sure that looking at balances would give any new insights into the abc conjecture. But nevertheless, I would be curious about their distribution, and of course there might be unexpected patterns.
Generally speaking, they tend to be very small (i.e. $a\ll b$), so:

Conjecture: For $0<\varepsilon<1$, there are only finitely many abc triples with balance $a/b>\varepsilon$.

I would not expect this to be weaker (in the sense of implied by) or stronger than the abc conjecture itself. But maybe those among the millions of known abc triples with a "not too bad" balance, say $>.8$, could exhibit some patterns?
Note that this conjecture may be completely wrong. There might be plenty of abc triples of quality just above $1$ with a high balance, though intuitively I would doubt that. At this point, I'm wondering if somebody with access to the corresponding computer power could look (or has already looked) for the most balanced abc triples and/or provide some statistics about the general distribution of balances. But I am aware that this is not supposed to be the goal of a question to be asked here, so I have a slightly related and more feasible question:

Is there a known infinite sequence of abc triples $(a_n,b_n,c_n)_{n\in\mathbb N}$ such that $a_{n+1}>a_n$?

I am only aware of such sequences with $a_n\equiv1$, i.e. the worst possible balance at all.

Answer 1

I assume by abc triple you mean good abc triple.

It is known that a single good abc triple gives rise to infinite sequence of good abc triples.

Let $a,b,c=a+b$ be good abc triple.

Then $A=4ab,B=(b-a)^2,C=(A+B)=(a+b)^2=c^2$ is good abc triple too, and it is twice bigger than the original, but in general of lower quality. The radical of $AB(A+B)$ is at most $(b-a)$ times the radical of $ab(a+b)$.

This construction is used in Bart de Smit high merit triples: https://www.math.leidenuniv.nl/~desmit/abc/?set=3 check for common factors of the high merit triples.

Maybe we need to swap A,B and clear a common factor of four.

I suspect the conjecture is false.

Treating $A,B$ as polynomials in $a,b$, $\deg(A(a,b)) = \deg (B(a,b))$ which shows they are not necessarily very unbalanced as in your construction with $a=1$. Likely a single balanced solution will give infinite many balanced ones.