abc streams (sequences of creek stones) A sequence of natural numbers $\ (c_n: n=1\ 2\ \ldots)\ $ is called a sequence of creek stones $\ \Leftarrow:\Rightarrow\ \forall_{n=1\ 2\  \ldots}\,c_{n+1}\ge c_n^2\ $.
Given natural $\ a\ b,\ $ such that $\ \gcd(a\ b) = 1,\ $ define $\ S(a\ b)\ :=\ \frac{L(a\ b)-1}{L(a\ b)+1},\ $ where
$$\ L(a\ b) := \frac{\log(a+b)}{\log(rad(a\cdot b\cdot(a+b)))} $$
is the Browkin-Brzeziński flavor of the abc coefficient.
A sequence of natural triples $\ ((a_n\ b_n\ c_n):n=1\ 2\ \ldots)\ $ is called an abc stream $\ \Leftarrow:\Rightarrow\ \forall_{n=1\ 2\ \ldots}\ \gcd(a_n\ b_n) = 1\ $ and $\ L(a_n\ b_n) > 1\ $ and $\ c_n=a_n+b_n,\ $ and
$\ (c_n: n=1\ 2\ \ldots)\ $ is a sequence of creek stones. With each such abc stream we associate a sequence of coefficients $\ C_n\ $ such that
$$ L(a_{n+1}\ b_{n+1})\,\ =\,\ 1\ +\ C_n\cdot S(a_n\ b_n) $$
for every $\ n=1\ 2\ \ldots\ $.
QUESTION 1:   Does there exist an abc stream
$\ ((a_n\ b_n\ c_n):n=1\ 2\ \ldots)\ $ such that the associated
coefficients $\ C_n\ $ approach infinity,
$\ \lim_{n\rightarrow\infty} C_n =\infty\ $ ?
There are abc streams such that $\ C_n>1\ $ for every $\ n =1\ 2\ \ldots$.
QUESTION 2:   Does there exist an abc stream and a constant
$\ \Gamma > 1\ $
such that $\ \forall_{n=1\ 2\ \ldots}\ C_n\ge \Gamma\ $?  
What is the supremum of such constants, $\ \sup\ \Gamma\ $ ?
 

REMARK  Many mathematicians (overwhelming majority?) believe that the abc conjecture is true, i.e. that $\ \limsup L(a\ b) = 1.\ $ Then the question is how fast/slow this limit is approached. My Question is addressing this issue.

 A: Instead of an actual answer (which I may get soon, I hope), I'll provide a minimal foundation and justification for the question. I'll do it by under-answering Question 2 by allowing $\ \Gamma=1\ $ (instead of $\ \Gamma > 1$). My text below is within the well-known scope but--for more or less--details.
Let $\ (x\ y\ z)\ $ be a reduced Pythagorean triple, i.e. $\ x\ y\ z\ $ are positive integers, $\ \gcd(x\ y) = 1\ $ and
$$ x^2 + y^2 = z^2 $$
We may also assume that $\ y\ $ is even (indeed, $\,\ x-y\equiv 1 \mod 2\ $ anyway). Thus we may start with
$$ a:=x^2\quad\qquad b:=y^2\quad\qquad c:=z^2 $$
so that $ a + b = c,\ $ and
$$ L(a\ b)\ =\ \frac{2\cdot \log(z)}{\log(rad(x\cdot y\cdot z))} $$ 
There are plenty of examples (as above) for which $\ L(a\ b) > 1,\ $ for instance $\ L(7^2\,\ 24^2) > 1.\ $  In a moment we will get infinitely many of such examples (see Theorem 2 below).
Now, let
$$ A := (x^2-y^2)^{\,2}\quad\qquad B:=(2\!\cdot\! x\!\cdot y)^2\quad\qquad 
          C := z^4 $$
hence
$$ A+B\ =\ C $$
represents a respective reduced Pythagorean triple $\ (X\ Y\ Z).\ $
Observe that
$$\ rad(B) = rad(2\cdot x\cdot y) = rad(x\cdot y) $$
it follows that:
$$ rad(A\!\cdot\! B\!\cdot\! C)\ =\ rad(A)\cdot rad(B)\cdot rad(C)\  
     \le\ A\cdot rad(x\!\cdot\! y)\cdot rad(z)\ =\ 
          A\cdot rad(x\!\cdot\! y\!\cdot\! z) $$
hence
$$     rad(A\!\cdot\! B\!\cdot\! C)\,\ <\,\ 
             z^2\cdot rad(x\!\cdot\! y\!\cdot\! z) $$
and
$$ L(A\ B)\ =\ \frac{\log(C)}{rad(A\!\cdot\! B\!\cdot\! C)}\ >\  
 \frac{4\cdot\log(z)}{2\!\cdot\!\log(z)+\log(rad(x\!\cdot\! y\!\cdot\! z))} $$
Thus,
$$ L(A\ B)-1\ >\ \frac{2\cdot \log(z) - \log(rad(x\!\cdot\! y\!\cdot\! z))}
    {2\cdot \log(z) + \log(rad(x\!\cdot\! y\!\cdot\! z))}\ 
      =\ \frac{L(a\ b)-1}{L(a\ b)+1}\ =\ S(a\ b) $$
THEOREM 1   $L(A\ B)\ > 1 + S(a\ b).\ $ Furthermore, if
$\ L(a\ b) \ge 1\ $ then $\ L(A\ B) > 1$.
Let's iterate the above step:
$$ (a\ b) \mapsto\ (A\ B)\ \mapsto\ (A'\ B')\ \mapsto\ (A''\ B'')\ \mapsto\ \ldots $$
We obtain
THEOREM 2   Every reduced Pythagorean triple $\ (x\ y\ z)\ $ leads to an abc stream $\ (A\ B)\ \mapsto\ (A'\ B')\ \mapsto\ (A''\ B'')\ \mapsto\ \ldots $ with $\ \Gamma=1\ $$ (see Q2 of the Question) appearing in a sharp way:
$$\forall_{n=1\ 2\ \ldots}\ \ \
       L(A^{(n+1)}\ B^{(n+1)})\ \ >\ \ 1\ +\ S(A^{(n)}\ B^{(n)}) $$
