Reference request for $(1,2^n-1,2^n)$ example related to abc-conjecture The $abc$-conjecture states that if $a,b,c$ are positive, relatively prime integers satisfying $a+b=c$, then the product of the primes dividing $abc$ (the radical of $abc$) is $\gg_\varepsilon c^{1-\varepsilon}$ for every $\varepsilon>0$.
We know several examples of $abc$-triples that refute the stronger assertion that the radical of $abc$ is $\gg c$. One example I have seen involves taking $a=1$, $b=2^n-1$, and $c=2^n$, so that the radical of $abc$ is twice the radical of $2^n-1$. By taking $n$ highly composite - say the least common multiple of the first $k$ integers - one forces $2^n-1$ to be divisible by lots of squares of primes (those not exceeding $k$, in this case), which implies that the radical of $2^n-1$ is $\ll 2^n/n$. That is, the radical of $abc$ is $\ll c/\log c$.
I'd like to cite this example in a paper I'm writing. Can someone tell me where to find it in the literature? I'd love the original citation, but even an accessible source that explicitly works out the upper bound $\ll c/\log c$ for the radical would suffice.
(Note that I don't need to be pointed to other examples of good $abc$-triples. I've put a couple of phrases above in boldface to emphasize the specific example I'm hoping to cite.)
 A: An argument along these lines for the triple $(1,2^n-1,2^n)$ with $n=p(p-1)$ and $p$ prime is given by Granville & Tucker, It’s as easy as abc (see first paragraph on page 1227). The radical of $abc$ in this case is bounded from above by $2b/p\simeq c/\sqrt{\log c}$, so less strong than your $c/\log c$ bound, but perhaps this does qualify as an "accessible source".
The source for Frits Beukers's proof of the $c/\log c$ bound for the triple $(1,3^{2^k}-1,3^{2^k})$, mentioned by @joro, is here.
A: I searched a bit harder, and think I found the proof for the specific triple $(1,2^k-1,2^k)$ that you are looking for, in a thesis from my own university: J.P. van der Horst, Finding ABC-triples using Elliptic Curves, page 10.
The abc triple $(1,2^{k_n}-1,2^{k_n})$ is considered, with $k_n$ chosen such that $2^{k_n}=1$ (mod $q^n$). The number $q$ can be any prime number $\neq 2$.
Van der Horst proves that the integer $k_n$ exists for any $n$ and that for $n\rightarrow \infty$ the radical of $abc$ is bounded from above by $e^\delta c/\log c$, with $\delta>0$ a constant that is independent of $n$.
A: The ABC example in Martin question was given by Lang: 
BULLETIN (New Series) OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 23, Number 1, July 1990
