More ABC thoughts

Last time I talked about the abc conjecture. I would like to mention it a little more today. Just to recap here is the conjecture again,

let $\epsilon>0.$ The abc conjecture states that there are only finitely many abc triples such that
\[rad(abc)^{1+\epsilon}<c. \]
What has always bugged me is that $epsilon.$ I could never see why it was needed. If I had bothered going to wikipedia then I would of found out why much quicker than I did.

What I decided to do was to look for a family of abc triples such that

\[rad(abc)<c. \]

To do this I thought I should look for numbers with pretty small square free part, such as powers of 2. Let $n\in\mathbb{N}, a=2^{6n}-1, b=1$ and $c=2^{6n}.$ Note that $rad(2^{6n})=2.$ $2^{6n}-1$ is divisible by 9. Therefore $rad(2^{6n}-1)\leq \frac{2^{6n}-1}{3}.$ Then

\[rad(abc)=rad(2^{6n}(2^{6n}-1)\leq\frac{2}{3}(2^{6n}-1)<2^{6n}=c.\]

Therefore infinitely many abc triples exist when we take $\epsilon=0.$



This quest led me to think about $rad(2^{n}-1).$ I decided to search the OEIS for it. There is where i saw the "$6n$" pattern. But what I found interesting is the name the sequence goes under. It seemingly has nothing to do with $rad(2^{n}-1).$

http://oeis.org/A090633

I might one day get round to writing up the connection between the two.