First I will explain why a weaker form is needed. And then I formulate the conjecture (more precisely, the formulation will be clear).

It is related to the question https://math.stackexchange.com/questions/40945/triangular-factorials and several Mathoverflow questions from the comments to that question. A number $m$ is called a triangular factorial if $m=\frac{n(n+1)}{2}=k!$ for some $n,k$. It is an open problem whether the set of triangle factorials is finite. Moreover the only known such numbers are $1, 6, 120$.

But (somewhat surprisingly for me) it can be shown that the ABC conjecture implies that there are only finitely many triangular factorials. Indeed, suppose that for arbitrary large $k,m$ we have $ \frac{n(n+1)}{2}=k!$. Then $n+1=\frac {2k!}{n}$. Let $a=n, b=1, c= \frac {2k!}{n}$. Then by the ABC conjecture $\frac {2k!}{n}<rad(2k!)^2$ where $rad(x)$ is the product of primes dividing $x$. Note that $n\sim \sqrt{2k!}$ and $rad(2k!)=rad(k!)$ is the product of all primes $\le k$ which, by Erdos theorem $\sim e^{k}$. Thus we have $\sqrt{2k!}< e^{2k}$ which is impossible for big enough $k$. Recall that $2k!\sim 2\sqrt{2\pi k}\, e^{k\log k-k}$.

**Question:** In the proof above what seems to be a very weak version of the ABC conjecture is used (instead of $rad(abc)^{1+\epsilon}$ one can take a much bigger function in $rad(abc)$). Perhaps that version can be proved easier than the original ABC conjecture?

**Edit:** It is easy to see that in the version of ABC conjecture used here, $b=1$. Perhaps that makes the conjecture easier? So we can formulate

**A conjecture** For every constant $d<\frac 12$ there are only finite number of natural $a$ such that $$a>rad(a(a+1))^{d\log\log a}.$$
Note that the exponent in the right hand side may have to be a little different.

has square$\sim e^{2k}.$ $\endgroup$ – Aaron Meyerowitz Jun 15 '17 at 5:10