Finding a Carmichael number $n$ such that $\prod_{p|n} \frac{p} {p-1} > 2$, where $p$ are primes, if it exists

Question

I am trying to find a Carmichel number $n$ with this property: $$\prod_{p|n} \frac{p} {p-1} > 2$$ where $p$ are primes.

I am unable to find it, the maximal value of the product, I have found, is $1.937\ldots$ for $n=62745=3\times5\times47\times89$.

Could someone to help me? Does it even exist?

Answer 1

We have $$\prod_{p|n} \frac{p} {p-1}=\frac{n}{\phi(n)},$$ and this is unbounded for $n\to\infty$, since ${\displaystyle \lim \inf {\frac {\varphi (n)}{n}}\log \log n=e^{-\gamma },}$ see here:

Is $n_{\rightarrow\infty}\overline{\frac{n}{\phi (n)}}$ unbounded?

So there should exist a (very large) Carmichael number $n$ with $\frac{n}{\phi(n)}>2$. Since Carmichael numbers are squarefree, and there are infinitely many of them, the number of prime factors should grow infinitely.

Edit: And it is indeed known that there are Carmichael numbers with arbitrarily many prime factors, see the link given below.

Edit: There is a page with Carmichael numbers with Lehmer index $\ge 2$, see here, which solves the problem! The first three Carmichael numbers $n$ with index $I'(n)=\frac{n}{\phi(n)}>2$ (a bit different from the Lehmer index) given there are \begin{align*} 3852971941960065 & = 3\cdot 5\cdot 23\cdot 89\cdot 113\cdot 1409\cdot 788129 \\ 655510549443465 & = 3\cdot 5\cdot 23\cdot 53\cdot 389\cdot 2663\cdot 34607 \\ 13462627333098945 & = 3\cdot 5\cdot 23\cdot 53\cdot 197\cdot 8009\cdot 466649 \end{align*} The first one has index $$ 2.0016266373563409120639854342870218296 $$ and the others even bigger index.

Answer 2

1886616373665 = 3 * 5 * 17 * 23 * 83 * 353 * 10979 is a smaller example than mentioned in the other answer; I found it by looking through A258801.