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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?

Dietrich Burde
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zhrd
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  • I recommend looking at this paper of Alford, Grantham, Hayman, and Shallue: they construct Carmichael numbers with huge numbers of prime factors, and they might say something about the particular statistic you're looking at. – Greg Martin Oct 02 '22 at 17:08
  • Thank you for the link. To be honest, I tried already search among the Carmichael number with more number of prime factors. I made a program, which generates a next Carmichael number by adding a prime factor to the previous. For example C1=483285864348001=11×13×19×31×71×73×97×101×113, C2=C1252001, C3=C2×504001, C4=C3×1008001, C5=C416128001, C6=C532256001, C7=C664512001, C8=C7*193536001 etc. The result after a lot my work? The maximum is 1.937… for C=62745. – zhrd Oct 03 '22 at 06:09

2 Answers2

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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.

Dietrich Burde
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  • I am not sure, that unbounded $n/\phi(n)$ for some $n$ is unbounded for Carmichael number too. – zhrd Oct 02 '22 at 15:44
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    It is in fact known that there are Carmichael numbers with arbitrarily many prime factors—see page 708 of Alford, Granville, and Pomerance's paper. – Greg Martin Oct 02 '22 at 17:07
  • OK, there are Carmichael numbers with arbitrarily many prime factors, but not with arbitrary combination of prime factors. Carmichael numbers are very rare! I cannot generally claim that what is true for some set is also true for a subset. – zhrd Oct 02 '22 at 18:28
  • You could say where the problem comes from. Otherwise it might be just an open problem, and we cannot say more. Do you know that there is a solution? By the way, have a look at the computational results on Carmichael numbers. I think there are examples, where this product is already close to $2$, or even bigger. – Dietrich Burde Oct 02 '22 at 18:46
  • No, it is not any homework. The problem has origin in "Lehmer's totient problem" see https://en.wikipedia.org/wiki/Lehmer%27s_totient_problem – zhrd Oct 02 '22 at 18:49
  • This would support the suspicion that it is an unsolved problem, right? Perhaps it is still possible to find a Carmichael number explicitly, with the product $>2$, and not arbitrarily big. Did you check the computational articles already? – Dietrich Burde Oct 02 '22 at 19:07
  • I didn't find any (not only) computational article about the so specific problem. This is also the reason, why I am asking here. – zhrd Oct 03 '22 at 03:18
  • Have you see the article by Löh? They construct very large Carmichael numbers with a large number of prime factors. They should work. – Dietrich Burde Oct 03 '22 at 08:14
  • I gave a look to this article. But there are not Carmichael numbers, only the way to find them. "Unfortunately, this form of presentation leaves it to the reader to recompute the whole set S if he wants to get hold of all the factors." It is beyond my ability and time to repeat their work and generate Carmichael numbers with a large number of prime factors. The only number in the article 11·13·17·19·29·31·37·41·43·61·71·73·97·101·109·113·151·181·193·641 has the product 1.70816. – zhrd Oct 03 '22 at 08:56
  • I have found it. For example n=3852971941960065=3×5×23×89×113×1409×788129 has the product $\prod_{p|n} \frac{p} {p-1} = 2.0016..$ Thanks for your help. – zhrd Oct 03 '22 at 09:22
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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.

Charles
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