Let $n(C)$ be the number of prime factors of the Carmichael-number $C$.

I Conjecture $\lim sup_{C\rightarrow \infty} n(C)=\infty$

In other words, the sequence $n(C)$, $C$ running over the Carmichael-numbers, is unbounded.

I learnt that Dickson's conjecture implies that this is the case. There are arbitary long strictly increasing vectors $v_1,...,v_n$ ($n\ge 3$) with $\sum_{j=1}^n \frac{1}{v_j}=1$. If $L\ :=\ lcm(v_1,...,v_n)$, then $\prod_{j=1}^n (\frac{L^2}{x_j}\times m+1)$ is a Carmichael-number if $\frac{L^2}{x_j}\times m+1$ is prime for $j=1,...,n$, and Dickson's conjecture implies that such a number $m$ always exists.

I also learnt that it is not known, whether there are infinite many Carmichael numbers with $k$ prime factors for any fixed number $k\ge 3$. But maybe my conjecture can be proven (or disproven).

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    $\begingroup$ A vote up for your conjecture, good luck! $\endgroup$ – user243301 Nov 3 '15 at 19:52

I don't believe this is currently known. Sequence A006931 references Alford, Grantham, Hayman, & Shallue who, among other things, construct a Carmichael number with 10,333,229,505 prime factors, but don't prove that there are Carmichael numbers with arbitrarily many prime factors.

  • $\begingroup$ I only heard of a construction of a Carmichael number with about a million of prime factors. Totally left in the dust by this number :) $\endgroup$ – Peter Nov 3 '15 at 20:36
  • $\begingroup$ @Peter: That's [9] in the cited paper: Günter Löh and Wolfgang Niebuhr, A new algorithm for constructing large Carmichael numbers, Math. Comp. 65 (1996), no. 214, 823–836. $\endgroup$ – Charles Nov 3 '15 at 20:55

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