5
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ A vote up for your conjecture, good luck! $\endgroup$ – user243301 Nov 3 '15 at 19:52
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.