# Does $C(n)$ grow exponential versus $n$?

Let $$λ(m)=n$$ be the Carmichael Function of $$m$$. For each (even) number $$n$$, there is a largest number $$m$$ such that $$λ(m)=n$$. Let $$C(n)$$ denote the largest integer $$m$$ such that $$λ(m)=n$$. For instance, $$C(2)=24$$ and $$C(4)=240$$ since $$24$$ and $$240$$ are the largest numbers with Carmichael Function equal to $$2$$ and $$4$$ respectively. In these cases, $$n$$ is really small, and $$C(n)$$ is rather large compared to $$n$$. However for other $$n$$, $$C(n)$$ and $$n$$ are about the same size so there is no significant growth rate of $$C(n)$$ compared to $$n$$. Suppose this is not the case. So as $$n$$ gets larger and larger, is it possible that $$C(n)$$ will grow exponential compared to $$n$$? If not, what is the maximum growth rate of $$C(n)$$ compared to $$n$$?

• Hmm, for the special case of $n$ being a power of $2$, the growth of $C(2^k)$ depends on the distribution of Fermat primes, and we don't even know if there are more than $5$ of those ... so this sounds pretty intractable. – Henning Makholm May 3 at 2:49
• You could calculate a few values of $C(n)$ and then look it up in the Online Encyclopedia of Integer Sequences, oeis.org – Gerry Myerson May 3 at 3:56
• – Robert Israel May 3 at 11:54

According to the Wikipedia page, there is a positive constant $$c$$ such that for sufficiently large $$A$$, there is $$n > A$$ with $$\lambda(n) < (\ln A)^{c \ln \ln \ln A}$$. That says if $$m = \lambda(n)$$, $$C(m) \ge n > A$$. Note that $$m = \lambda(n) < \exp(c (\ln \ln A)(\ln \ln \ln A)) < \exp(c (\ln \ln A)^2)$$ so $$C(m) > \exp(\exp(\ln(m)^{1/2}/c^{1/2}))$$ This is not quite as good as exponential growth, but I guess it is the best result known (or was at the time of writing).