# Can a carmichael number have consecutive prime factors?

Is there a set $[p_1,p_2,...,p_n]$ of consecutive primes , such that $\prod_{j=1}^n p_j$ is a carmichael number ?

For $3$ and $4$ prime factors, I checked upto $p_1\le 10^{10}$ without success.

The number $N\ :=\ 3\times 5\times 7\times 11\times 13\times ...\times p=\frac{p\#}{2}$ can never be a carmichael number because for $p\ge 7$, the condition $6|N-1$ cannot hold ($N$ is divisble by $3$) and $3$ and $3\times 5$ are obviously not carmichael numbers.

Assuming some widely believed bounds on prime gaps, this should be impossible at least in the case of 3 prime factors. Suppose $p, q = p+a,$ and $r = p+b$ are consecutive primes, and $N = pqr$ is a Carmichael number. This implies in particular that $p-1$ must divide $N-1$. Writing $N-1 = pqr-1$ modulo $p-1$, we get $N - 1 \equiv (p - (p-1))(q - (p-1))(r - (p-1)) - 1 = (a+1)(b+1)-1 \pmod{p-1}$. So it must be that $p-1$ divides $(a+1)(b+1)-1$. But it is believed that prime gaps are asymptotically $o(\sqrt n)$, which would imply that at least for large $p$, the positive number $(a+1)(b+1)-1$ is much smaller than $p-1$, contradiction.
• I do not understand the last claim. Why must $(a+1)(b+1)-1<<p-1$ hold assuming the gaps are of order $o(\sqrt{p})$ – Peter Oct 14 '15 at 9:57
• If $p$, $q$, and $r$ are consecutive primes, then $q - p = a$ and $r - q = (b-a)$ are prime gaps and therefore of order $o(\sqrt p)$. Then $a + 1$ and $b + 1$ are also $o(\sqrt p)$, so their product is $o(p)$. – Ravi Fernando Oct 14 '15 at 18:09
• But $o(p)$ is not necessarily much smaller than $p$. – Peter Oct 14 '15 at 18:11
• It is much smaller than $p$. I'm using little-oh notation here, which means exactly that. – Ravi Fernando Oct 14 '15 at 18:17
• Also note that the more commonly quoted conjecture is that prime gaps are just $O(n)$, or something like that. But in reality it's believed that the largest prime gaps grow on the order of $O(\log^2 n)$, which is microscopically smaller. – Ravi Fernando Oct 14 '15 at 18:41