Why are Carmichael Numbers less common with an arithmetic progression

Question

I have looked at the list of Carmichael numbers less than $1.4*10^{15}$, observing most of them are of the form $12n+1$. That is up to a certain number, $N$, there are more Carmichael numbers of the form $12n+1$ than $12n-1$ (or any other form such as $12n+3,5,7,9$) does anyone know why this is? That is comparing forms $12n+1$, $12n-1$ and all the Carmichael numbers less than $1.4*10^{15}$, the Carmichael of the form $12n+1$ outnumber the ones of the form $12n-1$ easily. I would suspect that is because a Carmicheal number $N$ $=$ $pqr$ with two prime factors $p$ $=$ $5$ $\pmod {12}$ and $q$ $=$ $7$ $\pmod {12}$ must be $1$ $\pmod {12}$. In fact, if $p$ $=$ $1$ $\pmod {12}$, no matter what $q$ is except for $2$ and $3$, $N$ will still be $1$ $\pmod {12}$. Can anyone come of with more heuristics and or arguments to support this? Again, there are infiniely many Carmichael numbers of the form(s) $12n+1, 3, 5, 7, 9, 11$. (The first Carmichael number of the form $12n+1$ is $1105$ followed by $1729$.) Conjecture is that for numbers $N$ $>$ $1729$, there will always be more Carmichael numbers of the form $12n+1$ than any other form $12n+k$ less than $N$.

Answer 1

It is because of Korselt's criterion on Carmichael numbers, namely that for any prime p dividing N, p-1|N-1. Therefore, if N is not equal to 12n+1, then no prime of the form 12n+1, can divide n. More specifically, either 3 or 4 does not divide N, therefore either no prime of the form 3n+1 can divide N, or no prime of the form 4n+1 can divide N. Either way, this rules out half of all possible prime divisors (three quarters if N=12n-1), making Carmichael numbers not of the form 12n+1 much more rare.