Define
$$C_k = \{n \in \mathbb{Z}: n > \max(k, 0) \text{ }\text{and}\text{ }a^{n - k + 1} \equiv a \pmod{n} \text{ }\text{for all} \text{ } a \in \mathbb{Z}\}$$
Thus it's easy to see that when $k = 1$, $C_1$ consists of all of the prime numbers and Carmichael numbers.
Now for squarefree $k \geqslant 1$, we define the generalized Carmichael numbers as
$$N_k = C_k - \left\{kp: p \equiv 1 \mod{\left(\dfrac{\lambda(k)}{\gcd(\lambda(k), k)}\right)} \right\} $$
where $\lambda(k)$ is the Carmichael function. For the other values of $k$, we let $N_k = C_k$.
Notice that $N_1$ coincides with the set of Carmichael numbers. We interpret $N_k$ as a type of generalized Carmichael numbers since they do not exhibit a nice pattern with respect to the set of primes.
Next, we consider the growth rate of the counting functions
$$C_k(X) \text{ } = \text{ }\mid C_k \text{ } \cap \text{ } (0,X\text{}] \text{ } \mid \quad \text{and} \quad N_k(X) \text{ } = \text{ }\mid N_k \text{ } \cap \text{ } (0,X\text{}] \text{ } \mid $$
Now my conjecture is as follows: For all $k \in \mathbb{Z}$, we have
$$\lim_{X \to \infty} \dfrac{N_{-k}}{N_{k}} = 1$$
I have tested this conjectures by Graphing. I have also observed that for the same value of $X$, $N_k(X)$ tends to be bigger when k is squarefree and has many prime factors. I don't have any rigorous proof for the same. Any Idea how to prove it? Thanks.
