A variation of Fermat's little theorem in the form $a^{n-d}\equiv a$ (mod $p$).

Solution 1:

If $d+1$ is not squarefree, then the multiples of it are not squarefree, therefore your value of $n$ is not squarefree, so then by the Generalized Korselt's criterion it is not a valid value of $n$, unless it equals $d+1$ itself, but that is just one case.

If $d+1$ is squarefree, we note that for every prime $p \geq d+1$ we have that $n=p(d+1)$ is squarefree. Then we want that for every prime divisior $q_i$ of $n$, that $q_i-1 \mid (p-1)(d+1)$. It surely holds for $q_i=p$. Note that since $d+1$ is squarefree, we have that $$\varphi(d+1)=(p_1-1)(p_2-1)(p_3-1) \cdots(p_t-1)$$

Here $\varphi(x)$ is the Euler-phi function.

Therefore if $\varphi(d+1) \mid p-1$, then our $n$ is a possibility. In other words, if $p = k \varphi(d+1) + 1$ for some $k$. Note that we always have $\varphi(d+1) \leq d$, so there are quite a few values.

Since $\gcd(\varphi(d+1),1)=1$, we have by a stronger version of Dirichlet's theorem that the primes in this arithmetic progression are distributed just as in the natural numbers, i.e. there are about $\frac{x}{\ln(x)}$ primes when $k<x$ ($k$ as in $p = k \varphi(d+1) + 1$). This gives some information on the second part of the conjecture.

For example, when $d=4$, $13\cdot5$, $17\cdot5$, $29 \cdot 5$ and $31 \cdot 5$, are all in your sequence, and they indeed have the form I described above.

Note: One could use $\lambda(x)$, the Carmichael function, instead of $\varphi(x)$.