Sophie Germain Star prime number Let us define following prime number :

Let $~S_p~$ be Sophie Germain Star prime number of the form :
$$S_p=12\cdot p \cdot (2p+1)+1,$$
where $p$ is a Sophie Germain prime .

Note that since $~p \equiv 5 \pmod 6 \Rightarrow S_p \equiv 5 \pmod 8$, we have:
$$
S_p \nmid 2^p-1 ~~\text{and}~~ S_p \nmid 2^{2p+1}-1 \Rightarrow ord_{S_p}(2) \neq p ~~\text{and}~~ ord_{S_p}(2) \neq 2p+1
$$
Heuristic results indicate that the percentage of SG Star prime numbers whose primitive root is $2$ among all SG Star prime numbers up to arbitrary upper bound $n$  is approximately $66$ %
What would be expected number (in percent) of composite numbers of the form 
$S_p=12\cdot p \cdot (2p+1)+1~$ such that $2^{S_p-1} \equiv 1 \pmod {S_p}$ 
up to some arbitrary upper bound $n$  ?
 A: Let $S_p$ not be a perfect square. Notice that if $2p+1|ord_{S_p}(2)$, then $2p+1|q-1$ for some prime factor $q$ of $S_p$. If $S_p$ is composite, then $S_p/q$ is not $1$ and further
$Q=S_p/q\equiv 1/1\mod 2p+1$.
Since we assumed $S_p$ is not a perfect square, WLOG $Q\geq 6(2p+1)+1$, and $q\geq 2(2p+1)+1$, so $S_p=Qq>(6(2p+1)+1)(2(2p+1)+1)=48p^2+64p+21>S_p$, a contradiction. Thus $S_p$ is prime.
Conversely assume that $p|q-1$, we also hav ethat $q\equiv 1\mod p$ and $Q\equiv 1\mod p$, in the notation above. But note that $q,Q$ are not equal to $2p+1$ since $S_p\equiv 1\mod 2p+1$. Thus wlog $q\geq 6p+1$ and since $S_p$ is squarefree, $Q\geq 8p+1$, and so $S_p=Qq\geq 48p^2+14p+1>S_p$, a contradiction, and thus $S_p$ is prime.
So if $S_p$ is a squarefree composite such that $2^{S_p-1}\equiv 1\mod S_p$, then $2^{12}\equiv 1\mod S_p$. Let $S_p>2^{12}$, and this cannot be true.
Thus we have for $S_p>2^{12}$, $0\%$ of non-perfect-square composite $S_p$ satisfy $2^{S_p-1}\equiv 1\mod S_p$. By providing an additional probably unnecessary bound (something like $p>12$), we can show by direct calculation that $S_p\neq (2(2p+1)+1)(2(2p+1)+1)$ and $S_p\neq (6p+1)(6p+1)$, and thus get:
for $S_p>2^{12}$, $0\%$ of composite $S_p$ satisfy $2^{S_p-1}\equiv 1\mod S_p$.
By direct computation one can find the smaller cases and whether they satisfy the identity.
