Here is what I observed :
Let $n$ be a natural number greater than 2.
Let $A = 2\cdot\binom{3n+1}{n}-\binom{3n}{n-1}+\binom{3n-1}{n-2}$
Let $p=2n+1$
$p$ is prime iff $A \equiv 2 \pmod{p}$
You can run this test here.
I tried with some prime and composite numbers below 350000 and I didn't find any counterexample.
Is there a way to prove it ? I think this is related to Wilson's theorem but I'm not sure.
Peter’s keen observation and idea is a Credit to the following partial answer.
I agree with what Peter said that the prime case is not difficult to prove using an interesting lemma below.
Lemma:
For any prime number $p$ and natural number $k$ satisfying $p<k<2p,$ $$ \left(\begin{array}{c} k \\ p \end{array}\right) \equiv 1 \quad (\bmod p) $$
Proof:
By definition, $$ \left(\begin{array}{l} k \\ p \end{array}\right)=\frac{k(k-1)\cdots(p+2)(p+1)}{(k-p)!} $$
Rearranging gives $$ (k-p)!\left(\begin{array}{l} k \\ p \end{array}\right)=k(k-1) \cdots(p+2)(p+1) $$
As $p\not |k,$ taking both sides in modulo $p,$ yields $$ \begin{aligned} (k-p) !\left(\begin{array}{c} k \\ p \end{array}\right) & \equiv (k-p) !\quad (\bmod p) \\ \left(\begin{array}{l} k \\ p \end{array}\right) & \equiv 1 \quad (\bmod p) \end{aligned} $$
Rewriting $A$ gives $$ \begin{aligned} A &=2\left(\begin{array}{c} 3 n+1 \\ 2 n+1 \end{array}\right)-\left(\begin{array}{c} 3 n \\ 2 n+1 \end{array}\right)+\left(\begin{array}{c} 3 n-1 \\ 2 n+1 \end{array}\right) \\ &=2\left(\begin{array}{c} 3 n+1 \\ p \end{array}\right)-\left(\begin{array}{c} 3 n \\ p \end{array}\right)+\left(\begin{array}{c} 3 n-1 \\ p \end{array}\right) \end{aligned} $$
If $p=2n+1$ is prime, then using the lemma yields $$ A \equiv 2-1+1 \equiv 2 \quad \pmod p $$
Wish someone can help solve the converse!