Can you prove why this recurrence relation holds for some (a subset of the) prime numbers? And does it go on infinitely? (I think it does)
$$3P_{n} - 2P_{n + 1} = P_{n - 1}$$
First few values for which it applies:
$5, 7, 13, 20, 25, 26, 28, 45, 49, 60, 64, 78, 88, 89, 93, 95, 113, 136, 144, 148, 152, 159, 182, 212, 225, 229, 230, 236, 243, 247, 249, 262, 265, 277, 286, 288, 291, 302, 315, 323$
E.G: For $n = 5, P_{n} = 11, P_{n + 1} = 13, P_{n - 1} = 7:$
$3(11) - 2(13) = 7$
$3P_{5} - 2P_{6} = P_{4}$
...&c for the others
Using the standard prime gap notation gives that
$$g_n = p_{n+1} - p_{n} \implies p_{n+1} = p_{n} + g_{n} \tag{1}\label{eq1A}$$
$$p_{n} = p_{n-1} + g_{n-1} \tag{2}\label{eq2A}$$
Next, \eqref{eq2A} in \eqref{eq1A} gives
$$p_{n+1} = p_{n-1} + g_{n-1} + g_{n} \tag{3}\label{eq3A}$$
Substituting \eqref{eq2A} and \eqref{eq3A} into your relation results in
$$\begin{equation}\begin{aligned} & 3p_{n} - 2p_{n + 1} = p_{n - 1} \\ & 3(p_{n-1} + g_{n-1}) - 2(p_{n-1} + g_{n-1} + g_{n}) = p_{n - 1} \\ & 3p_{n-1} + 3g_{n-1} - 2p_{n-1} - 2g_{n-1} - 2g_{n} = p_{n - 1} \\ & p_{n-1} + g_{n-1} - 2g_{n} = p_{n - 1} \\ & g_{n-1} = 2g_{n} \end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, your relation holds whenever any prime gap is double that of the next prime gap. With your example of $n = 5$, we have $g_{n-1} = 11 - 7 = 4$ and $g_{n} = 13 - 11 = 2$.
As for whether or not there's an infinite number of primes where \eqref{eq4A} holds, I'm quite certain that is currently unknown, but it's conjectured to be the case. Note there are infinitely many prime $k$-tuples which are admissible and, in the cases where they also represent a prime constellation, then \eqref{eq4A} is true, e.g., your example of $(0, 4, 6)$ (which is called a prime triplet), $(0, 8, 12)$, $(0, 12, 18)$, etc. The "Admissibility" section also states
In particular, the prime $k$-tuple conjecture states
This conjecture implies there are an infinite number of primes which satisfy \eqref{eq4A}.