Earlier, I asked a question; this is a question regarding an answer I received to it.
Apparently when $u^2 - 12p^2 = -3$, with $p$ a three digit prime, there is some sort of recursion of the values of $p$, causing $P(x+2) = 14P(x+1) - P(x)$, causing the first few values to be $1,13, 181,..$
Where did this recursion come from?
Note : $P(i)$ is referring to the $i$-th answer for $p$
Let $u_1=3$ and $p_1=1$. Then $u_1^2-12p_1^2=-3$. I.e., $(u_1-2\sqrt3p_1)(u_1+2\sqrt3p_1)=-3$.
Now $7^2-12\cdot2^2=1$; i.e., $(7-2\sqrt3\cdot2)(7+2\sqrt3\cdot2)=1$.
Therefore, if $u_n^2-12p_n^2=-3$, then $(u_n^2-12p_n^2)(7^2-12\cdot2^2)=-3$.
I.e., $(u_n-2\sqrt3p_n)(7-2\sqrt3\cdot2)(u_n+2\sqrt3p_n)(7+2\sqrt3\cdot2)=-3$.
I.e., $([7u_n+24p_n]-2\sqrt3[7p_n+2u_n])([7u_n+24p_n]+2\sqrt3[7p_n+2u_n])=-3$.
I.e., $u_{n+1}^2-12p_{n+1}^2=-3$ with $u_{n+1}=7u_n+24p_n$ and $p_{n+1}=7p_n+2u_n$.
Furthermore, $p_{n+2}=7p_{n+1}+2u_{n+1}$ and $p_{n+1}=7p_n+2u_n$ and $u_{n+1}-7u_n=24p_n$ imply
$\dfrac{p_{n+2}-7p_{n+1}}2-7\dfrac{p_{n+1}-7p_n}2=24p_n$ or $p_{n+2}=14p_{n+1}-p_n$
(a recurrence involving $p$ and not $u$, which you seemed to want).