The quick version is $n_0 = 0, \; \; n_1 = 40,$ then
$$ \color{magenta}{ n_{k+2} = 98 n_{k+1} - n_k + 40}. $$
Given an $(x,y)$ pair with $3x^2 - 2 y^2 = 1$ we then take $n = (x^2-1)/ 2 = (y^2 - 1)/ 3. $
The first few $x,y$ pairs are
$$ x=1, \; y= 1 , \; n=0 $$
$$ x=9, \; y=11, \; n=40 $$
$$ x= 89, \; y=109, \; n=3960 $$
$$ x=881, \; y=1079, \; n= 388080 $$
$$ x=8721, \; y=10681, \; n= 38027920 $$
$$ x=86329, \; y=105731, \; n= 3726348120 $$
and these continue forever with
$$ x_{k+2} = 10 x_{k+1} - x_k, $$
$$ y_{k+2} = 10 y_{k+1} - y_k. $$
$$ n_{k+2} = 98 n_{k+1} - n_k + 40. $$
People seem to like these recurrences in one variable. The underlying two-variable recurrence in the pair $(x,y)$ can be abbreviated as
$$ (x,y) \; \; \rightarrow \; \; (5x+4y,6x+5y) $$ beginning with
$$ (x,y) = (1,1) $$
The two-term recurrences for $x$ and $y$ are just Cayley-Hamilton applied to the matrix
$$ A \; = \;
\left( \begin{array}{rr}
5 & 4 \\
6 & 5
\end{array}
\right) ,
$$
that being
$$ A^2 - 10 A + I = 0. $$
Best Answer
HINT.-The fundamental unit of $\mathbb Q(\sqrt3)$ is $u=2+\sqrt3$ so the solutions $(x_n,y_n)$ of $b^2-3a^2=1$ is given by $$x_n+y_n\sqrt3=(2+\sqrt3)^n$$