why is $ 2 = \frac{5}{1+\frac{8}{4+\frac{11}{7 + \frac{14}{10 + \dots}}} } $ Why is $ 2 = \cfrac{5}{1+\cfrac{8}{4+\cfrac{11}{7 + \cfrac{14}{10 + \ddots}}} } $
where the sequences $5,8,11,14,\dots$ and $1,4,7,10,\dots$ are of the form $5 + 3 n$ and $1 + 3n$.
(This converges on both even and uneven iterates)
I was surprised this is an integer.
Maybe it would help to rewrite this generalized continued fraction into a "normal" simple continued fraction.
But I believe that would give us coefficients that generalized the double factorial to a sort of " triple factorial " meaning $ f(n) = n \cdot f(n-3) \cdot f(n-6) \cdot f(n-9) \cdots $ and I have almost no skills or understanding of those.
Maybe some transformation formula's make this easy, but Im not seeing it.

Assume the RHS of the equation as $A_n$ so we have the following recurrence relation:
$$
A_n = \frac{3n + 2}{3n - 2 + A_{n + 1}}.
$$
Claim:
$$A_n = \frac{n + 1}{n}.$$
Proof:
Step of the induction:
$$A_{n + 1} = \frac{n + 2}{n + 1} \Rightarrow A_n = \frac{3n + 2}{3n - 2 + \frac{n + 2}{n + 1}} = \frac{n + 1}{n}.$$
Base of the induction:
Limit of the $A_n$ in infinity according to our closed form solution:
$$A_{\infty} = \lim_{n \to \infty}{\frac{n + 1}{n}} = 1.$$
Limit of the $A_n$ in infinity according to the given recurrence relation:
$$ x =  \lim_{n \to \infty}{A_n} = \lim_{n \to \infty}{\frac{3n + 2}{3n - 2 + A_{n + 1}}} = \lim_{n \to \infty}{\frac{3n + 2}{3n - 2 + x}} = 1.$$
So the base of the induction is correct as well.
So $A_n = \frac{n + 1}{n}$.
Your question asks about the value of $A_{n = 1}$ and we have $ A_{1} = \frac{n + 1}{n} = \frac{1 + 1}{1} = 2$
