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.
 A: A relative simple solution can be found based on a fews facts about continued fractions which I will present briefly.

Brief summary of (positive) general continued fractions: For any sequences of positive numbers $(a_n)$ and $(b_n)$ defined   the maps
$$M_n(x)=\tfrac{a_1}{b_1+\tfrac{a_2}{b_2+\ddots \begin{matrix}
  \\
  +\tfrac{a_{n-1}}{b_{n-1}+\tfrac{a_n}{b_n+x}}
\end{matrix}}}$$
The terms $M_n(0)$ are known as converts of the continued fraction
$$\tfrac{a_1}{b_1+\tfrac{a_2}{b_2+\ddots \begin{matrix}
  \\
  +\tfrac{a_{n-1}}{b_{n-1}+\ddots}
\end{matrix}}}$$
It is easy to check that
$$M_n(x)=M_{n-1}\Big(\frac{a_n}{b_n+x}\Big)$$
By induction it follows that
$$M_n(x)=\frac{p_{n-1}x+p_n}{q_{n-1}x+q_n}$$
where
\begin{align}
p_n&=a_np_{n-2}+ b_np_{n-1},\qquad p_0=0,\, p_1=a_1\\
q_n&=a_nq_{n-1}+b_nq_{n-1},\qquad q_0=1,\, q_1=b_1
\end{align}
and
$$\begin{align}
p_{n-1}q_n-q_{n-1}p_n=(-1)^na_1\cdot\ldots \cdot a_n\tag{0}\label{zero}
\end{align}$$
Hence
$$\begin{align}
\frac{p_{n-1}}{p_{n-1}}-\frac{p_n}{q_n}=(-1)^{n}\frac{a_1\cdot\ldots\cdot a_n}{q_{n-1}q_n}\tag{1}\label{one}
\end{align}
$$
and
$$\begin{align}
\frac{p_n}{q_n}-\frac{p_{n-2}}{q_{n-2}}=(-1)^nb_n\frac{a_1\cdot\ldots\cdot a_{n-1}}{q_nq_{n-2}}\tag{2}\label{two}
\end{align}$$
More generally
$$\begin{align}
M_n(x)-M_n(y)=(-1)^n(x-y)\frac{a_1\cdot\ldots\cdot a_n}{\big(q_{n-1}x+q_n\big)\big(q_{n-1}y+q_n\big)}\tag{3}\label{three}
\end{align}$$
Identities \eqref{one} and \eqref{two} imply that
$$\frac{p_{2n}}{q_{2n}}<\frac{p_{2n+2}}{q_{2n+2}}<\frac{p_{2m+1}}{q_{2m+1}}<\frac{p_{2m-1}}{q_{2m-1}}$$
for all $m, n$. Thus,  there are numbers $0<\alpha\leq \beta$ such that $p_{2n}/q_{2n}\xrightarrow{n\rightarrow \infty}\alpha$ and $p_{2n+1}/q_{2n+1}\xrightarrow{n\rightarrow\infty}\beta$.
The following result  gives sufficient conditions for convergence of the convergents $M_n(0)=\frac{p_n}{q_n}$.

Lemma: If $\liminf_n\frac{b_{n-1}b_n}{a_n}>0$, then $\frac{p_n}{q_n}$ converges.

A proof of this result (with limit instead of $\liminf$) can be found in the classic high shool textbook Hall, H.S. and Knight, S. R., Higher algebra, 4th edition, 1889. pp. 362. A short proof of the Lemma is also shown at the end of this posting.

Solution t the OP:
For the problem in the OP consider $a_n=3n+2$ and $b_n=3n-2$ for each $n\in\mathbb{N}$.  The key part of the whole solution  is the observation made by Ivan Kaznacheyeu in his comment. Notice that
$M_1(x)=\frac{5}{1+x}$,  and that $2=M_1(1+\frac12)$. Let $f(n)=1+\frac{1}{n+1}$. This sequence satisfies
$$ f(n-1)=\frac{3n+2}{(3n-2)+f(n)}$$
From this,
$$
M_n(f(n))=M_{n-1}\big(\frac{3n+2}{(3n-2)+f(n)}\big)=M_{n-1}(f(n-1))$$
Thus, if $M_{n-1}(f(n-1))=2$, we have that $M_n(f(n))=2$.
This was Ivan Kaznacheyeu's insight into the problem. The rest of the solution, as we will see is routine.
Since
$$\frac{b_{n-1} b_n}{a_{n-1}}=\frac{(3n-5)(3n-2)}{3n-1}\xrightarrow{n\rightarrow\infty}\infty,$$
The sequence $M_n(0)=p_n/q_n$ converges.
As Ivan Kaznacheyeu mentioned in his comment, for any $n\in\mathbb{N}$
$$2=M_n(f(n)),\qquad\text{where}\quad f(n)=1+\frac{1}{n+1}$$
Consequently, from \eqref{three}
$$\begin{align}
2-\frac{p_n}{q_n}=M_n(f(n))-M_n(0)=(-1)^nf(n)\frac{a_1\cdot\ldots\cdot a_n}{\big(q_{n-1}f(n)+q_n\big)q_n}\tag{4}\label{four}\end{align}
$$
Identity \eqref{four} yields
$$\frac{p_{2n}}{q_{2n}}<2<\frac{p_{2m+1}}{q_{2m+1}},\qquad \forall \,n,m\in\mathbb{N}
$$
Hence, $\frac{p_n}{q_n}$ converges to $2$.

Proof of Lemma: From \eqref{zero}
$$\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}=-\frac{a_nq_{n-2}}{q_n}\Big(\frac{p_{n-1}}{q_{n-1}}-\frac{p_{n-2}}{q_{n-2}}\Big)$$
Then
$$
\frac{a_nq_{n-2}}{q_n}=\frac{a_nq_{n-2}}{a_nq_{n-2} + b_nq_{n-1}}=\frac{1}{1+\frac{b_nq_{n-1}}{a_nq_{n-2}}}
$$
and further
$$\frac{b_nq_{n-1}}{a_nq_{n-2}} = \frac{b_n(a_{n-1}q_{n-3}+b_{n-1}q_{n-2})}{a_nq_{n-2}}=\frac{b_ba_{n-1}q_{n-3}}{a_nq_{n-2}}+\frac{b_nb_{n-1}}{a_n}
$$
As the term $\frac{b_ba_{n-1}q_{n-3}}{a_nq_{n-2}}>0$ for all $n$,  the assumption in the Lemma implies that for some $c>0$
$$0<\frac{a_nq_{n-2}}{q_n}<\frac{1}{1+c}$$
for all $n$ large enough. From this, it follows that the difference $\Big|\frac{p_n}{q_n}-\frac{p_{n-1}}{q_{n-1}}\Big|\xrightarrow{n\rightarrow\infty}0$, that is $\alpha=\beta$.

A: 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$
A: Too long for a comment:
It appears that by cutting of the head at the top, one gets:
$1+1,1+1/2,1+1/3,1+1/4,1+1/5,...,1+1/r$
(*Mathematica start*)
r = 20;
n = Table[5 + 3*(n - 1), {n, r, 100}];
d = Table[1 + 3*(n - 1), {n, r, 100}];
f[y_, {m_, d_}] := m/(d + y);
continuedFraction = 
 Fold[f, Last@n/Last@d, Reverse@Most@Transpose@{n, d}]
"The output appears to be equal to r"
(N[continuedFraction, 20] - 1)^-1
(*end*)


$r=1$ gives: continuedFraction $= 2$
for sequences $n$ and $n-2$
(*Mathematica start*)
r = 10;
n = Table[n, {n, r, 100}]
d = Table[(n - 2), {n, r, 100}]
f[y_, {m_, d_}] := m/(d + y);
continuedFraction = 
 Fold[f, Last@n/Last@d, Reverse@Most@Transpose@{n, d}]
N[continuedFraction, 20]
"The output appears to be equal to r"
(N[continuedFraction, 20] - 1)^-1
(*end*)

A: Tommy1729 aka my mentor Tom Raes wrote to me :
Proof (sketch)
We want to prove cutting of the head at the top gives the result $1+1/n$ from which the case ' equals $2$ ' follows.
Now if we get close to $1+1/n$ then it must really be that ($1+ 1/n$) because the starting value for continued fractions is irrelevant just like $\dfrac{1}{1+\dfrac{1}{1+\dfrac{...}{1+x}}}$ converges to the golden mean independant of any fixed positive $x$.
Lets take $n$ and $k$ positive integers.
$ \dfrac{5 + 3(n-1)(1 + e_1)}{1+ 1/n} - [1 + 3(n-1)] = 1 + 1/(n+1) + e_1[\dfrac{5 + 3(n-1)}{1 + 1/n}]$
going down in the continued fraction $k$ times clearly gives us
$e_k$ follows $ (1+e_k)^{-1} (1 + 1/(n+k-1)) = 1 + 1/(n+k-1) + e_{k-1}[\frac{5 + 3(n+k-3)}{1 + 1/(n+k-2)}]$
From then we can conclude that for sufficienly large $n+k = v$ we get in the limit
$(1 + e_v)^{(-1)} = "1" + e_{v-1} * "\infty"$.
where " means approximating.
Notice $e_v$ going towards (positive or negative) $\infty $ cannot keep on satisfying the equations. So that never happens.
Similar with $e_v$ not converging to $0$ nor $-1$; that cannot keep on satisfying the equations.
If $e_v$ converges to $-1$ we get in the limit
$(1+ 1/v)^{-1} (1 + e_v) = 0/1 = 0$
so
$ (1 + 1/v + w)^{-1} = 0$
Therefore the starting value converges to infinity because $w$ does.
But that is not valid.
( compare with 3 = $\dfrac{1}{1+\dfrac{1}{1+\dfrac{...}{1+x_n}}}$ then $x_n$ diverges to infinity because the value 3 should be the golden mean. )
So the only solutions $1 + e_v$ that converge are being $1$ or converging to $1$ ( at rate $O ( 1/v^3)$  ).
SO we get closer to $1+1/n$ as desired.
And thus it actually IS $1+1/n$.
QED
Ps: the other continued fractions I sent you are harder to prove.
