2 conjectured recursion limits for $e$ and $\pi$. Consider the following recursions
$$ x_{n+2} = x_{n+1} + \frac{x_n}{n} $$
$$y_{n+2} = \frac{ y_{n+1}}{n} + y_n $$
I have been toying around with different starting values ( complex Numbers ) , divergeance etc. But was not able to conclude much.
However I noticed when
$$ x_1 = 0 $$
$$y_1 = 0 $$
$$ x_2 = 1 $$
$$ y_2 = 1 $$
We get the following limit recursions 
$$ \lim_{n \to \infty} \frac{n}{x_n} = e $$
$$ \lim_{n \to \infty} \frac{2 n}{y_n ^2} = \pi $$
How to prove these ??
And how about the divergeance / convergeance for other complex initial values ?

Edit : a partial answer occurs here 
Mirror algorithm for computing $\pi$ and $e$ - does it hint on some connection between them?
http://www.pi314.net/eng/miroir.php
But the issue of other starting values is not resolved yet.
( so this is not a complete duplicate )
For the first recursion we have an answer ( see below ) but at the time of posting , the second has no answer with respect to variable initial conditions yet.
 A: Here is a solution for the second case: Let $(y_n : n \geq 1)$ satisfy the recurrence relation
$$ y_{n+2} = \frac{y_{n+1}}{n} + y_n, \qquad y_1 = a, \quad y_2 = b. \tag{1}$$
Let $y$ be the generating function of $(y_n)$, i.e.,
$$ y(x) = \sum_{n=1}^{\infty} y_n x^n. $$
The recurrence relation $\text{(1)}$ translates to the following differential equation:
$$ x(x^2 - 1) y'(x) + (x+2)y(x) = ax(x+1) $$
Solving this equation under the constraint $y(x) = ax + bx^2 + \mathcal{O}(x^3)$ gives
$$ y(x) = \frac{ax}{1-x} +  \frac{x^2}{1-x}\left( \frac{a \arcsin x}{\sqrt{1-x^2}} + \frac{b-a}{\sqrt{1-x^2}} \right). $$
Now the following results are useful for our computation:

Fact. We have the following Taylor expansions:
  $$ \frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{(2n)!!} x^{2n}
\quad \text{and} \quad
\frac{\arcsin x}{\sqrt{1-x^2}} = \sum_{n=0}^{\infty} \frac{(2n)!!}{(2n+1)!!} x^{2n+1}. $$

From this, we find that
$$ y(x) = a\left( \sum_{n=1}^{\infty} x^n \right) + x^2 \sum_{n=0}^{\infty} \Bigg( a \sum_{0 \leq 2k+1 \leq n} \frac{(2k)!!}{(2k+1)!!} + (b-a) \sum_{0 \leq 2k \leq n} \frac{(2k-1)!!}{(2k)!!} \Bigg) x^n$$
and hence we have
$$ y_{n+2} = a + a \sum_{0 \leq 2k+1 \leq n} \frac{(2k)!!}{(2k+1)!!} + (b-a) \sum_{0 \leq 2k \leq n} \frac{(2k-1)!!}{(2k)!!}, \qquad n \geq 0. $$
Finally, from the Stirling's formula it is easy to see that
$$ \frac{(2n)!!}{(2n+1)!!} \sim \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{n}}
\quad \text{and} \quad
\frac{(2n-1)!!}{(2n)!!} \sim \frac{1}{\sqrt{\pi n}} $$
as $n \to \infty$. Therefore, by the Cesàro-Stolz theorem we have
$$ y_n \sim \left( \sqrt{\frac{\pi}{2}} a + \sqrt{\frac{2}{\pi}} (b-a) \right) \sqrt{n}, $$
or equivalently,
$$ \lim_{n\to\infty} \frac{y_n^2}{2n} = \frac{1}{\pi}\left( b + \left(\frac{\pi}{2}-1\right) a \right)^2 . $$
A: This is not an answer since obtained using a CAS.
For the first recursion $$x_{n+2} = x_{n+1} + \frac{x_n}{n}\qquad,\qquad x_1=a \qquad,\qquad x_2=b$$ what was obtained after simplifications of the results given by a CAS is $$x_n=a n +(b-2a)\frac {!n}{(n-1)!}$$ where appears the subfactorial function which makes $$\frac n {x_n}=\frac 1 {a+(b-2a) \frac{!n}{n!}}$$ Since $$\lim_{n \to \infty} \left(\frac{!n}{n!}\right)=\frac 1e$$  then $$\lim_{n \to \infty} \left(\frac{n}{x_n}\right)=\frac e {a e+(b-2a) }$$ 
For the second recursion, I have not been able to get anything. 
