Prime Foias constant It is a known fact that there is the unique constant for which the terms of recursion
$$x_{n+1}=(1+\frac{1}{x_n})^n, x_1=\alpha$$
tend to infinity (for other values we start jumping around). The value is known as the Foias constant $\alpha=1.1874523511...$
However, on examining some of the similar expressions, I cannot see that this is in any sense a special feature of this particular recursion.
Just as an example, if we take $p_n$ to be the $n^{th}$ prime number, the recursion in form of
$$x_{n+1}=(1+\frac{1}{x_n})^{p_n}, x_1=\alpha_p$$
has a special value for which $x_n$ tends to infinity as well, because as far as I could calculate it is $\alpha_p \approx 2.5365538743$ and the process can continue as much as I like.
The proof of this, as it seems, is arriving from the calculation itself, by increasing or decreasing $\alpha_p$ by a little amount, we can slow down $x_{n}$ reaching infinity as much as we want, thus we can find the constant in question and once we've reached certain precision this depends on the next term.
Is there really something special about the original Foias' recursion that you know of which would prevent the other cases like the one mentioned here?
 A: True, it does have its own constant.
First, we need to notice that the equation:
$$x=(1+\frac{1}{x})^n$$
has a solution for any $n$ and that these are rising with rising $n$ which is easy to see if we write the above as:
$$\frac{n+1}{n}=\frac{\ln(x+1)}{\ln(x)}$$
In order to understand what is happening, let us introduce a substitution $ y=1+\frac{1}{x} $ getting the recursion:
$$ y_{n+1}=1+\frac{1}{y_n^{p_n}} $$
For positive values, superficially we would expect that this starts jumping between $1$ and $2$ (if $y$ is close to $2$, power will bring the next value close to $1$, and the loop continues by bringing $y$ back to close to $2$). However, if $y$ can increase slowly in a specific sense, then this jumping can be postponed to infinity, giving a clear limit to the series.
Let us find the first derivative
$$ y_{n+1}'(x)=(-1)^{n+1}\frac{\prod\limits_{k=1}^{n+1}p_k}{\prod\limits_{k=1}^{n+1}y_{k-1}(x)^{p_{k}+1}}$$
where $y_0(x)=x$. Obviously, if the denominator's terms start switching between $1$ and $2$, the first derivative would become close to $0$.
However, if we manage to have $y_k(x)^{p_k}$ limited, for example by keeping it as close as possible to the solution of $ y_k(x)=1+\frac{1}{y_k(x)^{p_k}} $ then the first derivative would go to infinity suggesting a breaking point: before that point the recursion would start jumping between $1$ and $2$ and above that point the opposite, between $2$ and $1$.
If this breaking point exists then we would have for the primorial function $\prod\limits_{k=1}^{n}p_k \sim e^{n\ln{n}}$ and $\prod\limits_{k=1}^{n}y_k(x)<2^{n}$ and the first derivative would tend to infinity.
Let us then revert the recursion:
$$ y_n=\frac{1}{(y_{n+1}-1)^{\frac{1}{p_n}}} $$
This recursion for most values $ m > 10 $ and $ 1 < y_m < 2 $ reaches the value close to $ y_1 \approx 1.4 $.
If we start $y_m$ from the positive zero of 
$$y^{p_n+1}-y^{p_n}-1=0$$
we will similarly reach the value close to $1.4$ regardless of the value of $p_n$ and all other values will have the above requirement that avoids having the first derivative equal to $0$.
This is because if for some $k<n$ we would get anywhere close to $2$, $y_{k} = 2+\Delta$, it is easy to see that the next one $y_{k+1}$ would be close to $1$ and then again $y_{k+2}$ close to $2$ and so on, thus we could not have the future $y_{n+1}$ and $y_{n}$ ever equal. For our story, it is sufficient to prove that there is this particular selection proces for which if we set $y_{n+1}(x)=y_{n}(x)$ then for all $k<n$ it is $y_{k}(x)<2$.
The uniqueness follows from the fact that if we force $y_{n+1}(x)=y_{n}(x)$ the series that follows cannot have the limit because for $k>n$ the terms start having two accumulation points: $1$ and $2$. So it is strictly necessary to start the reverse recursion from the positive zero of $y^{p_n+1}-y^{p_n}-1=0$.
This gives a simple algorithm for calculating $\alpha_p$
Find the solution of $x=(1+\frac{1}{x})^{p_n}$ and go in reverse by
$$x_{n}=\frac{1}{\sqrt[p_n]{x_{n+1}}-1}$$
until reaching $x_1$.
For first $50$ primes we got:
$$\alpha_p \approx 2.53655387425979$$
