Is there any way to solve this recurrence relation? This is my first time posting here so sorry if I mess up anything, but I was wondering if there were any way to solve this recurrence relation:
$g(n)^n-n=g(n+1)$
I was trying to solve this problem here for $g(1)$:
$g(n)=\sqrt[n]{n+\sqrt[n+1]{n+1+\sqrt[n+2]{n+2+...}}}$
 A: The value of $g(1) = 2.91163921624582$ (for the square root formula) is highly unstable and cannot be used in the forward direction to calculate more than a few steps, losing precision rapidly. 
Effectively you are trying to track a point that when raised to power $n$ produces an increment of almost exactly $n$, just enough less that the same rule applies indefinitely.
By contrast taking the root after $n+10$ at value $1$ will give you a very stable value. The error due to the rest of the sequence gets shunted out of any noticeable precision, since the value is in any case close to $1$.
\begin{array}{|c|c|} \hline
n & g(n) \\ \hline
1 & 2.911639216\\
2 & 1.911639216\\
3 & 1.654364493\\
4 & 1.527866372\\
5 & 1.449309679\\
6 & 1.394490732\\
7 & 1.353494872\\
8 & 1.321398182\\
9 & 1.295433322\\
10 & 1.273905731 \\ \hline
\end{array}
A: Maybe as a first approximation, turn it into a differential equation.
$$g(n)^n − g(n) - n=g(n+1)-g(n)$$
Which is approximately the same as the differential equation,
$$g^x − g - x=\frac{dg}{dx}$$
If you could find a solution for that, it might give a hint at how to solve the difference equation. But the differential equation is not separable.
My guess is that there is no easy solution to the differential or the difference equation.
