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### Closed form solution to an iterative equation.

$f(n+1) = f(n) + f(n)^{a}$ where $a \in (0,1)$ and $n \ge 1$ with $f(1) = m$.
If $a=0$, we see $f(n) = m + n - 1$ and if $a=1$, we see $f(n) = 2^{n-1}m$. So the recursion seems to interpolate ...