TITLE: Is there a closed form for this recursion equation?
QUESTION [7 upvotes]: Given
$$f(a)=\frac{1}{6} \sum _{i=1}^6 \left[1-(1-f (a-1)) \left(\frac{i}{6}\right)^a\right]$$
with $f(1)=0$, $a\in\mathbb{N}$ and $f:\mathbb{N}\to\mathbb{Q},$

is there a way to write a closed form solution for the above recursion function. I am conducting a research and obtained the above result, but was interested if a function of $f(a)$ only in terms of $a$ exists, for arbitrary $a$. This would save a lot of computational work for my research. This in fact, is a small result which will be used in more complex calculations, therefore a closed form would be optimal. Thank you.

REPLY [4 votes]: Your recurrence can be written, with $v(a)=1-f(a)$,
$$v(a)=v(a-1)\frac16\sum_{i=1}^6\left(\frac{i}{6}\right)^a$$
And $v(1)=1$.
This leads immediately to:
$$f(a)=1-\frac{1}{6^{a-1}}\prod_{k=2}^{a}\left[\left(\frac{1}{6}\right)^{k}
+\left(\frac{2}{6}\right)^{k}+\left(\frac{3}{6}\right)^{k}+\left(\frac{4}{6}\right)^{k}+\left(\frac{5}{6}\right)^{k}+\left(\frac{6}{6}\right)^{k}
\right]$$
Not sure it qualifies as a "closed form", but I suspect you can't do much better. If the problem is performance, notice that you can improve it very much by keeping all preceding values of $f(a)$ in memory. Then computing the next one simply requires applying your recurrence, which is not difficult to compute. Notice also that $f(a)$ converges to $1$ linearly.