3 broken formula fixed; added 2 characters in body

How can one solve the following recurrence relation?

$$f(0) = 1$$ $$f(n) f(n) = 1 + \frac{1}{2^n} \sum_{k = 0}^n\frac{1}{2^n} (n \choose k) f(k)$$ for 0}^n {{n}\choose{k}} f(k)n \geq 1$f(0) = 0$

As it happens, I can show $f(n) = \Theta(\log n)$ through other means (see below). But I'd like to know how to solve the recurrence "directly".

The recurrence relation comes from the following coin flipping problem. There are $n$ independent, unbiased coins, and we toss all of then for a number of rounds. Let $T(n)$ be the first round when each coin has got head at least once (ie, $T(n) = \text{arg} \min_t \text{s.t.} H_t(i) \geq 1; \forall i \in [n]$, where $H_t(i)$ is the number of heads the $i^{th}$ coin has got in the first $t$ rounds). Then one can see that $E(T(n))$ fulfills the recurrence relation mentioned above.

To see that $E(T(n)) = O(\log n)$, note that we reduce the number of coins who haven't gotten head yet by a factor of 2, in expectation. The $\log n$ bound follows routinely from that. On the other direction (ie, $E(T(n)) = \Omega(\log n)$), let $S = \log n/20$. Then at time $S$ with high probability, a large number of coins will still be "headless", from which the lower bound follows.

2 changed the math, so it is slightly more readable, but still not great.

How can one solve the following recurrence relation?