Subsets whose size is a power of $2$ A set with $n$ elements contains $2^n$ subsets. What if we restrict to subsets whose size is a power of two? Does this quantity behave differently asymptotically? I.e. what is the asymptotic behaviour of:
$$f(n) = \sum_{k=0}^{\lfloor\log{n}\rfloor}\binom{n}{2^k}$$
 A: As shown in the comments, it's exponential.  Perhaps the best way to describe the asymptotics of such a sequence are to examine the $\limsup$ and $\liminf$ of $\frac{1}{n}\log f(n)$.  
Let's first notice that $$\max_{k \leq \log n} \binom{n}{2^k} \leq f(n) \leq \log n \max_{k \leq \log n} \binom{n}{2^k}.$$
This implies that $$\frac{1}{n} \log f(n) = \frac{1}{n}\max_{k \leq \log n} \log \binom{n}{2^k} + o(1)$$ as $n \to \infty$.  Note that this maximum is at most $\log \binom{n}{|n/2|} / n$ which approaches $\log(2)$; moreover, this maximum is achieved along powers of $2$ implying that the $\limsup$ is indeed $2^n$.
The lower bound isn't much harder: observe that for $n$ in the range $[2^m, 2^{m+1}]$, the maximum is smallest at $n = 3\cdot 2^{m-1}$.  We're thus interested in the quantity $$\lim_{m \to \infty} \frac{1}{3\cdot 2^{m-1}} \log\binom{3 \cdot 2^{m-1}}{2^{m-1}}\,.$$  Applying Stirling's formula shows that the limit along this sequence is $\log(3) - \frac{2}{3}\log(2).$  This shows that $$\limsup_{n \to\infty} \frac{1}{n}\log f(n) = \log(2) \approx .693,\qquad \liminf_{n \to\infty} \frac{1}{n} \log f(n) = \log(3) - \frac{2}{3}\log(2) \approx .637$$
