Let $f(n)$ be the number of ways to write $n$ as a sum of powers of 2, where we keep track of the order of the summation. For example, $f(4)=6$ because 4 can be written as $4, 2+2, 2+1+1, 1+2+1, 1+1+2$, and $1+1+1+1$. Find the smallest $n$ greater than 2013 for which $f(n)$ is odd.
The answer is 2047. For convenience, we agree that \( f(0)=1 \). Then by considering cases on the first number in the representation, we derive the recurrence 
\[ f(n)=\sum_{k=0}^{\left\lfloor\log _{2} n\right\rfloor} f\left(n-2^{k}\right) \]
We wish to understand the parity of \( f \). The first few values are 
\[ \begin{aligned} 
& f(0)=1 \\ 
& f(1)=1 \\ 
& f(2)=2 \\ 
& f(3)=3 \\ 
& f(4)=6 \\ 
& f(5)=10 \\ 
& f(6)=18 \\ 
& f(7)=31 . 
\end{aligned} \]
Inspired by the data we make the key claim that Claim - \( f(n) \) is odd if and only if \( n+1 \) is a power of 2. 

This only takes a few cases:
- If \( n=2^{k} \), then \((\odot)\) has exactly two repetitive terms on the right-hand side, namely 0 and \( 2^{k}-1 \).
- If \( n=2^{k}+2^{\ell}-1 \), then \((\odot)\) has exactly two repetitive terms on the right-hand side, namely \( 2^{\ell+1}-1 \) and \( 2^{\ell}-1 \).
- If \( n=2^{k}-1 \), then \((\bigcirc)\) has exactly one repetitive terms on the right-hand side, namely \( 2^{k-1}-1 \).
- For other \( n \), there are no repetitive terms at all on the right-hand side of \((\Omega)\).

Thus the induction checks out. So the final answer to the problem is 2047.