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A recursive solution for this problem should be derived by induction as follows.

Base Case:

$$$f(1) = 1$$$

$$$f(2) = 2$$$

Inductive Case: $$$m \geq 1$$$

$$$f(2m+1) = f(2m)$$$

$$$f(2m+2) = f(2m)+f(m+1)$$$

Proof:

By definition, $$$f(n) = |S(n)|$$$, where the sumset $$$S(n)$$$ is the set of different multisets of powers of 2 that sum to $$$n$$$. Let $$$T(N) = [S(n) | 1 \leq n \leq N]$$$ be the set of all sumsets up to N. It can be shown sumsets in $$$T(N)$$$ are mutually disjoint, i.e. $$$S(i) \cap S(j) = \emptyset$$$ when $$$i \neq j$$$. Therefore, $$$|S(i) \cup S(j)| = |S(i)|+|S(j)|$$$ when $$$i \neq j$$$.

1. For $$$n = 1$$$, $$$S(1) = [1]$$$.

2. For $$$n = 2$$$, $$$S(2) = [1+1,2]$$$.

3. For $$$n = 2m+1$$$, $$$m \geq 1$$$, $$$S(2m+1) = [~1+s~|~\forall s \in S(2m)]$$$.

4. For $$$n = 2m+2$$$, $$$m \geq 1$$$, $$$S(2m+2) = [~1+1+s~|~\forall s \in S(2m)] \cup [2s|~\forall s \in S(m+1)]$$$.

Example:

When $$$n = 7$$$:

$$$S(7) = [~1+s~|~\forall s \in S(6)]$$$

$$$S(6) = [~1+1+s~|~\forall s \in S(4)] \cup [2s|~\forall s \in S(3)]$$$

$$$S(4) = [~1+1+s~|~\forall s \in S(2)] \cup [2s|~\forall s \in S(2)]$$$

$$$S(3) = [~1+s~|~\forall s \in S(2)]$$$

Therefore,

$$$S(3) = [1+1+1,1+2]$$$

$$$S(4) = [1+1+1+1,1+1+2] \cup [2+2,4]$$$

$$$S(6) = [1+1+1+1+1+1,1+1+1+1+2] \cup [1+1+2+2,1+1+4] \cup [2+2+2,2+4]$$$

and so on.

It can be solved using a simple DP.

Consider $$$dp[i]$$$ is the number of ways you can make $$$i$$$. Now $$$dp[i] = \displaystyle\sum_{0}^{20} dp[i-x]$$$ where $$$x$$$ is a power of two. Now we have a problem of counting the same way several times, consider $$$i=3$$$, now we'll count $$$2,1$$$ and $$$1,2$$$ as two different ways while they are only one. So this needs a small modification.

Let's consider $$$dp[i][j]$$$ is the number of ways you can make $$$i$$$ while you are using $$$2^j$$$. Here, $$$dp[i][j] = dp[i-2^j][j] + dp[i][j+1]$$$. This means you can either subtract $$$2^j$$$ from $$$i$$$ and stay at the same power of two (so you can subtract it again in $$$dp[i-2^j][j]$$$), or let's now try the following power which is $$$j+1$$$. Like this we won't count the same way twice because we're subtracting the powers of two in order

Like this we'll consider the way only once. If things are not explained pretty well take a look at this problem: Coin Combinations II and icecuber's editorial: CSES DP section editorial.