For all integer $n$, let $b(n)$ be the number of partition of $n$ into power of two. For instance, $b(4)=4$, since \begin{align*} 4 &= 2^2 \\ &= 2^1+2^1 \\ &= 2^1+2^0+2^0 \\ &= 2^0+2^0+2^0+2^0. \end{align*}
I want to show the congruence $b(n) \equiv 0 \pmod 2$ for all $n \ge 2$ using combinatorial proof. Recently, I have proved this congruence by strong induction on $n$. But, I want to know about the combinatorial proof of this congruence, that is, using some rules or a bijection between partition to other partition.
But, I didn't have any idea to show it.
What I tried is as follows.
I was thinking for a rule that we should divide each binary partition of $n$ into two classes/sets with the same cardinality based on whether it contains $2^0$ or not. For example, consider $n=2,3,4,5$. The binary partition of them, respectively, are $$2^1, 2^0+2^0,$$ $$2^1+2^0, 2^0+2^0+2^0,$$ $$2^2,2^1+2^1,2^1+2^0+2^0,2^0+2^0+2^0+2^0,$$ and $$2^2+2^0,2^1+2^1+2^0,2^1+2^0+2^0+2^0,2^0+2^0+2^0+2^0+2^0.$$
We can divide each binary partitions of $2$ into two sets, namely, $A=\{2^1\}$ and $B=\{2^0+2^0\}$. Similarly for $n=3,4,$ and $n=5$. But, for $n=6$, this rule didn't work, since \begin{align*} 6 &= 2^2+2^1 \\ &= 2^2+2^0+2^0 \\ &= 2^1+2^1+2^1 \\ &= 2^1+2^1+2^0+2^0 \\ &= 2^1+2^0+2^0+2^0+2^0 \\ &= 2^0+2^0+2^0+2^0+2^0+2^0. \end{align*}
If we apply the same rule as above, we must have that all binary partitions of $6$ divided to two sets, namely, $$A=\{2^2+2^1,2^1+2^1+2^1\}$$ and $$B=\{2^2+2^0+2^0,2^1+2^1+2^0+2^0,2^1+2^0+2^0+2^0+2^0,2^0+2^0+2^0+2^0+2^0+2^0\}.$$ But then, $|A|=2\ne 4 =|B|$.
Any idea please? I really need an insight about this rule to show the congruence of $b(n)$. Many thanks in advanced.
Let $A$ be the set of partitions with an even number of parts and $B$ the set of partitions with an odd number of parts. Consider the function $f:A\to B$ which sends a partition $p = [2^{a_1},\dots,2^{a_k}]$ where $a_1 \leq a_2 \leq \dots \leq a_k$ to
$$p'=\begin{cases} [2^{a_1},\dots, 2^{a_k-1}, 2^{a_k - 1}], \text{ if } a_k\neq a_{k-1} \\ [2^{a_1}, \dots, 2^{a_{k-2}}, 2^{a_k+1}], \,\text{ if } a_k = a_{k-1}\end{cases}$$
It's easy to see that when defined that way, $f$ is bijective, so $|A|=|B|$. But $b(n) = |A|+|B| \equiv 0 \pmod{2}$.