Partition of Integer into Powers of 2 for Consecutive Integers

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $\map b n$ denote the number of ways $n$ can be partitioned into (integer) powers of $2$.

Then:
 * $\map b {2 n} = \map b {2 n + 1}$

Proof
We prove the theorem by establishing a bijection between the set of partitions of $2n$ with that of $2n+1$,

under the constraint where each partition is an integer power of $2$.

For each partition of $2n+1$ into integer powers of $2$,

since $2^k$ is even for all $k>0$ and $2n+1$ is even, there is a part of size $1$.

Removing this part gives a partition of $2n$ into integer powers of $2$.

The inverse of this is adding a part of size 1 to a partition of $2n$.

This will give a partition of $2n+1$ into integer powers of $2$,

so both removing and adding are well-defined mappings, and so they are bijections.

Therefore $\map b {2n} = \map b {2n+1}$.