10. (Solved)

Let b(n) denote the number of partitions of n into non-negative powers of 2. Prove that:

$\displaystyle \sum^\infty_{n=0}b(n)q^n=\prod^\infty_{n=0}(1-q^{2^n})^{-1}$

OK as I said I have solved that question.

10. Using question 10 prove the following three identities:

a. $\displaystyle b(2n+1)=b(2n)$

b. $\displaystyle b(2n)=b(2n-1)+b(n)$

c. $\displaystyle b(n)\equiv0$ mod 2 for n>1

I am strugling with this. I think identity a is fairly obvious because for each partition of 2n we create a partition of 2n+1 by adding the only odd part that is available (2^0).

I think identity c followis from identities a. and b. by induction.

However, I have no idea how to tackle identity b.