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Math Help - Partitions of n into non-negative powers of 2.

  1. #1
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    Partitions of n into non-negative powers of 2.

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

    \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. b(2n+1)=b(2n)
    b. b(2n)=b(2n-1)+b(n)
    c. 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.
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  2. #2
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    your ideas for part a) and c) are correct. for part b) let 2n=\sum 2^{m_j}. if all m_j are > 0, then 2n=2 \sum 2^{m_j -1}, and \sum 2^{m_j - 1} is a partition of n. if at least one of m_j is 0, say m_1=0, then

    2n=1 + \sum_{j \neq 1} 2^{m_j} and \sum_{j \neq 1} 2^{m_j} is a partition of 2n - 1, and you're done. this also tells us how to find partitions of 2n when the partitions of n and 2n - 1 are given. for example for n = 3:

    partitions of 3: 1 + 1 + 1, 1 + 2.

    partitions of 5: 1 + 2 + 2, 1 + 4, 1 + 1 + 1 + 2, 1 + 1 + 1 + 1 + 1.

    now to find the partitions of 2n = 6, we multilpy every partition of n = 3 by 2, and add 1 unit to every partition of 2n - 1 = 5 to get all 6 partitions of 6:

    2 + 2 + 2, 2 + 4, 1 + 1 + 2 + 2, 1 + 1 + 4, 1 + 1 + 1 + 1 + 2, 1 + 1 + 1 + 1 + 1 + 1.
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  3. #3
    Super Member PaulRS's Avatar
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    b) We have: <br />
\sum\limits_n {b\left( n \right) \cdot x^{2n} }  = \prod\limits_n {\left( {\tfrac{1}<br />
{{1 - x^{2^{n + 1} } }}} \right)}  = \left( {1 - x} \right) \cdot \prod\limits_n {\left( {\tfrac{1}<br />
{{1 - x^{2^n } }}} \right)}  = \left( {1 - x} \right) \cdot \sum\limits_n {b\left( n \right) \cdot x^n } <br />

    i.e. <br />
\sum\limits_n {b\left( n \right) \cdot x^{2n} }  = \sum\limits_n {\left[ {b\left( n \right) - b\left( {n - 1} \right)} \right] \cdot x^n }

    Thus <br />
b\left( n \right) - b\left( {n - 1} \right) = \left\{ \begin{gathered}<br />
  0{\text{ if }}n{\text{ is odd}} \hfill \\<br />
  b\left( {\tfrac{n}<br />
{2}} \right){\text{ if }}n{\text{ is even}} \hfill \\ <br />
\end{gathered}  \right.<br />
which proves both, (a) and (b)
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