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Thread: 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:

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

    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.
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  2. #2
    MHF Contributor

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

    $\displaystyle 2n=1 + \sum_{j \neq 1} 2^{m_j}$ and $\displaystyle \sum_{j \neq 1} 2^{m_j}$ 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: $\displaystyle
    \sum\limits_n {b\left( n \right) \cdot x^{2n} } = \prod\limits_n {\left( {\tfrac{1}
    {{1 - x^{2^{n + 1} } }}} \right)} = \left( {1 - x} \right) \cdot \prod\limits_n {\left( {\tfrac{1}
    {{1 - x^{2^n } }}} \right)} = \left( {1 - x} \right) \cdot \sum\limits_n {b\left( n \right) \cdot x^n }
    $

    i.e. $\displaystyle
    \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 $\displaystyle
    b\left( n \right) - b\left( {n - 1} \right) = \left\{ \begin{gathered}
    0{\text{ if }}n{\text{ is odd}} \hfill \\
    b\left( {\tfrac{n}
    {2}} \right){\text{ if }}n{\text{ is even}} \hfill \\
    \end{gathered} \right.
    $ which proves both, (a) and (b)
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