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  "body": "\n## Proof\n\nWe build a bijection, due to Glaisher, between the two sets of [[partitions|0404R9P9J3]].\n\n1. A partition of $n$ is determined by its multiplicity function $\\mu : \\Z_{\u003e0} \\to \\N$. Here $\\mu(k)$ is the number of parts equal to $k$. It has finite support and $\\sum_k k\\, \\mu(k) = n$.\n2. Conversely, each such function with finite support and $\\sum_k k\\, \\mu(k) = n$ is the multiplicity function of exactly one partition of $n$.\n3. Let $\\mathcal{O}$ be the set of these $\\mu$ with $\\mu(k) = 0$ for every even $k$. These are the partitions of $n$ into odd parts.\n4. Let $\\mathcal{D}$ be the set of these $\\mu$ with $\\mu(k) \\le 1$ for every $k$. These are the partitions of $n$ into distinct parts.\n5. Fact A: each positive integer $d$ is $2^a m$ for exactly one pair $(a, m)$ with $a \\in \\N$ and $m$ odd.\n6. Proof of Fact A. Let $a$ be the largest natural number with $2^a \\mid d$; it exists because $2^a \\le d$. Then $m = d / 2^a$ is odd, otherwise $2^{a+1} \\mid d$.\n7. If $2^a m = 2^b m'$ with $m, m'$ odd and $a \u003c b$, then $m = 2^{b-a} m'$ is even, a contradiction. So $a = b$, and then $m = m'$.\n8. Fact B: each $k \\in \\N$ is $\\sum_{a \\in B} 2^a$ for exactly one finite set $B \\subseteq \\N$. We call $B$ the binary set of $k$. We use [[strong induction|0KMVFSHFGZ]] on $k$.\n9. For $k = 0$ the only such set is $B = \\varnothing$, because a nonempty $B$ gives a positive sum.\n10. Let $k \\ge 1$. In $\\sum_{a \\in B} 2^a$ all terms with $a \\ge 1$ are even. So $0 \\in B$ exactly when $k$ is odd.\n11. Let $\\varepsilon = 1$ if $k$ is odd and $\\varepsilon = 0$ otherwise, and let $k' = (k - \\varepsilon)/2 \u003c k$.\n12. Then $k = \\sum_{a \\in B} 2^a$ holds if and only if $0 \\in B \\Leftrightarrow \\varepsilon = 1$ and $k' = \\sum_{a \\in B, a \\ge 1} 2^{a-1}$.\n13. By the induction hypothesis, $k'$ has exactly one binary set $B'$. So $B = \\{a + 1 : a \\in B'\\}$, together with $0$ when $\\varepsilon = 1$, is the only binary set of $k$.\n14. Define $\\Phi : \\mathcal{O} \\to \\mathcal{D}$. For $\\mu \\in \\mathcal{O}$ and odd $m$, let $B_m$ be the binary set of $\\mu(m)$.\n15. For $d = 2^a m$ as in Fact A, put $\\nu(d) = 1$ if $a \\in B_m$, and $\\nu(d) = 0$ otherwise. Let $\\Phi(\\mu) = \\nu$.\n16. The support of $\\nu$ is finite, because only finitely many $m$ have $\\mu(m) \u003e 0$ and each $B_m$ is finite.\n17. We have $\\sum_d d\\, \\nu(d) = \\sum_{m \\text{ odd}} m \\sum_{a \\in B_m} 2^a = \\sum_{m \\text{ odd}} m\\, \\mu(m) = n$, by Fact A. So $\\nu \\in \\mathcal{D}$.\n18. Define $\\Psi : \\mathcal{D} \\to \\mathcal{O}$. For $\\nu \\in \\mathcal{D}$ and odd $m$, let $\\mu(m) = \\sum_{a \\in A_m} 2^a$, where $A_m = \\{a \\in \\N : \\nu(2^a m) = 1\\}$. Let $\\mu(k) = 0$ for even $k$.\n19. Each $A_m$ is finite, and only finitely many $A_m$ are nonempty, because $\\nu$ has finite support. So $\\mu$ has finite support.\n20. By Fact A, $\\sum_k k\\, \\mu(k) = \\sum_{m \\text{ odd}} \\sum_{a \\in A_m} 2^a m = \\sum_d d\\, \\nu(d) = n$. So $\\Psi(\\nu) = \\mu \\in \\mathcal{O}$.\n21. Let $\\mu \\in \\mathcal{O}$ and $\\nu = \\Phi(\\mu)$. Then $A_m = B_m$ for each odd $m$, so $\\Psi(\\nu)(m) = \\sum_{a \\in B_m} 2^a = \\mu(m)$. Thus $\\Psi \\circ \\Phi$ is the identity.\n22. Let $\\nu \\in \\mathcal{D}$ and $\\mu = \\Psi(\\nu)$. By Fact B, the binary set of $\\mu(m)$ is $A_m$. So $\\Phi(\\mu)(2^a m) = 1$ exactly when $a \\in A_m$, that is, when $\\nu(2^a m) = 1$.\n23. By Fact A every $d \\ge 1$ has the form $2^a m$, so $\\Phi(\\mu) = \\nu$. Thus $\\Phi \\circ \\Psi$ is the identity.\n24. A map with a two-sided inverse is a [[bijection|01VG2A0BHW]]. So $\\Phi$ is a bijection from $\\mathcal{O}$ to $\\mathcal{D}$.\n25. Both sets are finite, since there are finitely many partitions of $n$. By the [[bijection principle|0GV2573FP4]], $\\abs{\\mathcal{O}} = \\abs{\\mathcal{D}}$.\n\n## Idea\n\nGroup equal odd parts: $m$ appears $\\mu(m)$ times. Write $\\mu(m)$ in binary and merge the copies of $m$ into parts $2^a m$, one for each binary digit $1$.\nThe new parts are distinct, because each integer is a power of two times an odd number in only one way. Splitting each part $2^a m$ back into $2^a$ copies of $m$ undoes the map.\n",
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