# Equsum Partitions of Sets of Positive Integers +
## 4 Divisor Sets  
For any \(n\in\mathbb{Z}^{+}\) the divisor set of \(n\) is \(D(n):=\{d\in\mathbb{Z}^{+}:d|n\}\), and  
\[\Sigma D(n):=\sigma(n)=\prod\Biggl{\{}\sigma(p^{a})=\frac{p^{a+1}-1}{p-1}:\ p^{a}\in P(n) \Biggr{\}},\]  
where \(P(n):=\{p^{a}:p\in\mathbb{P},\ a\geq 1,\ p^{a}\parallel n\}\) is the set of maximal prime-power divisors of \(n\). We call \(\omega(n):=\bigl{|}P(n)\bigr{|}\) the _rank_ of \(n\) : it is simply the number of distinct prime factors of \(n\). For any \(b\in\mathbb{Z}^{+},\ b\geq 2\), a necessary condition for \(D(n)\) to have an equisum \(b\)-partition is \(b\bigl{|}\sigma(n)\) and \(s:=\sigma(n)/b\geq n\), since \(n\in D(n)\). Thus, we need \(\sigma(n)\geq bn\), so the \"boundary case\" \(\sigma(n)=2n\) requires \(n\) to be _perfect_[6]; in all other cases, \(\sigma(n)>2n\), so \(n\) is _abundant_. No prime-power is perfect or abundant, since \(\sigma(p^{a})<2p^{a}\) always holds: for \(D(n)\) to have an equisum \(b\)-partition with \(b\geq 2\), the rank of \(n\) must be at least 2. Indeed, order \(b\geq 3\) requires a rank of at least 3, order \(b\geq 5\) requires rank at least 6, and so on. If \(n\) is odd and has rank 2 then \(\sigma(n)<15n/8\), so an equisum 2-partition of \(D(n)\) for \(n\) of rank 2 is only possible if \(2|n\). If \(n=2^{a}p\) for some \(a\in\mathbb{Z}^{+},\ p\in\mathbb{P},\ p\geq 3\), then  
\[\sigma(n)=\big{(}2^{a+1}-1\big{)}(p+1)\geq 2n\Longleftrightarrow 2^{a+1}\geq p+1\]  
so \(D(2^{a}p)\) can have an equisum 2-partition only if \(2^{a+1}\geq p+1\). **Lemma 5**. : _Let \(a\in\mathbb{Z}^{+},\ p\in\mathbb{P},\ 2^{a+1}-1\geq p\geq 3\). For every \(m\in\mathbb{Z}^{+},\ m\geq a\), the divisor set \(D(2^{m}p)\) has an equisum 2-partition._  
Proof. : Let \(k:=(p+1)/2\). Choose any \(m\in\mathbb{Z}^{+},m\geq a\), and let  
\[B_{1}:=D\big{(}2^{m-1}p\big{)},\ B_{2}:=\{2^{m},\ 2^{m}p\}.\]  
Then \(\mathcal{B}(m):=\{B_{1},B_{2}\}\) is a 2-partition of \(D(2^{m}p)\), with block sums  
\[s_{1}:=\Sigma B_{1}=(2^{m}-1)(p+1),\ \ s_{2}:=\Sigma B_{2}=2^{m}(p+1)=s_{1}+2k.\]  
To begin, suppose \(k=2^{a}\) and \(m=a\). (This is the case in which \(2^{a}p\) is perfect.) Then the block sums of \(\mathcal{B}(a)\) are  
\[s_{1}=2k(k-1),\ \ s_{2}=2k^{2}\]  
so the transfer  
\[B_{1}^{\prime}:=B_{1}\cup\{2^{a}\}=B_{1}\cup\{k\},\ B_{2}^{\prime}:=B_{2}\backslash \{2^{a}\}=\{2^{a}p\}=\{k(2k-1)\}\]  
produces a new 2-partition \(\mathcal{B}^{\prime}(a):=\big{\{}B_{1}^{\prime},B_{2}^{\prime}\}\) of \(D(2^{a}p)\) which is equisum, since its block sums are  
\[s_{1}^{\prime}:=\Sigma B_{1}^{\prime}=s_{1}+k=k(2k-1),\ s_{2}^{\prime}:= \Sigma B_{2}^{\prime}=s_{2}-k=k(2k-1).\]  
Note that \(B_{1}^{\prime}=D(2^{a}p)\backslash\{2^{a}p\}\), \(B_{2}^{\prime}=\{2^{a}p\}\) and \(s_{1}^{\prime}=s_{2}^{\prime}=2^{a}p\). Now suppose \(k=2^{a}\) and \(m>a\), or \(\ k<2^{a}\) and \(m\geq a\). (In this case \(2^{m}p\) is abundant.) Recall that the block sums of \(\mathcal{B}(m):=\{B_{1},B_{2}\}\) satisfy \(s_{2}=s_{1}+2k\). Let \(r:=2^{m}-k\). Then \(0<r<2^{m}\) because either \(k=2^{a}<2^{m}\) or \(k<2^{a}\leq 2^{m}\). Since \(B_{1}\) contains \(\big{\{}2^{i}:0\leq i\leq m-1\big{\}}\), there is some set \(R\subseteq B_{1}\) such that \(\Sigma R=r\) matches the binary representation of \(r\). Now the exchange  
\[B_{1}^{\prime}:=(B_{1}\backslash R)\cup\{2^{m}\},\ \ B_{2}^{\prime}:=(B_{2} \backslash\{2^{m}\})\cup R\]  
defines a 2-partition \(\mathcal{B}^{\prime}(m):=\big{\{}B_{1}^{\prime},B_{2}^{\prime}\big{\}}\) of \(D(2^{m}p)\) which is equisum, since  
\[s_{1}^{\prime}:=\Sigma B_{1}^{\prime}=s_{1}-r+2^{m}=s_{1}+k,\ \ s_{2}^{\prime}:= \Sigma B_{2}^{\prime}=s_{2}-2^{m}+r=s_{2}-k=s_{1}+k.\]  
This construction explicitly satisfies the claim. When \(a,m\in\mathbb{Z}^{+}\), the divisor set \(D(2^{a}p^{m})\) contains exactly \(m+1\) odd factors of \(2^{a}p^{m}\), so the block sums of any 2-partition of \(D(2^{a}p^{m})\) have the same parity if and only if \(m\) is odd. Hence, an equisum 2-partition of \(D(2^{a}p^{m})\) is only possible when \(m\) is odd. Suppose \(m\) is odd and \(2^{a+1}\geq\sigma(p^{m})\).