# Equsum Partitions of Sets of Positive Integers +
## 2 Initial Intervals of \(\mathbb{Z}^{+}\)
: _For \(V=[1,\ 14]\) the following is a consesum partition of order 10, with defect \(\delta=1\):_  
\[\{1,\ 5\},\{7\},\{2,\ 6\},\{9\},\{10\},\{3,\ 8\},\{12\},\{13\},\{14\},\{4,\ 11\}.\]  
_Its direct sum with the reversed trivial consescum 10-partition of \([1,\ 10]\) yields an equisum 10-partition of \([1,\ 24]\) with defect \(\delta=1\):_  
\[b=10,\ s=30:\ \{1,\ 5,\ 24\},\{7,\ 23\},\{2,\ 6,\ 22\},\{9,\ 21\},\{10,\  
To generalise this example, it is notationally convenient to denote the set of _triangular numbers_ by \(T:=[t_{k}:=(k-1)k/2:k\in\mathbb{Z}^{+}]\). Note that \(t_{k}+k=t_{k+1}\).  
**Lemma 4**.: _For \(k,\ m\in\mathbb{Z}^{+},\ k\geq 3\), let \(b:=t_{k}\). The set \([1,(k-1)(mk+1)]\) has an equisum \(b\)-partition with defect \(\delta=1\)._  
Proof.: For any \(k\in\mathbb{Z}^{+},\ k\geq 3\), let \(\mathcal{F}:=\{F_{i}|1\leq i\leq b:=t_{k}\}\) be the \(b\)-partition of \(I_{k+1}:=[1,\ k_{k+1}-1]=[1,\ k-1+b]\) with blocks  
\[F_{i} :=\{k+i\},\ 1<i<t_{k},\ i\
otin T;\] \[F_{t_{j}} :=\{j,\ k+t_{j}\},\ \ 1\leq j\leq k-1.\]  
Since \(\Sigma F_{t_{j}}=k+t_{j+1}\) for \(1\leq j\leq k-1\), it follows that \(\{\Sigma F_{i}|i\in I_{k}\}=[k+1,\ t_{k+1}]\), so \(\mathcal{F}\) is consecsum with defect \(\delta=1\). Let \(\mathcal{E}:=\{E_{i}:=\{i\}|i\in I_{k}\}\), so \(\mathcal{F}^{*}:=\mathcal{F}\oplus\mathcal{E}^{R}\) is a defect \(\delta=1\) equisum \(b\)-partition of \([1,\ k-1+2b]= \begin{bmatrix}1,\ k^{2}-1\end{bmatrix} \). Let \(\mathcal{A}:=\mathcal{E}\oplus\mathcal{E}^{R}\) and \(\mathcal{A}^{(1)}:=\mathcal{A}\), \(\mathcal{A}^{(m+1)}:=\mathcal{A}^{(m)}\oplus\mathcal{A}\) for all \(m\in\mathbb{Z}^{+}\). Then \(\mathcal{A}^{(m)}\) is a uniform b-partition of \([1,\ 2 mb]=[1,\ mk(k-1)]\). Hence \([1,\ (k-1)(mk+1)]\) has the defect \(\delta=1\) equisum \(b\)-partition \(\mathcal{F}^{*}\oplus\mathcal{A}^{(m-1)}\), for any \(m\geq 2\). Lemma 4 describes a family of sets \([1,n:=(k-1)(mk+1)]\) with an equisum partition of order \(b\) such that \(2b=(k-1)k\) and \(k-1|n,\ k|n+1\). In particular, \(k=4\) leads to an order \(b=6\) equisum partition of \([1,\ 3+12m]\) with defect \(\delta=1\), starting with the order \(6\) partition of \([1,\ 15]\) given in Example 2. Similarly, \(k=5\) leads to an order \(b=10\) equisum partition of \([1,\ 4+20m]\) with defect \(\delta=1\), starting with the order \(10\) partition of \([1,\ 24]\) in Example 3. **Example 4**. : _Here is an order \(6\) consecsum partition for \(V=[1,\ 14]\), with defect \(\delta=1\):_  
\[n=14,\ b=6:\ \{2,\ 4,\ 9\},\{3,\ 13\},\{5,\ 12\},\{1,\ 7,\ 10\},\{8,\ 11\},\{6,\ 14\}\]  
_and an order \(10\) consecsum partition for \(V=[1,\ 25]\), with defect \(\delta=1\):_  
\[n=25,\ b=10:\ \{4,\ 11,\ 13\},\{1,\ 8,\ 20\},\{2,\ 9,\ 19\},\{6,\ 7,\ 18\},\{15,\ 17\},\] \[\{3,\ 5,\ 25\},\ \{10,\ 24\},\{12,\ 23\},\{14,\ 22\},\{16,\ 21\}.\]  
_Each of these partitions yields a defect \(\delta=1\) equisum \(b\)-partition when we form its direct sum with the reversed trivial consecsum \(b\)-partition of \([1,b]\). With Lemma 1 it now follows for all \(m\in\mathbb{Z}^{+}\) that \([1,\ 8+12m]\) has an equisum \(6\)-partition, and \([1,\ 15+20m]\) has an equisum \(10\)-partition, with defect \(\delta=1\), in all instances._  
As 12 is not a triangular number, we cannot use Lemma 4 for the case \(b=12\), but the method used in Example 4 can be applied. **Example 5**. : _Here are defect \(\delta=1\) order \(12\) consecsum partitions for the intervals \(V=[1,\ 20]\) and \(V=[1,\ 27]\):_  
\[n=20,\ b=12:\ \{12\},\{13\},\{14\},\{5,\ 10\},\{1,\ 15\},\{8,\ 9\},\]  
\[\{7,\ 11\},\{19\},\{2,\ 18\},\{4,\ 17\},\{6,\ 22\},\{3,\ 20\}.\]  
\[n=27,\ b=12:\ \{4,\ 22\},\{11,\ 16\},\{3,\ 25\},\{10,\ 19\},\{6,\ 24\},\{1,\ 12,\ 18\},\]  
\[\{9,\ 23\},\{7,\ 26\},\{14,\ 20\},\{5,\ 13,\ 17\},\{15,\ 21\},\{2,\ 8,\ 27\}.\]  
_Equisum \(12\)-partitions with defect \(\delta=1\) result by forming direct sums with the reversed trivial consecsum \(12\)-partition of \([1,\ 12]\).