When are all pairwise sums consecutive? What finite ascending sequences of integers $(a_1, \cdots, a_n)$, with $a_1 = 0$, are such that the sequence obtained by sorting all the pairwise sums $a_i + a_j\;\;(j > i)$ consists of ${n \choose 2}$ distinct consecutive integers?
One example is the sequence $(0, 2, 3, 4)$, whose ${4 \choose 2} = 6$ pairwise sums (after sorting) make up the sequence of consecutive integers $(2, 3, 4, 5, 6, 7)$.
Are there longer sequences with this property?
 A: I think we can do it by exhaustion.
Suppose $a_2=1$. Then it must be that $a_3=2$ for the sequence must include the successor of $0+1$. At this point, the pairwise sums are $(1,2,3)$, so we would need to fill in the gap with $a_4=4$, yieding pairwise sums $(1,2,3,4,5,6)$. We would need to fill in the gap with $a_5=7$, but then our pairwise sums become $(1,2,3,4,5,6,7,8,9,11)$, which are not in sequence. Notice that one may not solve this by setting $a_6=10$ for then $11=1+10=4+7$ has two sums.
We can do a similar reasoning for other values of $a_2$. Consider the table:
\begin{array}{ccc}
 a_2  & \text{Sequence} & \text{Pairwise sums}\\
1&(0,1,2,4)&1\text{ to } 6\\
2&(0,2,3,4)&2\text{ to } 7\\
3&(0,3)&3\\
4&(0,4)&4
\end{array}
And I think you can see where this is going.

Claim: If $a_2\geq 3$, the longest such sequence is $(0,a_2)$.

Indeed, for one to continue the sequence we'd need to see $a_3=a_2+1$,and at this point the sequence of sums tentatively looks like $(a_2,a_2+1,2a_2+1)$. Since the sequence is ascending and $2a_2+1$ is the smallest sum of nonzero terms, if the sequence could be extended we'd need to fill in all the gaps, setting $a_3=a_2+2$, $a_4=a_2+3$, etc.
But then $3a_2=a_2+2a_2=(a_2+1)+(2a_2-1)$ has multiple sums, so the sequence cannot be extended. Notice this works because $a_2+1 \geq 4$ terms are defined this way, so that the minimum four distinct terms required for the multiple sums are always available.
