Now also posted to MathOverflow.
Consider a set of the form $\{a,b,a+b\}$ where $a$ and $b$ are positive integers with $b > a$. I will refer to such a set as a triplet. Consider now the problem of constructing $L$ disjoint triplets with as small a maximum element across them as we can manage. I will refer to a collection of $L$ disjoint triplets having the smallest maximum element across them as optimal. At best, the triplets will partition $\{1,2,\dots,3L\}$ giving a lower bound of $3L$ on the maximum element.
Here are some examples:
For $L = 1$, we have $\{1,2,3\}$ achieving the lower bound of $3 = 3L$.
For $L=2$, we have $\{1,3,4\}$, $\{2,5,7\}$ achieving $7 = 3L+1$.
It appears that this is optimal for $L=2$.
For $L=3$, we have $\{1,4,5\}$, $\{2,6,8\}$, $\{3,7,10\}$ achieving $10 = 3L+1$.
For $L=4$, we have $\{1,8,9\}$, $\{2,10,12\}$, $\{3,4,7\}$, $\{5,6,11\}$ achieving the lower bound of $12 = 3L$.
By random computer search, I was able to verify the existence of constructions with maximum element $3L+1$ for $L = 5,6,7$.
I am broadly interested in anything that would help with understanding this problem.
Can I always achieve $3L$ or $3L + 1$? If not, is there a good upper bound on the maximum element across $L$ disjoint triplets?
Ideally, I would hope that there is an explicit optimal construction hiding in the examples I've listed but I can't quite see the pattern.
For $L=5$, optimal is $3L=15$, achieved by $$\{\{1,14,15\},\{2,10,12\},\{3,8,11\},\{4,5,9\},\{6,7,13\}\}.$$
For $L\in\{1,\dots,100\}$, I found via integer linear programming that optimal is $3L$ if $L \pmod 4\in\{0,1\}$ and $3L+1$ if $L \pmod 4\in\{2,3\}$.
By request, here's the ILP formulation I used. Let $U$ be an upper bound on the elements. For $L\le 100$, you can take $U=3L+1$. Let $$T=\{(a,b,a+b): a \in \{1,\dots,U-1\}, b \in \{a+1,\dots,U-a\}\}$$ be the set of triplets. Let binary decision variable $x_{a,b,c}$ indicate whether triplet $(a,b,c)$ is selected. Let integer decision variable $z$ represent the maximum element used. The problem is to minimize $z$ subject to linear constraints \begin{align} z &\ge c\ x_{a,b,c} && \text{for $(a,b,c) \in T$} \tag1\label1 \\ \sum_{(a,b,c) \in T:\ i \in \{a,b,c\}} x_{a,b,c} &\le 1 && \text{for $i\in\{1,\dots,U\}$} \tag2\label2 \\ \sum_{(a,b,c) \in T} x_{a,b,c} &= L \tag3\label3 \end{align} Constraint \eqref{1} enforces the minimax objective. Constraint \eqref{2} selects each element at most once. Constraint \eqref{3} selects exactly $L$ triplets.
Because of \eqref{2}, you can strengthen \eqref{1} as $$z \ge c\sum_{(a,b,c)\in T} x_{a,b,c} \quad \text{for $c\in \{3,\dots,U\}$ such that $(a,b,c) \in T$}$$
Alternatively, you can perform a bisection search for the minimum $U$, solving a sequence of feasibility problems (no objective) defined by \eqref{2} and \eqref{3}.