An Interesting Property Concerning a Sequence of Integers A non-decreasing sequence of positive integers $a_1,a_2,\dots a_n\ (n\geq 3)$ is good if for each $3\leq k\leq n$ there are $1\leq i\leq j<k$ such that $a_i+a_j=a_k$.
Let $\ell,m$ be positive integers, and consider the set $[\ell]=\{1,2,\dots,\ell\}$. We say that $[\ell]$ is of type $P(m,1)$ if for any partition of $[\ell]$ into $m$ non-empty disjoint subsets $S_1,\dots,S_m$, there exists an $i\leq m$ such that one can choose, possibly with repetition, some elements in $S_i$ to form a good sequence. Otherwise $[\ell]$ is of type $P(m,0)$.
Let $f(m)$ denote the smallest positive integer of type $P(m,1)$. My question is, what is $f(3)$?
I managed to prove that
$\bullet$ if $\ell$ is of type $P(m,0)$, then so are $1,2,\dots,\ell-1$
$\bullet$ if $\ell$ is of type $P(m,1)$, then so are $\ell+1,\ell+2,\dots$
$\bullet$ $f(1)=2$
$\bullet$ $f(2)=5$
Indeed, note that $[4]=\{1,4\}\cup \{2,3\}$. So $4$ is of type $P(2,0)$, and therefore so are $1,2,3$. Now assume that some $n\geq 5$ is of type $P(2,0)$ and that $[n]=S_1\cup S_2$, where $S_1,S_2$ are non-empty disjoint subsets of $[n]$. WLOG, $1\in S_1,2\in S_2$. (Note that if $1,2$ are in the same $S_i$, then there would be a good sequence $1,1,2$.) Let $r$ be the largest element in $S_1$. If $r=1$, then $2,4\in S_2$ but $2,2,4$ is a good sequence. So $r>1$. If $r<n$, then $r+1\leq n$ and $r+1\in S_2$. Thus, $r-1=(r+1)-2\not\in S_2$, meaning that $r-1\in S_1$. Now $1,r-1,r$ is a good sequence in $S_1$, a contradiction. Therefore, $r=n$. But then $n-1=r-1\not\in S_1$. So $n-1\in S_2$, and $n-3=(n-1)-2\in S_1$. Finally, $3=n-(n-3)\in S_2$, and $5=2+3\in S_1$. Then $4=5-1\in S_2$, creating a good sequence $2,2,4$. We conclude that any $n\geq 5$ is of type $P(2,1)$.
$\bullet$ $12<f(3)$ (because $[12]=\{1,3,10,12\}\cup \{2,5,8,11\}\cup \{4,6,7,9\}$)
$\bullet$ $f(m)\leq \left[\sum_{j=0}^m\frac{1}{j!}\right]m!\ \forall m\geq 1$ (which follows from a repeated use of pigeonhole principle)
In particular, $13\leq f(3)\leq 16$
But I'm not able to see if $13,14,15$ are of type $P(3,1)$. So any help is appreciated. Thanks!
I'm also wondering if the following holds.
$$f(m)=\left[\sum_{j=0}^m\frac{1}{j!}\right]m!\ \forall m\geq 1$$
Equivalently, is it true that
$$(m+1)f(m)+1\leq f(m+1)\ \forall m\geq 1$$

Ah! Actually this problem is basically Schur's theorem. The upper bound given by the finite sum is far from tight. For instance, when $n=5$, the Schur number is shown to be $160$.
And it can be shown by graph coloring that when $n=3$, the Schur number is $13$.
