Say you take the integers $(1,\cdots,N)$ and color them with $c$ different colors, where $c < N$ and there is no constraint to use all colors. For a given $m < N$, is it possible to determine a lower bound on the longest coloring appearing in $(1, \cdots, m)$ that re-appears at least once more in $(1,\cdots, N)$?
E.g. if we have c = R,G,B and N = 10, and m = 4, then the sequence RRRGGGBBBR repeats "R", while RGRGRGBBBB repeats "RGRG". So a definite lower bound in this case would be 1 (due to a trivial application of the pigeonhole principle). I was thinking maybe this is covered by Ramsey theory, but could not find results that would cover this question.
Thanks
Edit: As Peter has pointed out, the lower bound for $m < c$ is $0$. So I'll add the additional constraint that $m > c$. Thank you for the answers so far.
We claim that the required minimax length of the reappearing colored sequence equals the largest $l$ such that $c^l<m-l+2$. Indeed, if $c^l<m-l+2$ then by the pigeonholes principle among the first $m-l+2$ sequences of length $l$ we find two matching. The first of them appears in $(1,\dots,m)$. On the other hand, if $c^l\ge m-l+2$ then we can first color the discrete segment $(1,\dots, m+1)$ according to a de Brujin sequence of order $l$ on a size-$c$ alphabet ending with $1^l$ and then continue to color $11\dots$.