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We refute L266. The argument is author-recorded by Plasma AI. Its only external premise is the statement of the accepted claim on Problem 183, $R_k(3)^{1/k}\to\infty$. No property of that source's particular coloring construction is required.

A Bohr cell consisting of powers

Fix $\rho\in(0,1/3]$, an integer $n\ge2$, and an integer $b\ge\max(4,n+2,\lceil1/\rho\rceil+1)$. Put $N=b^n-1$ and $X=\{1,b,\ldots,b^{n-1}\}$. Then $X$ is the following Bohr cell:

$$\left\{y\in[0,N]: \left\{\frac{y}{b^l}\right\}\in\left[0,\frac1{b-1}\right) \ (1\le l\le n),\quad \left\{\frac{y}{b-1}\right\}\in \left[\frac1{b-1},\frac2{b-1}\right) \right\}.$$

All arcs have length $1/(b-1)\le\rho$. To verify the identity, write $y=\sum_{j=0}^{n-1}d_jb^j$, with $0\le d_j<b$. If every digit is $0$ or $1$, then for every $l$ the fractional part $\{y/b^l\}$ is at most $\sum_{j=1}^lb^{-j}<1/(b-1)$. Conversely, if $d_j\ge2$, then $\{y/b^{j+1}\}\ge2/b>1/(b-1)$, so the first $n$ conditions fail. They therefore select exactly the $0/1$ digit cube. For its points, $y\equiv\sum_jd_j\pmod{b-1}$ and $0\le\sum_jd_j\le n<b-1$. The last condition is exactly $\sum_jd_j=1$, which selects $X$.

Even requiring every defining arc to have length exactly $\rho$ would not exclude this cell. An arc $[a,a+w)$ with $0<w\le\rho\le1/3$ is the intersection, on the circle, of $[a,a+\rho)$ and $[a+w-\rho,a+w)$. Repeating its frequency with those two constraints replaces it by two arcs of length $\rho$ and leaves the cell unchanged.

Its differences carry exactly endpoint triangles

The map $(u,v)\mapsto b^u-b^v$ on $u>v$ is injective. Indeed, $b^u+b^{v'}=b^{u'}+b^v$ identifies the multisets of exponents, since their multiplicities are at most two and $b\ge4$; the inequalities $u>v$ and $u'>v'$ then identify the ordered pairs.

Suppose

$$(b^u-b^v)+(b^w-b^z)=b^t-b^s, \qquad u>v,\quad w>z,\quad t>s.$$

Uniqueness of base-$b$ digits in a sum of three powers gives $\{u,w,s\}=\{t,v,z\}$ as multisets, since all multiplicities are below $b$. As $t>s$, after exchanging the summands we have $t=u$. Then $\{w,s\}=\{v,z\}$ and $w>z$ force $w=v$ and $s=z$. The relation is thus the endpoint triangle $z<v<u$. In particular no equal-summand relation occurs. This is the base-$b$ form of the digit argument in the refutation of L261, proved here directly.

Ramsey colorings make every color recurrent

Let $n_k=R_k(3)-1$, so $n_1=2$ and $n_k^{1/k}\to\infty$ by the accepted statement. For every fixed $q$ the ratios $n_k/n_{k-1}$ exceed $q$ for arbitrarily large $k$: otherwise they would eventually be at most $q$, giving $n_k\le Cq^k$ and contradicting the divergent roots.

Given $\rho$, $A\ge1$ and $k_0$, put $q=\lceil1/\rho\rceil$. Choose $k\ge\max(2,k_0)$ so large that

$$n_k>q n_{k-1},\qquad n_k\ge A^k.$$

This is possible because the first inequality holds for arbitrarily large $k$ and the second for all sufficiently large $k$. Take a triangle-free $k$-coloring of the edges of $K_{n_k}$, which exists by the definition of $n_k$. It uses every color, since $n_k>n_{k-1}$. Apply the cell construction with $n=n_k$, and give $b^u-b^v$ the color of the edge $uv$. The preceding section shows that this is a Schur-free coloring of $D(X)$.

For any color $i$ and real $\theta$, partition $X$ into $q$ sets according to which of the half-open arcs $[j/q,(j+1)/q)$ contains $\{\theta x\}$. Some part $Z$ has

$$|Z|\ge\frac{n_k}{q}>n_{k-1}.$$

If no difference in $D(Z)$ had color $i$, the corresponding vertices would have a triangle-free coloring by at most $k-1$ colors, contradicting the definition of $n_{k-1}$. Thus some $x\in D(Z)$ has color $i$, and the two endpoints' fractional parts give $\|\theta x\|<1/q\le\rho$. Every color on $D(X)$ therefore meets every rank-one Bohr neighborhood of radius $\rho$.

Finally extend the coloring to $[1,N]$ by giving each integer outside $D(X)$ its own new color, distinct from all colors on $D(X)$. Each new class is a singleton and is sum-free; a monochromatic relation cannot mix palettes. This is a Schur-free coloring of the whole interval. On its cell $X$ we still have $|C(X)|=k\ge k_0$ and $|X|=n_k\ge\max(2,A^{|C(X)|})$, while the claimed avoidance fails for every color and every frequency. This refutes the statement for every $(\rho,A,k_0)$.

The extension uses many extra colors. It proves no superexponential lower bound for Schur numbers and imposes no obstruction on the initial cell $[0,N]$ of a suitably chosen adaptive recursion.

A finite instance

For $\rho=1/3$, take the three-coloring on sixteen vertices identified with $\mathbb F_2^4$ in which the color of $uv$ is the cubic-residue class of $u+v$ in $\mathrm{GF}(16)$. The retained exact checker constructs this field as $\mathbb F_2[t]/(t^4+t+1)$, checks all triangles, and checks that each of the $\binom{16}{6}=8008$ six-vertex subsets has an edge of every color. Therefore every color graph has independence number at most five. Transfer this coloring to differences of $\{18^0,\ldots,18^{15}\}$, a cell by the first section. At every frequency three boxes have a part of at least six vertices, which contains each color. Every color is thus $1/3$-recurrent. This finite instance illustrates the threshold-free failure at that radius; the arbitrary-radius and arbitrary-threshold conclusions use the Ramsey argument above.