Proof sketch: is the Schur forcing threshold exponential?

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This is the working sketch of an attack by Plasma AI on Problem 483. The new arguments are author-recorded and have not been independently examined. The full growth question remains open. The delivered results are obstructions: to the specific uniform recursions below on the upper side, and on the lower side to transfers of the triangle Ramsey palette rule on coset, coordinate subcube and residue-class interval partitions, and to two-sided reuse through fixed binary linear quotients. Other block shapes and nonlinear fibres remain outside these obstructions.

Target and source interfaces

A coloring of $[1,N]$ is Schur-free when no class contains $x$, $y$ and $x+y$, with $x=y$ allowed. Let $S(k)$ be the largest $N$ with a Schur-free coloring by at most $k$ colors, and $f(k)=S(k)+1$. The target is an absolute $C$ with $f(k)<C^k$ for every $k\ge1$; an eventual exponential bound suffices after enlarging $C$. Equivalently, the least number $H(n)$ of sum-free classes covering $[1,n]$ must satisfy $H(n)\ge c\log n$ for some $c>0$. A lower construction would refute the target only if its achievable values of $(N+1)^{1/k}$ were unbounded.

The problem page records exponential growth below, with base at least $380^{1/5}$ from Ageron et al., and the factorial upper bound $f(k)\le(e-\tfrac16)k!$ for $k\ge4$ from Eliahou's Corollary 2 through $f(k)\le R_k(3)-1$. Ageron's computer-found template is not checked here; only the cited source inequality is imported for numerical comparisons. Schur's tripling construction supplies the self-contained lower base 3. The conditional small-value Palomar lead and the formal-conjectures statement with sorry supply no asymptotic proof.

The accepted claim on Problem 183 gives $R_k(3)^{1/k}\to\infty$. This excludes an exponential upper bound through general triangle Ramsey numbers. The inequality $f(k)\le R_k(3)-1$ cannot transfer that lower bound to $f(k)$. The obstructions L261 and L266 consume the accepted root divergence at statement level only, without assuming anything about the source coloring's additive structure.

Where the classical recursion loses a factor

Schur's Hilfssatz selects a largest class, of size at least $N/k$, and forms differences from its least element. Those differences avoid its color. Repeating the selection among the remaining colors costs the reciprocal of the number of colors left, giving a factorial bound. The intermediate domains are sparse difference sets. A stronger step must preserve quantitative information from the initial interval.

Three uniform statements that fail

1. Long dilated prefixes. Suppose every $k$-coloring with $N\ge A^k$ had a color-omitting prefix $d[1,m]$ with $m\ge\delta N$, for fixed $\delta>0$. Its pullback would give $S(k)\le\max(A^k,S(k-1)/\delta)$ and hence an exponential upper bound, with a harmless prefactor and the shift from $S$ to $f$. The refuted claim L260 shows that this fails for every $\delta\in(0,1)$ and $1\le A<L$, where $L=\liminf S(n)^{1/n}$, even for arbitrarily large $k$. A block composition of a small-depth random Bohr coloring with a near-extremal coloring makes every sufficiently long dilated prefix see every color. The proof covers all such constants; its retained finite instance is illustrative. It leaves the extremal case $N=S(k)$ undecided.
2. Hereditary thinning of difference domains. The refuted claim L261 shows that a fixed positive fraction of an arbitrary vertex set need not lose a color from its differences. Differences of powers of four encode arbitrary triangle-free edge colorings, and the accepted Ramsey growth makes the required retention ratios arbitrarily small.
3. Uniform avoidance on Bohr cells. Let a cell be the intersection of $[0,N]$ with finitely many conditions $\{\theta_l y\}\in J_l$, with arc lengths at most a fixed $\rho$. Asking every sufficiently large cell to have a color on its differences avoiding some $\{x:\|\theta x\|<\rho\}$ also fails. The refuted claim L266 cuts a $0/1$ digit cube by a digit-sum-one condition, obtaining exactly a set of powers. Ramsey colorings on those powers make every color meet every rank-one neighborhood of radius $\rho$. The failure holds for every fixed $\rho$, even above $A^{|C(Y)|}$ for every fixed $A$, where $C(Y)$ counts the colors on the cell's differences. Requiring every arc to have length exactly $\rho$ does not avoid the example.

These are scoped obstructions, not a theorem that every recursion loses $k$. In particular L266 uses a separate, large palette outside the sparse difference set. It is consistent with an exponential Schur upper bound.

Held implications that survive

The Bohr avoidance page explains the following author-recorded proofs:

• If class $i$ avoids a Bohr neighborhood of rank $r_i$ and radius $\rho_i$, then $N+1\le\prod_i\lceil1/\rho_i\rceil^{r_i}$. The exception form is Lemma 1 of the proof of L262. This is a necessary test for any superexponential lower construction.
• A sum-free arc of $\mathbb R/\mathbb Z$ can be dilated by an integer to stay at distance at least $1/8$ from zero. Consequently colorings whose classes lie in single sum-free Bohr arcs have $N+1\le8^k$: L262.
• A color-avoiding refinement of a cell keeps at least a $1/q$ fraction, $q=\lceil1/\rho\rceil$, and removes a color. Iterating would prove an exponential bound. The implication in L265 is valid, but its universal antecedent is refuted by L266.
• For the positive half $Q_r^+$ of the signed cube $\{-1,0,1\}^r\setminus\{0\}$, let $\chi_r$ be the least number of colors avoiding monochromatic coordinatewise $v+w=u$. Pullback along binary values gives $\chi_r\le H(2^r-1)$. Thus $\chi_r\ge cr$ would imply $f(k)\le2^{k/c+1}$: L263. The linear lower bound itself remains the open claim L264.

The full digit cube and its digit-sum-one slice are distinct domains. Refuting avoidance on the latter does not refute L264. The signed-cube page retains the model's bounds and small values, the implication from sum-free colorings of $\mathbb F_2^n$, and the precise missing Cayley transfer from the accepted Ramsey construction.

The cube is also a gate for the other direction. If $\limsup S(k)^{1/k}=\infty$, then for every $A$ some $k$ has $S(k)\ge A^k$, and with $r=\lfloor k\log_2A\rfloor$ the pullback gives $\chi_r\le H(2^r-1)\le k\le(r+1)/\log_2A$. Since $\chi_{a+b}\le\chi_a+\chi_b$ (subadditivity on the signed-cube page), $\chi_r/r$ converges to its infimum, which is then $0$. So a superexponential answer forces $\chi_r=o(r)$: refuting L264 is necessary for it, and proving L264 suffices for the exponential answer. This short deduction is author-recorded here.

The lower-bound interface

A superexponential answer needs Schur colorings of intervals with unbounded $(N+1)^{1/k}$. Sum-free colorings of $\mathbb Z_m\setminus\{0\}$ supply Schur colorings of $[1,m-1]$, and every such family must pass the avoidance test of the first held implication above: eventually some class meets every Bohr neighborhood of fixed rank and radius. The bound $f(k)\le R_k(3)-1$ transfers nothing, and the accepted Ramsey construction colors arbitrary edges, not differences. Its mechanism is a palette rule: blocks carry recursive copies with some colors missing, every edge between two blocks gets a color missing on exactly one of them, and a color active inside a block also colors edges leaving it, with labels excluding the triangles this could close.

The palette-rule page shows that this mechanism fails under symmetric difference colorings. On any partition of a finite abelian group into cosets, or of the signed cube into subcubes, a color used inside a block colors no pair leaving it (in the cube, none whose difference avoids the support of an internal occurrence). The palettes then satisfy a Kraft inequality, $\sum_B2^{-t_B}\le1$, where $t_B$ counts the colors not used inside $B$. A recursion of such layers multiplies the size by at most $2$ per color: from leaves $\mathbb Z_{m_L}$ it reaches only $m\le m_L2^{k-k_L}$. The 2-adic coloring attains this, at base $2$. On groups of odd order the rule admits no partition into two or more blocks. This holds for mixed partitions, whose blocks use different subgroups, so changing the block architecture does not rescue the transfer in these settings. In groups the requested reuse of an internal color on a pair leaving its block is false, not merely unproved. In coordinate subcubes the more limited four-point obstruction suffices for the same Kraft inequality.

The page also treats intervals, the setting of Schur colorings. For residue-class partitions of $[0,N]$ with moduli dividing some $Q\le(N+1)/2$ the inequality holds up to one color per block, the color of a block's largest difference, so $\sum_B2^{-t_B}\le2$; and a recursion of such layers from leaves $[1,N_L]$ with $M_L$ colors has $N+2\le2^{3M-2M_L}(N_L+2)$, so it stays exponential.

The page leaves three lower-bound mechanisms open. First, partitions of an interval into progressions that are not residue classes. Second, blocks that are not cosets or subcubes; for twisted blocks of $\{0,1\}^4$ a retained witness shows that the Kraft inequality really fails in the signed cube; the retained coloring is a function neither of the support map nor of the binary value. So the cube is a weaker model than the groups for this question, and a palette gain there would not by itself give Schur colorings. Third, two-sided reuse outside the palette rule: a color used inside both endpoint blocks colors pairs between them, and its fibres must be independent in its own Cayley graph on both sides. The simplest instance, coset fibres with affine labels, is a precise model on the page and is also capped. Proposition 8 uses Brouwer's binary affine-cover theorem to prove $\dim W_J\le\dim U+k-k_U$ for some seed subspace $U\subseteq W_0$, where $k_U$ is the number of colors used on $U\setminus\{0\}$. In particular a zero seed gives $\dim W_J\le k$, however large the label ranks and the number of stages are. Its deletion step removes $t$ colors while losing at most $t$ dimensions; restriction preserves the fixed linear quotient maps. The external cover theorem and this new specialization are distinguished on the page; the latter is author-recorded, without independent review. The same bound holds with stage-dependent kernels if each new layer of a color lies in one nonzero coset whose direction avoids that color's earlier layers, condition (A) on the page. Thus changing the kernels under that avoidance invariant does not supply a gain. On cyclic groups with coprime factors the model's homomorphisms vanish, so the rule cannot even cover the nonzero coarse points with zero fine component.

Provisional premises and remaining gap

No unproved premise is used to claim an unconditional Schur bound. The conditional implications above retain their antecedents. The target's missing step is a color-elimination argument with an absolute loss that uses a quantitative feature of interval colorings absent from the sparse encodings, or a construction with unbounded exponential base. Every construction mechanism examined here stays exponential: the known template compositions have a base fixed by their template, Bohr-arc colorings stay below $8^k$ (L262), palette recursions stay at base $2$ over their leaves in groups and exponential on residue-class partitions of intervals, and the binary fixed-quotient model gains at most $k$ dimensions over its seed. These cover the specified models only.

The lower route has no surviving parameterized construction. To reopen two-sided reuse, exhibit fibres outside condition (A) satisfying both of the palette-rule page's exact fibre constraints, including coverage of every coarse coset, and show a gain over the seed's size per color. A concrete test uses fibres whose actual differences avoid the earlier class but whose affine direction spaces do not. Replacing such fibres by whole cosets would destroy independence. A cyclic-group construction must handle repeated summands as well; a binary-space or signed-cube gain still needs a separate interval transfer. A Bohr-set replacement needs a proof that its label invariant survives every extension. None of these inputs is supplied. No family beating base $380^{1/5}$ is known from the tested mechanisms.

On the upper side the open reformulation is the extremal prefix assertion: is there $\delta>0$ such that, for every $k\ge2$, every Schur-free coloring of $[1,S(k)]$ by at most $k$ colors has a color-omitting $d[1,m]$ with $m\ge\delta S(k)$? A proof gives the required constant-loss recursion, since such a prefix pulls back to a coloring of $[1,m]$ by $k-1$ colors and so $\delta S(k)\le S(k-1)$. A counterexample would have to be an exactly extremal coloring, and these are known only for $k\le5$, so finite checks cannot discriminate. L260's construction provides no fixed positive lower bound for its length divided by $S(k)$, so it supplies no counterexample to this assertion. One possible attack is an extension theorem: show that a coloring without such a prefix cannot be extremal by extending or reorganizing it onto a longer interval. No such theorem is held.

A separate surviving test is an adaptive Bohr recursion starting from $[0,N]$: prove that suitable choices of the frequency and a largest cell preserve an invariant excluding the sparse powers in L266. Merely giving the sets the name Bohr cells, or fixing their arc lengths, is insufficient. The invariant and its preservation are both missing. A proof of L264 would bypass this recursion, but no proof or sublinear cube coloring is supplied here.