Limits of progression and rearrangement transfers

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The arguments here are author-recorded formulation checks by Plasma AI. External results retain the standing at their linked homes. A failed implication below does not refute the corresponding Erdős problem.

Root growth does not control consecutive quotients

For Problem 138, the accepted bound recorded on the claim page is $W_r(k)>k^{k\lfloor\log_2r\rfloor/100000}$ for all sufficiently large $k$. At $r=2$ it gives $W(k)^{1/k}>k^{1/100000}\to\infty$, answering the displayed example question. This uses the recorded bound, not a new examination of its proof.

The formal-conjectures file also asks whether $W(k+1)/W(k)\to\infty$ in variants.quotient (statement read on 2026-10-14). Root growth alone does not supply that assertion, even together with monotonicity and growing differences. For $k\ge2$ put

$$b_k=2^{4^{\lfloor\log_2k\rfloor}}+k^2.$$

The sequence is increasing and $b_{k+1}-b_k\ge2k+1$. If $2^j\le k<2^{j+1}$, then $\log_2(b_k)/k\ge4^j/2^{j+1}=2^{j-1}$, so $b_k^{1/k}\to\infty$. But at $k=2^j$ the two exponential terms in $b_k$ and $b_{k+1}$ agree, and $b_{k+1}/b_k=1+(2k+1)/b_k\to1$. This is a counterexample to the numerical implication, not a proposed model of van der Waerden numbers. Any route to the quotient question needs additional information comparing adjacent values. A formulation check should stop once it has separated that question from the root limit.

A natural-number permutation needs a new integer extension

Problem 195 concerns a bijection $\mathbb N\to\mathbb Z$. The known alternatives for the largest unavoidable monotone progression length are $3$ and $4$. A construction of a four-term-progression-free permutation of the natural numbers cannot be used as an integer permutation without an additional extension argument.

The Conjectures.io exposition of the natural-number construction uses positivity in Lemma 2.1 and finite saturation in Lemma 3.3. The former bounds the next terms of a progression by $c=2b-a\le2b$ and $d=2c-b\le2c$ for nonnegative inputs. For the integer progression $(-3,0,3,6)$, already $c\le2b$ fails.

The direct integer version of the saturation condition has a separate obstruction. A condition requiring

$$b\in A,\quad \rho(a)<\rho(b),\quad a\le2b \quad\Longrightarrow\quad a\in A$$

cannot always be imposed on a finite integer set containing a prescribed element. Set $\rho(0)=1$ and $\rho(a)=0$ for every negative $a$, and require $0\in A$. The condition forces every negative integer into $A$. The weighted upper cutoff used for nonnegative integers likewise leaves all negative integers below a positive cutoff. These examples rule out literal transfer of those two steps. They do not rule out a different two-sided extension theorem or an integer construction. Reopening this route requires a finite-prefix extension that covers both signs and preserves all future four-term constraints.

An extremal common point is still missing

For Problem 272, let

$$B(N)=\binom N2+1+\left\lfloor\frac{N-1}{4}\right\rfloor.$$

The source account on [[problems/additive_combinatorics/E0272/claims/2026_07_25_yang|Yang's claim page]] gives a lower construction of this size and the upper bound $B(N)$ for families with a common element. These two facts would determine the unrestricted maximum if some maximum family had a common element. The assertion that every maximum family has one is stronger than needed, but neither assertion follows from the accepted linear-error estimate.

The $N\ge10000$ results in Section 7 of Yang's paper (Theorem 7.5 and Proposition 7.6) retain the additional hypothesis that every member with at least four elements is itself an arithmetic progression, and leave a counting endgame. They are not an unrestricted exact formula above $10000$, so they do not justify a finite complement $N<10000$.

A private triple cannot simply replace a larger member. In the family $\{\{1,2,4,6\},\{6\}\}$ the only intersection is the nonempty progression $\{6\}$. The triple $\{1,2,4\}$ belongs only to the first member, but replacing that member by the triple makes the intersection empty. Thus private-witness counting is not by itself a compression preserving admissibility. A useful next theorem must either remove the AP-member hypothesis or prove existence of a maximum common-point family; finite computations at larger $N$ do neither.

Eventual rearrangement is not a numbered finite range

For Problem 475, the recorded range theorems cover all subset sizes for sufficiently large primes. One way to compose them uses the small range $t\le\log p/\log\log p$, the medium range $C_\alpha\le t\le p^{1-\alpha}$, and the large range $t\ge p^{1-c}$ for an absolute $c>0$. Fix $0<\alpha<\min(c,1/2)$. The medium and large ranges overlap, and the small range eventually reaches $C_\alpha$. This proves eventual coverage conditional on the three source theorems recorded on the [[problems/additive_combinatorics/E0475/claims/2026_02_17_pham_sauermann|range-chain claim page]].

It does not provide a numerical prime cutoff. Section 5 of the [[../library/additive_combinatorics/pham_2026_graham_s_rearrangement_conjecture/_index|medium-range paper]] chooses $D=\lceil3/\alpha\rceil$ and constrains $C_\alpha$ using anticoncentration constants $C_1,C_D$ that are not numerically extracted in that account. The large-range paper still requires sufficiently large group order, including in Theorem 9.11 with its displayed $\gamma=2^{-20}$. An explicit exponent is not an explicit order threshold. No claim of inherent ineffectivity is intended.

If a numerical $P$ were proved sufficient, an exact residual certificate would specify every prime $p<P$ and every subset $A\subseteq\{1,\ldots,p-1\}$ not otherwise covered, together with a permutation of each $A$. A checker would verify bijectivity and pairwise distinctness modulo $p$ of the nonempty prefix sums. It must not compare them with the empty prefix: the target permits final sum zero. Without $P$, this describes a conditional checker interface, not a finite test that has been run or authorized by the theorem. The route stops at constant extraction unless those constants are supplied.

Two different sum restrictions

Problem 787 forbids the sum of two distinct elements of the selected set $B$ from landing anywhere in the ambient set $A$. In Problem 792, the sum is tested only against $B$, but equal summands are also forbidden. Neither restriction can be substituted for the other in an imported bound.

The real-to-integer reduction used for Problem 787 can be made explicit without a quantitative loss. For finite $A\subset\mathbb R$, choose a basis over $\mathbb Q$ for its rational span. After clearing all coordinate denominators, associate to each $a\in A$ an integer polynomial $P_a(T)$. For each pair $a\ne b$, $P_a-P_b$ is a nonzero polynomial. For each triple with $a+b\ne c$, $P_a+P_b-P_c$ is nonzero. There are finitely many such polynomials, so choose a positive integer $T$ outside their finite sets of roots. The map $a\mapsto P_a(T)$ is injective and preserves both directions of $a+b=c$ on $A$. It therefore preserves exactly the feasible subsets in Problem 787, including distinctness of the two summands. This proves the reduction to integers, not to positive integers: translating the image need not preserve $a+b=c$. It does not close the gap between the known growth bounds.

For Problem 792, zero belongs to no sum-free set because $0+0=0$. Deleting zero costs at most one ambient element, so asymptotic leading terms transfer between conventions allowing and excluding zero, but exact finite bounds require the corresponding change from $n$ to $n-1$. The existing problem account already records the main term $n/3$ and the unresolved error term.