RESULTS: Erdős #126, the size-dichotomy arm (support run 95bb5cb7)

Hunt, not a result. Nothing here is evidence for or against RH (docs/08).

Verdict: the size dichotomy cannot close, and the place it fails is exact. The "controlled interval" horn works if and only if $\max A = \mathrm{rad}(S)^{o(1)}$. That threshold is now pinned from both sides. The "spread out" horn cannot supply it. For every set size strictly below the extremal one, the height of a primitive admissible set is unbounded, with explicit witnesses, and even a best-possible $abc$-shaped height bound would land at $\mathrm{rad}(S)^{1+o(1)}$, one full power of the radical above what the counting horn needs.

Reproduce with python3 hunts/support_95bb5cb7/probe.py (~20 s, stdlib only).

Notation follows hunts/r_186989: $S$ is a set of $k$ primes, $A$ a set of distinct positive integers with every off-diagonal sum $a+b$ $S$-smooth, $g(k) = \max |A|$, and the target is $\log g(k) = o(k)$. Write $\mathrm{rad}(S) = \prod_{p\in S} p$ and $\theta = \log \mathrm{rad}(S)$.

1. Normalisation, first, because it decides what "small diameter" can mean

Two facts, both proved, that fix the coordinates the dichotomy has to work in.

(N1) $S$-smooth scaling is a symmetry. If $A$ is admissible and $\lambda$ is $S$-smooth then $\lambda A$ is admissible, with the same cardinality. Conversely if $d = \gcd(A)$ then every off-diagonal sum is $d\cdot(\text{sum of }A/d)$, and a product is $S$-smooth iff both factors are, so $A/d$ is admissible and $d$ is $S$-smooth. Hence WLOG $\gcd(A) = 1$, and $g(k)$ is unchanged. Without this normalisation "diameter" is meaningless: $\lambda A$ has diameter $\lambda\cdot\mathrm{diam}(A)$ for every smooth $\lambda$.

(N2) Translation is not available. $A + t$ shifts every sum by $2t$, which destroys smoothness. So the only scale freedom is (N1), and after (N1) the height $\max A$ is a genuine invariant of the set. Everything below is about that invariant.

2. The counting horn, pinned from both sides

Lemma 1 (the correct counting step). Let $a_n = \max A$, $|A| = n$. The $n-1$ sums $a_j + a_n$ ($j<n$) are distinct $S$-smooth integers lying in $(a_n,\,a_n + a_{n-1}] \subseteq (N, 2N]$ with $N = \max A$. Hence

$$|A| \;\le\; 1 + \#\{\,m \le 2N : m \text{ is } S\text{-smooth}\,\} \;=\; 1 + \Psi(2N, S).$$

(The dyadic-block refinement $\Psi(2N)-\Psi(N)$ and the short-interval refinement $(N, N+a_{n-1}]$ are both available and neither changes anything on the logarithmic scale, which is the scale the target lives on. That is the whole of what the "large gaps" idea buys on this horn.)

Lemma 2 (threshold). Write $\log 2N = c\,\theta$. Then, uniformly in $k$,

$$\frac{\log \Psi(2N,S)}{k} \;=\; \Theta_c(1),\qquad \text{and } \to 0 \iff c \to 0 .$$

Both directions are rigorous and measured, with $S$ = the first $k$ primes (the choice that maximises $\Psi$, so this is the worst case, i.e. the statement is uniform over all $S$ with $|S| = k$):

• upper, Rankin: $\Psi(x,S) \le x^{\sigma}\prod_{p \in S}(1-p^{-\sigma})^{-1}$ for every $\sigma>0$, minimised numerically;
• lower, every lattice point of $\{e \ge 0: \sum e_i \log p_i \le \log x\}$ is an $S$-smooth integer $\le x$, and the point count is at least the simplex volume $(\log x)^k / (k!\prod_i \log p_i)$.

Read the table in both directions. Above $c \approx 1/2$ the lower bound is bounded away from zero, so at $\max A \ge \mathrm{rad}(S)^{1/2}$ the counting horn is not merely unproven, it is false: $\Psi$ really does have $e^{\Omega(k)}$ elements to hide a large $A$ in. Below $c \to 0$ the Rankin bound goes to zero, so the horn works. There is no third regime.

Consequence. The controlled-interval horn delivers $\log g(k) = o(k)$ iff it is handed $\max A = \mathrm{rad}(S)^{o(1)}$, equivalently $\log\max A = o(\theta(p_k)) = o(k\log k)$. Optimising the choice of $S$, the scale of $A$, the interval split, or the smooth-count estimate cannot move this: the threshold is a property of $\Psi$, not of the argument.

3. The other horn cannot supply it. Two independent reasons.

3.1 Measured: admissible sets of sub-extremal size have unbounded height

Exhaustive enumeration of primitive ($\gcd = 1$) admissible triples, through their three sums rather than through their elements: the sums are what is constrained, so bounding them is the honest universe. Every witness re-verified from scratch by full trial division.

$S=\{2,3\}$ (so $k=2$, $\mathrm{rad}=6$, $g(2)=4$):

The height tracks the cutoff linearly over eleven orders of magnitude. There is no height bound for $|A| = 3$, even at $k=2$, even after the only available normalisation. The same holds one size up: for $S=\{2,3,5\}$ the largest primitive 4-set element is 7 213 at cutoff $10^4$ and 98 099 at cutoff $10^6$.

This is what kills any descent. A descent argument reaches a smaller admissible set and asks for interval control there; at sub-extremal sizes there is none to be had.

3.2 Structural: the clique supplies no three-term $S$-unit relation

Height bounds for $S$-smooth numbers (Baker–Győry effectively, $abc$ conjecturally) are theorems about $x + y = z$ with all three terms $S$-smooth (or two smooth and one fixed). An admissible clique produces no such relation. The relations it does produce are

$$(a+b) + (c+d) \;=\; (a+c) + (b+d),$$

a four-term $S$-unit equation. For four terms there is no height theorem at all, only the Evertse–Schlickewei–Schmidt bound on the number of non-degenerate solutions, which is $\exp(O(k))$: the right order for the classical $f(n) \gg \log n$, and provably not improvable to $\exp(o(k))$ in general, since Erdős–Stewart–Tijdeman exhibit $S$ for which $x+y=1$ already has more than $\exp(c(k/\log k)^{1/2})$ $S$-unit solutions. A count is not a height, and the target needs a height.

And even granting the best imaginable outcome, a three-term relation plus $abc$, one gets $\max A \ll_\varepsilon \mathrm{rad}(S)^{1+\varepsilon}$, which is $c \to 1$ in Lemma 2, exactly where the table's lower bound reads $1.01\,k$. Not close: one full power of the radical, permanently.

4. Where the measured optima actually sit

$\varepsilon$ with $\max A = \mathrm{rad}(S)^{\varepsilon}$, over the witnesses of hunts/r_186989 §3 (all seven re-verified here, see §5):

$\varepsilon$ falls, which is the only encouraging number in this file, and it is not encouraging enough: Lemma 2 needs $\varepsilon \to 0$ and these are seven points of a staircase whose witnesses are not known to be the extremal ones (see §5, where $\varepsilon$ at $k=3$ turns out to be 1.81, not 1.13).

5. Audit of hunts/r_186989/RESULTS.md

Verified.

• All seven witnesses of §3 are admissible: distinct, positive, every off-diagonal sum $S$-smooth by full trial division. The lower bounds $g(k) \ge 2,4,5,6,8,10,11$ for $k=1..7$ stand.
• §2, $f(n-1)\le f_0(n)\le f(n)$: the proof is correct, and $f(2)=1$, $f_0(2)=0$ check out.
• §4, the Fekete argument: correct, and it is the most useful thing in that file. $g(1)=2$ (their 2-adic argument for $S=\{2\}$ is right, and parity handles odd $p$), so supermultiplicativity would force $\lim g(k)^{1/k} = \sup_k g(k)^{1/k} \ge 2$ and refute the conjecture. A composition law is a refutation route, not a proof route.

Corrected. §3 and door 2 assert "Every optimal witness has all elements $< 50$" and rank "the smooth-sum graph is locally starved" on that evidence. That is false. Growing all admissible triples with sums $\le 10^5$ for $S=\{2,3,5\}$ gives exactly six primitive extremal ($|A|=5$) sets:

$\{5,11,25,245,475\}$: the sums are $16, 30, 250, 480, 36, 256, 486, 270, 500, 720$, all $\{2,3,5\}$-smooth, $\gcd = 1$. Their branch-and-bound reported one witness per $k$ and the file generalised from it. The correct statement is that the smallest optimal witness is tiny; the optimal witnesses are not. Door 2's ranking should be re-read with that in mind.

Added. A small upper bound in a better universe. Their searches bound the elements; bounding the sums is exhaustive for the property being tested. For $S=\{2,3,5\}$ there is no 6-element admissible set all of whose pairwise sums are $\le 10^5$ (every such set contains a triple with sums $\le 10^5$, and the enumeration of those is complete). Likewise for $S=\{2,3\}$ there are exactly 8 primitive 4-sets with sums $\le 10^9$, all inside $[1,47]$. These are still bounded-universe statements and are not upper bounds on $g(k)$.

6. What we could not settle

• Whether extremal sets have bounded height. The data hints yes, at $k=3$ the six extremal 5-sets are identical at cutoffs $10^4$ and $10^5$, while 3-sets and 4-sets keep growing, but a proof would be circular for the dichotomy's purposes: it would presuppose knowing $g(k)$. And $475 = \mathrm{rad}^{1.81}$ is on the wrong side of Lemma 2 anyway.
• The ladder for $k=4$. The set-growing enumeration is $O(|{\rm sets}| \cdot \Psi)$ per level and exceeded ten minutes at $S=\{2,3,5,7\}$. A real clique solver over the sum-bounded universe would settle it.
• Nothing improves $f(n) \gg \log n$. This arm produces no upper bound on $g(k)$ and was never going to; it produces the reason.

The doors

This run measured a ceiling (the threshold in Lemma 2), so it owes the list.

1. Active constraints at the optimum.

2. Frozen-constant inventory.

3. Information class. Every door above stays inside the data this family reads, integers and the smoothness of their pairwise sums, and therefore cannot produce an upper bound on $g(k)$. Moving Erdős #126 requires reading more: a height theorem for four-term $S$-unit equations, which does not exist and is not a parameter of anything here. That is the same conclusion r_186989 reached from the other side, and this run makes it quantitative: the missing object is not "a better estimate", it is one power of $\mathrm{rad}(S)$.

Loose threads

• Four-term $S$-unit heights. The clique's only relation is $(a+b)+(c+d)=(a+c)+(b+d)$. Why it might matter: it is the exact obstruction, stated as a request to the literature rather than as a wall. First step: search for any height result (even conditional on $abc$) for $x_1+x_2 = x_3+x_4$ in $S$-units with a non-degeneracy hypothesis that a clique can supply.
• Extremal-height boundedness. Why it might matter: if extremal sets provably sit in $[1, \mathrm{rad}(S)^{C}]$ then $g(k)$ becomes computable exactly, which is the missing upper-bound half of r_186989's thread 2. First step: run the sum-bounded ladder at $k=4,5$ with a real clique solver and see whether the extremal sets stop moving as the cutoff grows.
• The $\varepsilon$ staircase. 1.58, 1.34, 1.81, 0.48, 0.44, 0.37, 0.30, and the $k=3$ entry only rose because we looked harder. Why it might matter: $\varepsilon \to 0$ is the conjecture, restated in the one coordinate that Lemma 2 says matters. First step: recompute $\varepsilon$ at $k=4..7$ over all extremal witnesses rather than one, as done here for $k=3$; the entries will rise and the shape of the rise is the datum.