RESULTS: Erdős #126, bounded reconnaissance (r_186989)

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

Verdict: the scout is dead, on a kill condition the brief pre-registered. Arm 1 produced suggestive finite data and no iterable lemma; arm 2 produced no $\exp(o(k))$ clique bound; arm 3 turned out to point the wrong way. Three things are worth keeping anyway, and they are the reason this file is longer than "no".

Reproduce with python hunts/r_186989/probe.py (~2 min, stdlib only, writes results.json).

1. The restatement, which is what made the arms comparable

$f$ and $g$ are inverse non-decreasing staircases:

$$g(k) = \max\{n : f(n) \le k\}, \qquad f(n)/\log n \to \infty \iff g(k) = \exp(o(k)) \iff g(k)^{1/k} \to 1 .$$

$g(k)$ is finite for every $k$ precisely because the classical $f(n) \gg \log n$ holds; the conjecture is the statement that its exponential growth rate is exactly $1$. Everything below is a measurement of $g$.

2. Arm 0: the Formal Conjectures positivity mismatch. Settled.

Let $f$ be the minimum over $n$-sets of positive integers and $f_0$ the minimum over $n$-sets of non-negative integers (the Finset ℕ version). Then for $n \ge 2$:

$$f(n-1) \;\le\; f_0(n) \;\le\; f(n).$$

Upper: every positive set is a non-negative set. Lower: let $A$ attain $f_0(n)$. If $0 \notin A$ then $A$ is a positive $n$-set and $|P(A)| \ge f(n) \ge f(n-1)$. If $0 \in A$, then $A' = A \setminus \{0\}$ is a positive $(n-1)$-set and every off-diagonal sum of $A'$ is an off-diagonal sum of $A$, so $P(A') \subseteq P(A)$ and $|P(A)| \ge f(n-1)$. $\square$

So the two staircases differ by at most one step, and

$$f(n)/\log n \to \infty \iff f_0(n)/\log n \to \infty .$$

The mismatch is harmless for the limit statement and is not harmless for pinned finite values. Exhaustive search over subsets of $[0,40]$:

$f(2)=1$ (best is $\{1,2\}$, sum $3$); $f_0(2)=0$ (take $\{0,1\}$, sum $1$, empty prime support). A formalisation that pins a finite value of $f$ from the Finset ℕ statement is wrong at $n=2$. The asymptotic statement is unaffected, so the Formal Conjectures entry is a faithful formalisation of the limit and need not be repaired for that purpose.

(The table entries are exhaustive over $[0,40]$ and are therefore upper bounds on the true $f$, $f_0$; the inequality above is proved, not measured.)

3. Arms 1 and 2: $g_N(k)$, exact inside a bounded universe

Branch-and-bound clique search on $a \sim b \iff a+b$ is $S$-smooth, $S$ = the first $k$ primes. Each row completed its search exhaustively inside $[1,N]$ (no timeout), and each witness was re-verified from scratch by full trial-division prime support.

These are lower bounds on $g(k)$, and only that. The search universe is bounded, so a witness with a larger element would raise any row. No upper bound on $g(k)$ is established anywhere in this hunt.

Two things are worth reporting about the run itself.

The optima are tiny and the universe does not matter. A first pass used $N = 4000, 1200, 700, 400, 260, 200$. Widening to the table's $N$, a factor of 30 to 60, changed not one row. Every optimal witness lives well below 50. Whatever is capping these sets is not the size of the search box.

Arm 2: the choice of $S$ buys something, and it is a constant. At $k=3$, $N = 20\,000$, exhaustive:

Dropping $2$ from $S$ is catastrophic: $\{3,5,7\}$ admits no 3-element set at all inside $20\,000$, because among any three integers two have the same parity and their sum is even. That is the whole of the $S$-dependence we could see: $2 \in S$ or not, then a wobble of one. No $\exp(o(k))$ clique bound came out of it. Arm 2 is a constant-level observation, which is itself a pre-registered kill condition.

4. Arm 3: the composition gadget points the wrong way

This is the finding that changes how the problem should be briefed.

$g$ is non-decreasing and $g(1) = 2$. The witness is $\{1,3\}$, sum $4$. For $g(1) \le 2$: if $a<b<c$ are positive with $a+b=2^x$, $a+c=2^y$, $b+c=2^z$ then $x<y<z$, so $2a = 2^x + 2^y - 2^z \le 2^x + 2^y - 2^{y+1} = 2^x - 2^y < 0$, impossible. (Any single-prime $S=\{p\}$ with $p$ odd gives $g \le 2$ by parity: two of any three integers share a parity class and their sum is even.) Suppose a rigorous multiplicative-size / additive-support composition law existed: $g(k_1+k_2) \ge g(k_1)\,g(k_2)$. Then $\log g$ is superadditive, and Fekete's lemma gives

$$\lim_k g(k)^{1/k} = \sup_k g(k)^{1/k} \;\ge\; g(1) = 2,$$

hence $g(k) \ge 2^k$ and $f(n) \le \log_2 n + O(1)$, the conjecture is false.

So arm 3, as briefed, is a refutation search wearing the clothes of a proof route. Erdős #126 is not asking for a composition law; it is asking, in this formulation, for an anti-composition theorem, a proof that prime supports cannot be combined multiplicatively. That is a structurally harder object than the brief's promotion criterion assumed, and it is why "find the gadget" is not a fundable direction unless the goal is to refute.

The measured data is consistent with no such gadget, in a bounded sense worth stating precisely. Supermultiplicativity at $k_1=1, k_2=2$ would need $g(3) \ge 8$: an 8-element set with all pairwise sums supported on 3 primes. Exhaustive search to $N = 60\,000$ found a maximum of 5. Every one of the 11 tested $(k_1,k_2)$ pairs fails the same way, by a wide margin ($g(3)\cdot g(3) = 25$ vs. $g_N(6) = 10$). This does not disprove supermultiplicativity, the products are lower bounds on both sides and a gadget could produce elements past $N$, but it does say that any gadget must leave the small-integer regime where all the measured optima live, which is a real constraint on its shape.

5. What we could not settle

• No upper bound on $g(k)$, therefore no contribution to $f(n) \gg \log n$ or to improving it. The bounded search is structurally incapable of one.
• No iterable lemma. The $p$-adic pruning in arm 1 is a smoothness table lookup, not an induction with a loss we could track across $k$. Nothing in it iterates.
• The trend in $g_N(k)^{1/k}$ decides nothing. It falls monotonically, $2.00 \to 1.41$ over $k=1..7$, which is exactly what a sequence converging to $1.3$ looks like as well as one converging to $1$. Seven points cannot separate those. This is the "suggestive finite data" the brief named as a kill condition, and we are calling it that rather than dressing it up.

6. Closing the loop

The brief's CLOSE THE LOOP section assumes this task was returned by a function reading a ledger under harness/departments/. It was not: the brief records Source: operator, Reference: prize:erdos-126:bounded-scout. We enumerated the ledger anyway. harness.review.standing_reasons reports exactly two claims whose review is not standing, rf-c003-window and k2-far-constant-depth1, both wanting a blind attack, and neither has anything to do with Erdős #126. Writing an AttackOutcome against either would be recording an attack that did not happen.

So no ledger was appended, and the generator's two open items are still open. That is a report, not a completion: this run did not close them and does not claim to.

The doors

This hunt measured a ceiling ($g_N(k)$, and the wall at $\log n$), so it owes the door list.

1. Active constraints at the optimum. Ranked by how hard they bind.

2. Frozen-constant inventory. Every chosen-not-optimised number:

3. Information class. Doors 1 and 3 (optimise $S$; push $k$) stay inside the data this family already reads: sets of integers and the smoothness of their pairwise sums. They are under the configuration ceiling and cannot produce an upper bound on $g(k)$ no matter how far they are pushed, because a bounded search never sees the whole set. Any door that could actually move Erdős #126 requires reading more, the $S$-unit equation machinery (Evertse–Győry style bounds on the number of solutions of $x+y=1$ in $S$-units) rather than enumeration. That is a different information class, and it is where the 92-year wall actually sits.

The top-ranked door is therefore not one of ours: it is the $S$-unit counting bound, and this hunt has no purchase on it at scout cost.

Loose threads

• $g(k)$'s exact values look like a sequence someone has already tabulated. 2, 4, 5, 6, 8, 10, 11 for $k=1..7$. Why it might matter: if this is a known OEIS sequence, the literature attached to it names the best known constructions and would immediately tell us whether our lower bounds are sharp or badly weak. First step: query OEIS for 2,4,5,6,8,10,11 with the smooth-pair-sum description; the repo already has an OEIS backend behind ontology/knownness.py.
• The optimal witnesses are all tiny, and nobody knows why. Every maximum is attained below 50 with $N$ up to $2\cdot 10^5$. Why it might matter: a proof that optima can be normalised into $[1, C^k]$ would turn the bounded search into an exact computation of $g(k)$, which is the missing upper-bound half. First step: test whether every maximum witness can be scaled/shifted into $[1, 2^{k+2}]$ for $k \le 6$ by re-running the search with $N = 2^{k+2}$ and comparing.
• $\{3,5,7\}$ giving $g_N = 2$ generalises. Any $S$ omitting 2 caps the set at 2 by parity. Why it might matter: it is the only clean iterable statement the run found, and the same argument mod small primes may cap sets whose $S$ omits 3, mod 3 residue classes. First step: prove $|A| \le p-1$ when $p \notin S$ is small, by pigeonhole on residues mod $p$ against the pairing $r \leftrightarrow -r$; check it against the measured table.
• The $S$-optimisation door was sampled at $k=3$ only. Why it might matter: it is the one frozen constant with real trade shape, per the doors table. First step: at $k=4$, search all $\binom{8}{4}=70$ subsets of the first 8 primes at $N=5\,000$ and see whether any beats 6.