This is a hunt. Nothing in hunts/ is a result, and nothing here is evidence for or against RH (docs/08).
The question
Erdős #126. For a set $A$ of $n$ distinct positive integers write
$$P(A) = \{\,p \text{ prime} : p \mid a+b \text{ for some } a \ne b \in A\,\}, \qquad f(n) = \min_{|A|=n} |P(A)|.$$
Classical work gives $f(n) \gg \log n$ and nothing better in 92 years. Erdős asks for $f(n)/\log n \to \infty$. The prize is \$250, paid only on a full solution published in a reputable journal.
This run is a parallel scout on a 30 minute cap, not the primary lane. It was launched to decide one thing: is there anything here worth funding, or is the wall exactly where the literature says it is?
The brief specified three capped arms and a preliminary:
- arm 0, resolve the Formal Conjectures positivity mismatch: its statement is over
Finset ℕ, which admits0, while Erdős assumes positive integers. - arm 1: blockwise $p$-adic pruning, seeking $\exp(o(k))$ retention.
- arm 2: an $S$-unit clique reduction, seeking an $\exp(o(k))$ clique bound.
- arm 3: exact composition search through $k \le 8$ for a multiplicative-size / additive-support gadget.
Promotion required an iterable lemma, an $\exp(o(k))$ reduction, or a rigorous composition law. Constant improvements, a Lean statement, or suggestive finite data kill the scout.
What this hunt actually measures
It does not attack $f$. It measures the inverse staircase
$$g(k) \;=\; \max\{\, n : \text{some } n\text{-set of positive integers has } |P(A)| \le k \,\},$$
because $|P(A)| \le k$ for some $n$-set iff $f(n) \le k$ iff $n \le g(k)$, and therefore
$$f(n)/\log n \to \infty \iff g(k) = \exp(o(k)) \iff g(k)^{1/k} \to 1 .$$
All three arms then speak about one integer sequence. probe.py computes $g_N(k)$, the exact maximum inside the bounded universe $[1,N]$, by branch-and-bound clique search on the graph $a \sim b \iff a+b$ is $S$-smooth. Every $g_N(k)$ is a rigorous lower bound on $g(k)$ with an explicit witness that is re-verified from scratch. No upper bound on $g(k)$ is established here, because the universe is bounded.
Sources: <https://www.erdosproblems.com/126>, <https://combinatorica.hu/~p_erdos/1934-03.pdf>.
id: r_186989
question: Does bounded reconnaissance on Erdos 126 find an iterable lemma, an exp(o(k)) reduction, or a rigorous composition law strong enough to promote the pair-sum prime-support problem to a funded lane?
frontier: f(n) >> log n classical, unimproved since 1934; the conjecture f(n)/log n -> infinity is open. Equivalently g(k)^(1/k) -> 1, where measured exact maxima inside bounded universes give g(k) >= 2, 4, 5, 6, 8, 10, 11 for k = 1..7.
proposed_attack: invert the problem to the growth rate of g(k) and measure g_N(k) exactly by branch-and-bound clique search on the S-smooth pair-sum graph inside [1, N]
dead_routes:
- composition/gadget search for a multiplicative-size additive-support law: it points the wrong way, since g(1) = 2 plus supermultiplicativity gives g(k) >= 2^k by Fekete and so refutes the conjecture rather than proving it
- reading the conjecture off finite data: the measured k-th roots decrease monotonically over the whole range searched, which is equally consistent with a limit above 1
required_oracles:
- exact branch-and-bound clique enumeration inside a stated bounded universe
- independent re-verification of every witness set by full trial-division prime support
- exhaustive small-n subset enumeration for the positivity comparison
kill_conditions:
- the run produces only suggestive finite data and no iterable lemma
- the only improvement available is to a constant
- the deliverable degenerates to a Lean restatement of the conjecture
- a witness set fails independent re-verification
agents_may:
- search
- derive
- code
- attack
agents_may_not:
- declare novelty
- declare theorem status
- promote their own claim
- claim an upper bound on g(k) from a bounded searchScope
Writes only hunts/r_186989/ plus one case-log entry in hunts/README.md. Touches no ledger under harness/departments/: see RESULTS.md, "Closing the loop", for why there was nothing there to close.