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Library · hunts/support_baf4cde6/RESULTS.md

RESULTS: Erdős #126, entropy/combinatorics arm (support run baf4cde6)

2,661 words · 280 lines · source

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

Reproduce with python hunts/support_baf4cde6/probe.py (~10 s, stdlib only, writes results.json).

Setting: $S$ is a set of $k$ primes; $A$ is a finite set of distinct positive integers with $a+b$ free of primes outside $S$ for all distinct $a,b \in A$ (call such an $A$ $S$-summable); $g(k) = \max|A|$. Erdős–Turán 1934 gives $g(k) < 3\cdot 2^{k-1}$; #126 asks for $g(k) = \exp(o(k))$.

Verdict

The lane is closed, and closed by a proof rather than by a failed attempt.

The natural signature model does not merely fail to beat $2^k$. It proves nothing at all: for every prime bound $P$, arbitrarily large sets satisfy every divisibility constraint the hypothesis imposes at primes $\le P$ (Theorem B). So no entropy, VC, container, dependent-random-choice or forbidden-pattern argument whose input is the $p$-adic signature of $A$ at a fixed finite prime set can bound $|A|$ by any function of $k$, because the input is consistent with $|A| = \infty$. The whole force of the hypothesis sits at primes larger than the elements of $A$, where "signature" is not local data but the cofinite smoothness condition itself.

Three things survive and are worth the parent's attention:

  1. Lemma A, the exact local obstruction (a residue-class cap, not an element cap), which corrects a thread the parent left open in a form that is false.
  2. Proposition C, the entropy accounting: pushing the residue constraint to all primes $\le P$ with CRT injectivity saves exactly one bit per prime outside $S$, and lands at $2^{53}$ where Erdős–Turán gives $2^{k+1.6}$. Quantified, not guessed.
  3. The $S$-unit door survives this arm. Each 4-subset of $A$ gives a non-degenerate solution of a three-variable unit equation. Today's counting theorem is worse than 1934 by a factor $\approx 10^{11}$ in the exponent, but nothing known forbids it being subexponential, so the parent's top-ranked door is not closed by anything I found. The gap in that route is the fiber problem, named in §6.

1. Audit of hunts/r_186989/RESULTS.md

Re-derived independently, not read off.

ClaimVerdict
$f(n-1) \le f_0(n) \le f(n)$, and $f(2)=1 \ne 0 = f_0(2)$Correct. Proof reads cleanly; the witness $\{0,1\}$ checks.
$g(1) = 2$ via $2a = 2^x+2^y-2^z < 0$Correct ($x<y<z$ forces $z \ge y+1$).
Supermultiplicativity $\Rightarrow$ $g(k) \ge 2^k$ $\Rightarrow$ conjecture falseCorrect, and the direction is right. Fekete gives $\lim g(k)^{1/k} = \sup_k g(k)^{1/k} \ge g(1) = 2$. A composition law refutes #126; it does not prove it.
$g_N(3) = 5$ with witness $\{1,3,7,17,47\}$; $g_N(4)=6$ with $\{1,2,3,5,7,13\}$Reproduced by an independently written branch-and-bound search at $N = 3000$ and $1500$.
All seven tabulated witnesses are $S$-summableRe-verified by full trial division (parent_witnesses_all_valid: true).
Loose thread: "prove $\A\\le p-1$ when $p \notin S$"False as stated. Refuted.

The refutation is explicit: $A = \{1,3,7,13\}$, $S = \{2,5,7\}$, with sums $4, 8, 14, 10, 16, 20$. Here $3 \notin S$ and $|A| = 4 > p - 1 = 2$. A small exhaustive sweep over $S$ with $2 \in S$, $3 \notin S$, $|S| \in \{3,4\}$, $N = 400$ reaches $|A| = 5$ (e.g. $S=\{2,5,11,17\}$, $A = \{1,9,31,79,241\}$), so the guess fails by a growing margin and not by an edge case.

What the parent was reaching for is true in a different shape, and it is Lemma A.

2. Lemma A: the exact local obstruction

Lemma A. Let $A$ be $S$-summable and let $p \notin S$ be prime. Then

  1. $A$ contains at most one element $\equiv 0 \pmod p$;
  2. $A$ never meets both classes $r$ and $-r \bmod p$ for $r \not\equiv 0$;
  3. hence $A$ meets at most $(p+1)/2$ of the $p$ residue classes mod $p$;
  4. and for $p = 2$ this forces $|A| \le 2$.

Proof. If $a \equiv -b \pmod p$ with $a \ne b$ in $A$, then $p \mid a+b$, so $p \in S$. That is (1) with $r=0$ and (2). Nonzero classes pair off as $\{r,-r\}$ into $(p-1)/2$ pairs for odd $p$, and $A$ uses at most one class from each, plus possibly class $0$, giving (3). For $p=2$ both classes are self-paired ($r \equiv -r$), so each holds at most one element. $\square$

Two things to notice, because they are the whole reason the naive guess fails. The lemma caps classes, not elements: for odd $p$ any number of elements may share one class $r$ with $2r \not\equiv 0$, since their sums are $\equiv 2r \not\equiv 0$. And prime powers add nothing: when $p \notin S$ the condition is $p \nmid a+b$, which lives entirely mod $p$.

Verified on all seven of the parent's witnesses at every prime $\le 37$ outside their $S$ (lemma_A_holds_on_all_witnesses: true), including the sharp cases: $\{1,3,7,13\}$ uses exactly $2 = (3+1)/2$ classes mod 3.

Part (4) is the clean proof of the parent's "parity is worth more than every other structural choice", uniform in $k$ and not tied to a search box.

3. Theorem B: the signature model has no ceiling at all

This is the extremal abstract pattern the brief asked for. It is not a pattern of size $2^k$; it is unbounded, which is a stronger statement about the model.

Fix $P$. Call $P$-local data everything an argument can read from the divisibility of the off-diagonal sums by primes $\le P$: the residue vectors $(a \bmod p)_{p \le P}$, the valuations $v_p(a+b)$ for $p \le P$, and any constraint derived from them (this is exactly what an entropy, VC, container or forbidden-pattern argument over prime-valuation signatures consumes).

Theorem B. Let $S$ be any prime set with $2 \in S$ and let $P$ be any bound. Then for every $n$ there is an $A$ with $|A| = n$ satisfying every constraint that $S$-summability imposes at primes $\le P$, namely $p \nmid a+b$ for all primes $p \le P$ with $p \notin S$ and all distinct $a,b \in A$.

Proof. Let $M = \prod \{p \le P : p \notin S,\ p \text{ odd}\}$ and $A = \{1 + iM : 0 \le i < n\}$. For distinct $a,b \in A$, $a+b \equiv 2 \pmod M$, so no odd $p \mid M$ divides $a+b$. The prime $2$ lies in $S$ and is unconstrained. $\square$

Checked computationally at $P = 60$, $S = \{2,3,5\}$: $|A| = 200$ against the Erdős–Turán bound $3 \cdot 2^{2} = 12$ for that $k$, with zero violations at any of the 14 absent odd primes below 60 (local_model_unbounded).

Consequence. Any argument of the briefed families, applied to $P$-local data alone, is vacuous: its hypotheses admit sets of unbounded size, so it cannot output a finite bound. To beat $2^k$, or to reach any bound whatsoever, an argument must use divisibility information at primes exceeding $\max A$. That is not a signature; it is the statement "$a+b$ is an $S$-unit", i.e. the original hypothesis. The model has no proper subset of the hypothesis to work with.

This also explains the parent's puzzled measurement that every optimal witness lives below 50 while the search box ran to $2\cdot 10^5$. Nothing local is binding. What binds is a condition about primes bigger than the elements, and that condition tightens as elements grow.

4. Proposition C: the entropy bound, and exactly how far short it falls

The one non-vacuous way to use Lemma A is to take $P$ large enough that CRT makes the residue vector injective on $A$. That is a genuine entropy argument and it does produce a theorem. It is also strictly worse than 1934.

Proposition C. Let $A \subseteq [1,N]$ be $S$-summable, $|S| = k$, and let $P$ satisfy $\prod_{p \le P,\, p \notin S} p > N$. Then $$|A| \;\le\; \prod_{p \le P,\ p \notin S} \frac{p+1}{2}.$$

Proof. By CRT the map $a \mapsto (a \bmod p)_{p \le P, p \notin S}$ is injective on $[1,N]$, and by Lemma A(3) each coordinate takes at most $(p+1)/2$ values. $\square$

In entropy form: with $a$ uniform on $A$, $\log_2|A| = H(a) \le \sum_p H(a \bmod p) \le \sum_p \log_2\frac{p+1}{2}$. Since $\sum_{p\le P}\log_2\frac{p+1}{2} = \sum_{p \le P}\log_2 p - \pi(P) + O(\log\log P)$, the method saves exactly one bit per prime outside $S$, against a modulus that must already exceed $N$. So it says $$|A| \;\lesssim\; N \cdot 2^{k}\cdot 2^{-(1+o(1))\log N/\log\log N},$$ a real saving over the trivial $N$, and a bound in $N$ rather than in $k$. Measured at $\log_2 N = 64$: the bound is $2^{53.1}$ ($k=5$), $2^{57.1}$ ($k=10$), $2^{54.2}$ ($k=20$), against Erdős–Turán's $2^{5.6}$, $2^{10.6}$, $2^{20.6}$ (entropy_accounting). It is never competitive, for any $N$: to save the $\log_2 N$ bits that would make it a $k$-only bound you would need $\pi(P) \approx \log_2 N$ primes, but those primes multiply to $e^{\theta(P)} \gg N$ long before that.

This is the quantitative reason the arm is dead, and it is more useful than the qualitative one: the residue constraint is worth one bit per prime, and the problem needs $\log N$ bits.

5. The other briefed tools, briefly and honestly

6. Where the constraint actually lives (for the parent's door ranking)

Four distinct $a,b,c,d \in A$ satisfy $(a+b) + (c+d) = (a+c) + (b+d)$. Dividing by $a+b$, $$x + y + z = 1, \qquad x = \frac{a+c}{a+b},\quad y = \frac{b+d}{a+b},\quad z = -\frac{c+d}{a+b},$$ with $x,y,z$ in the group of rationals supported on $S$ (rank $k$, plus sign). The solution is non-degenerate: $x+y = 0$ is impossible for positive elements, $x+z=0 \iff a=d$ and $y+z=0 \iff b=c$, both excluded. So every 4-subset of $A$ yields a non-degenerate solution of a three-variable $S$-unit equation.

Direction check, stated explicitly. Let $E_3(k)$ bound the number of non-degenerate solutions and $m$ bound the fibers of the map (4-subsets) $\to$ (solutions). Then $\binom{|A|}{4} \le m\, E_3(k)$, so $|A| \le (24\,m\,E_3(k))^{1/4}$. This is an upper-bound route, unlike the composition law the parent killed, which was a lower-bound route wearing the wrong clothes.

Two facts about it:

7. What I could not settle

The doors

This arm measured a ceiling (the signature model's), so it owes the list.

1. Active constraints at the optimum.

RankWhat bindsEvidence
1Cofiniteness. The hypothesis constrains all primes above $\max A$, and no finite-prime restriction of it constrains anything.Theorem B: unbounded sets satisfy every constraint at primes $\le P$, for every $P$.
2One bit per absent prime. Lemma A's saving is additive in $\pi(P)$ while the CRT cost is $\theta(P)$.Proposition C, measured: $2^{53}$ vs $2^{5.6}$ at $\log_2 N = 64$.
3Parity. $p = 2$ is the only prime whose absence caps $A$ rather than the class count.Lemma A(4), proved, uniform in $k$.

2. Frozen-constant inventory.

FrozenValueWhat relaxing it trades
the signature level (residues mod $p$, $p \le P$)primes onlyPrime powers give nothing when $p \notin S$ (§2), so this door is provably shut, not merely unexplored.
$P$ in Theorem BanyAlready universally quantified. No trade.
the identity used in §6the 4-point additive relationThe door with real trade shape. Longer relations ($5$- and $6$-point) give unit equations in more variables: worse ESS constants, but possibly smaller fibers. Nobody has looked at the trade.
$\log_2 N = 64$ in the accountingdisplay onlyProposition C is proved for all $N$; the table is an illustration.
the search box $N \le 400$ in the $3 \notin S$ sweep400Affects only the size of the refutation margin, not the refutation.

3. Information class. Doors 1 and 2 stay inside the data the signature family reads, and Theorem B proves they cannot produce a bound at any depth. Door 3 (the fiber count for the 4-point relation) requires reading more: the joint distribution of sums, which is exactly the information the local model discards. That is the only door in this arm with anything behind it, and it points at the parent's $S$-unit door rather than away from it.

Loose threads