GPT 6 Astra Ultra attempt on Math-647

Erdős Problem #25: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

Let $1\le n_1<n_2<\cdots$ and choose a residue $a_i$ modulo each $n_i$. Define \[ A=\{m\ge1:\ \forall i,\ m<n_i\text{ or }m\not\equiv a_i\pmod{n_i}\}. \] Must the logarithmic density of $A$ exist? The activation thresholds are essential.

All eleven supplied versions were considered in full: 25.tex, 25_v2.tex, 25_v3.tex, 25_v4.tex, 25_v5.tex, 25_v6.tex, 25_v7.tex, 25_v8.tex, 25_v9.tex, 25_v10.tex, and 25_v11.tex. Byte comparison confirms that versions 5 and 6 are identical, as are versions 9, 10 and 11; there are eight distinct texts.

The first version develops activated inclusion--exclusion. Versions 2 and 3 give reciprocal-summability and zero-stage-density criteria, but incorrectly assert global containment in every finite stage; version 7 correctly repairs this to tail containment. Version 4 gives affine padding, while versions 5 and 6 extract its compression of stage-density variation. Version 8 gives a correct first-hit decomposition, but proposes a universal statement about the thresholdless core that the later explicit example disproves. Versions 9--11 give that valid dense exceptional-set example, but also incorrectly claim the earlier exact decomposition itself was refuted. These are distinct issues. I correct the latter claim and strengthen the reciprocal-summability theorem to ordinary natural density, with a finite-stage error bound.

2) WORK

Write \[ A^{(k)}=\{m\ge1:m\not\equiv a_i\pmod{n_i}\ (i\le k)\},\quad \delta_k=d(A^{(k)}),\quad \delta=\lim_k\delta_k, \] and $L_k=\operatorname{lcm}(n_1,\ldots,n_k)$. The stage sets are periodic and decreasing, so the displayed limit exists. Let $r_i\in\{0,\ldots,n_i-1\}$ represent $a_i$.

The earlier exceptional-set formula is valid.

Put \[ B=\bigcap_k A^{(k)},\qquad E=\{r_i:1\le r_i<n_i,\ r_i\in A^{(i-1)}\},\qquad A^{(0)}=\mathbb N_{>0}. \] Then $A=B\sqcup E$, exactly as claimed in version 8. Here is an independent proof. Every point of $B$ is in $A$. If $m=r_i$ is counted in $E$, it avoids all earlier classes, while its own and all later constraints are inactive since $m<n_i$; hence $m\in A\setminus B$. Conversely, for $m\in A\setminus B$, choose the least index $i$ whose class contains $m$. All earlier classes miss $m$, and the class at $i$ must be inactive, so $m<n_i$ and $m=r_i\in E$.

A repeated representative can qualify only at its earliest occurrence, but it need not qualify at every occurrence. The later versions' example \[ (n_1,a_1)=(5,1),\qquad(n_2,a_2)=(7,1) \] therefore does not refute the set formula: $r_2=1\notin A^{(1)}$, but $r_1=1\in A^{(0)}$, so $1$ is still in $E$. It only refutes the unnecessary implication $r_i\in A\Rightarrow r_i\in A^{(i-1)}$ for a prescribed, possibly nonfirst index.

There is a separate harmless finite-point error in version 4: its exact formula always inserts $\{1\}$ into a thresholdless finite stage of the padded instance. If some embedded residue is $b_j=0\pmod{m_j}$, then the padded congruence $1\pmod{Mm_j}$ excludes $1$ from that stage. The stage-density formulas are unaffected by this singleton error.

Reciprocal summability actually gives natural density.

Assume \[ \sum_{i\ge1}{1\over n_i}<\infty,\qquad R_k=\sum_{i>k}{1\over n_i}. \] An activated class has an especially simple counting bound: \[ \#\{m\le x:m\ge n_i,\ m\equiv r_i\pmod{n_i}\} \le {x\over n_i}. \tag{1} \] Indeed its elements are $jn_i+r_i$ for integers $j\ge1$, including the case $r_i=0$; there are at most $x/n_i$ such integers. There is no additive one per modulus in $1$, precisely because the small representative is inactive.

For $m\ge n_k$, membership in $A$ implies membership in $A^{(k)}$. Any point in $A^{(k)}\setminus A$ must be caught by a later active class. Consequently, writing $A(x)=|A\cap[1,x]|$, \[ A^{(k)}(x)-xR_k\le A(x)\le A^{(k)}(x)+n_k. \tag{2} \] The periodic counting estimate is \[ |A^{(k)}(x)-\delta_kx|\le L_k. \] Moreover, a finite union bound on the later classes and then a limit give $0\le\delta_k-\delta\le R_k$. Combining with $2$ yields the explicit bound \[ \boxed{|A(x)-\delta x|\le xR_k+L_k+n_k\qquad(k\ge1).} \tag{3} \] For fixed $k$, divide by $x$ and let $x\to\infty$; then let $k\to\infty$. Since $R_k\to0$, this proves \[ \boxed{d(A)=\delta.} \] Partial summation also gives its logarithmic density. This strengthens the supplied logarithmic-only theorem, but is a special-case theorem, not a solution in the divergent-reciprocal regime.

In particular the source's family $n_i=Q^i+i$, $r_i=i$, $Q\ge3$, has empty thresholdless core $B$: the class at index $i=m$ catches every positive integer $m$ without activation. Nevertheless $3$ proves that $A=E$ has natural density \[ d(E)=\delta\ge1-\sum_{i\ge1}{1\over Q^i+i}>1-{1\over Q-1}. \] Thus the source's large-exceptional-set example is valid, even in the stronger natural-density sense. It disproves universal negligibility of $E$, not the decomposition $A=B\sqcup E$. The conditional implication $\delta_{\log}(B)=\delta\Rightarrow\delta_{\log}(A)=\delta$ also remains valid; its proposed universal premise fails on this example.

The ostensibly more general tail criterion has exactly the same scope.

The sources define \[ \Theta_k=\limsup_{x\to\infty}{1\over\log x} \sum_{\substack{i>k\\n_i\le x}}{\log(x/n_i)+1\over n_i}. \] In fact \[ \boxed{\Theta_k=\sum_{i>k}1/n_i}\tag{4} \] as an extended nonnegative real number, and the limit exists. Each fixed summand tends to $1/n_i$. If the reciprocal tail is finite, for $x\ge e$ it is dominated by $2/n_i$, so dominated convergence proves $4$. If that tail diverges, restricting to any finite subcollection shows that the liminf is at least its reciprocal sum; these finite sums are arbitrarily large. Thus every $\Theta_k$ is infinite in the divergent case. The condition $\Theta_k\to0$ does not reach beyond reciprocal summability.

3) ADVERSARIAL VERIFICATION

The earliest-index argument handles repeated residues rather than imposing an unwarranted condition on every occurrence. The bound $1$ uses activation and would generally fail for the thresholdless progression because of its first positive representative. The period $L_k$ may be enormous, but $3$ takes $x\to\infty$ before $k\to\infty$, so no uniform period assumption is hidden. The parameter $k$ may also be optimized in $3$ if an explicit rate is desired.

A small exhaustive checker in verification/batch_2_25_checks.py verifies the decomposition for all choices of three distinct moduli from $\{2,\ldots,7\}$, all their residue assignments, and all positive integers through 30. It also checks the repeated-representative example and the finite-stage singleton error. These computations are sanity checks; the proofs above cover unbounded parameters.

The padding constructions control the amplitude of a stage-density change, not its convergence rate as a function of the modulus or period. Accordingly they do not, by themselves, rule out every possible quantitative convergence-rate hypothesis.

4) FINAL

UNRESOLVED. The verified progress is an ordinary-density theorem with error bound $3$, the exact identification $4$ of the tail criterion's scope, and a correction of the latest versions' attempted refutation of a valid decomposition. None handles general positive stage-limit instances with divergent reciprocal moduli. An argument controlling their activated tail beyond the union bound, or a genuine example of logarithmic nonconvergence, remains missing.

5) SOURCES CHECKED

All eleven local versions were read or, for byte-identical copies, compared against the full text already read; equality was verified by SHA-256 and by direct byte comparison in the checker. The exact target and activation convention were checked on 2026-09-23 in https://raw.githubusercontent.com/google-deepmind/formal-conjectures/main/FormalConjectures/ErdosProblems/25.lean. That file contains a placeholder proof and is used only as a statement reference. The arguments in this attempt use no external density theorem.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a measure inherited from the earlier 88--98 percent estimates. COMPLETION: 15%