gpt pro 5.4 attempt on Math-647

Round-6 Objective

We pursue (A) a proof strategy. Round 5 gave a broad existence theorem based on finite-stage inclusion--exclusion, with an error term controlled by the cumulative compatible-subfamily count \[ M_k:=\#\left\{I\subseteq\{1,\dots,k\}: \exists n\in\mathbb Z,\ n\equiv a_i \pmod{n_i}\ \forall i\in I\right\}. \] The exact remaining weakness of that argument is that the same compatible subset is charged on every later activation interval, so the Round 5 hypothesis involved \[ \sum_{k\le K}\frac{M_k}{n_k}. \]

The new goal of this round is to remove that repeated charging. We do this by writing an exact global inclusion--exclusion formula for $A$ itself, indexed by the largest active modulus in each compatible subfamily. This yields a strictly sharper criterion in terms of the local counts \[ C_r:=\#\{I\text{ compatible finite}:\ \max I=r\}, \] namely \[ \sum_{r\le K}\frac{C_r}{n_r}=o(\log n_K) \quad\Longrightarrow\quad \delta_{\log}(A)\ \text{exists}. \] This is stronger than Round 5 because it counts each compatible subfamily exactly once, at the moment it first becomes active.

Round-5 Foundation Used

We rely on the following vetted Round 5 facts.

• Activation identity (Round 1). For \[ A^{(k)}:=\{n\in\mathbb N:\ n\not\equiv a_i\pmod{n_i}\ \forall 1\le i\le k\}, \] one has \[ \mathbf 1_A(n)=\mathbf 1_{A^{(k)}}(n)\qquad (n_k\le n<n_{k+1}). \]
• Monotone stage densities (Round 2). Each $A^{(k)}$ is periodic, its natural density \[ \delta_k:=d(A^{(k)}) \] exists, and $(\delta_k)$ is decreasing with limit \[ \delta:=\lim_{k\to\infty}\delta_k\in[0,1]. \]
• Single progression harmonic estimate (Round 5). For every modulus $m\ge 1$, every residue class $b\pmod m$, and integers $2\le X<Y$, \[ \sum_{\substack{X\le n<Y\\ n\equiv b\ (\mathrm{mod}\ m)}}\frac1n = \frac1m\log\frac{Y}{X}+O\!\left(\frac1X\right), \] with an absolute implied constant, uniform in $m,b,X,Y$.
• Weighted-average principle (Round 2 / Round 4). If a nonnegative weight system has total mass tending to $1$, then the weighted averages of the monotone sequence $\delta_k\downarrow\delta$ converge to $\delta$.

New Insight / Tool (Round-6)

The new tool is an exact global inclusion--exclusion expansion for $A$, rather than a separate stage-by-stage approximation on each interval $[n_k,n_{k+1})$.

For each finite compatible set of indices $I$, let $r(I):=\max I$. The corresponding congruence progression first becomes active only when $n\ge n_{r(I)}$. Therefore, when summing \[ S(x):=\sum_{\substack{1\le n\le x\\ n\in A}}\frac1n, \] each compatible $I$ should be charged once, from $n_{r(I)}$ onward, rather than being re-charged on every later interval.

This replaces the Round 5 cumulative parameter $M_k$ by the sharper local parameter \[ C_r:=\#\{I\text{ compatible finite}:\ r(I)=r\}. \] The resulting error term is \[ O\!\left(\sum_{r\le K(x)}\frac{C_r}{n_r}\right), \] instead of the Round 5 stagewise error \[ O\!\left(\sum_{k\le K(x)}\frac{M_k}{n_k}\right). \]

A second new insight is graph-theoretic: by the generalized Chinese remainder theorem, a finite family of congruences is compatible if and only if it is pairwise compatible. Thus the compatible families are precisely the cliques of a natural compatibility graph. The local count $C_r$ is exactly the number of cliques in the backward neighborhood of $r$.

Attack Plan (Round-6)

The exact gap left after Round 5 was that the inclusion--exclusion approximation was done separately at each stage, causing the same compatible subfamily to be counted on every later activation interval.

The Round 6 plan is:

1. Define the local compatible-subfamily counts $C_r$ by maximum index.
2. Prove an exact global inclusion--exclusion formula for $A(n)$ in which each compatible subfamily appears only once, activated at $n_{r(I)}$.
3. Use the Round 5 single-progression harmonic estimate on each such activated progression, obtaining a total error $O(\sum_{r\le K} C_r/n_r)$.
4. Rewrite the main term in terms of the stage densities $\delta_r$ and apply the Round 2 weighted-average principle.
5. Reformulate $C_r$ graph-theoretically and deduce a clean clique/degree-based corollary that rules out a broad class of counterexample strategies.

Work (Round-6)

6.1. Pairwise compatibility and the compatibility graph

For a finite nonempty set $I\subset\mathbb N$, call $I$ compatible if the system of congruences \[ n\equiv a_i\pmod{n_i}\qquad(i\in I) \] has an integer solution. Write $\mathcal C_\infty$ for the family of all finite compatible nonempty sets.

Lemma 6.1 (pairwise criterion).

A finite nonempty set $I$ is compatible if and only if for every $i,j\in I$, \begin{equation} a_i\equiv a_j\pmod{\gcd(n_i,n_j)}. \tag{6.1} \end{equation}

Proof. Necessity is immediate: any common solution must be congruent to both $a_i$ and $a_j$, hence their difference is divisible by $\gcd(n_i,n_j)$.

For sufficiency, argue by induction on $|I|$. The case $|I|=1$ is trivial. Suppose $I=J\cup\{r\}$ with $|J|\ge 1$ and assume the claim already proved for $J$. By induction, the subsystem indexed by $J$ has a solution \[ n\equiv b\pmod{L_J},\qquad L_J:=\mathrm{lcm}(n_j:\ j\in J). \] For each $j\in J$, we have $b\equiv a_j\pmod{n_j}$ and by hypothesis also $a_j\equiv a_r\pmod{\gcd(n_j,n_r)}$, so $b-a_r$ is divisible by every $\gcd(n_j,n_r)$. Hence $b-a_r$ is divisible by their least common multiple. Prime-by-prime, one checks that \[ \mathrm{lcm}\left(\gcd(n_j,n_r):\ j\in J\right)=\gcd(L_J,n_r). \] Thus \[ b\equiv a_r\pmod{\gcd(L_J,n_r)}. \] The two-modulus Chinese remainder theorem therefore gives a simultaneous solution to \[ n\equiv b\pmod{L_J},\qquad n\equiv a_r\pmod{n_r}. \] So $I$ is compatible. $\square$

Define the compatibility graph $G$ on the vertex set $\mathbb N$ by joining $i<j$ whenever the pair $\{i,j\}$ is compatible, equivalently whenever (6.1) holds. By Lemma 6.1, the compatible finite subsets are exactly the finite cliques of $G$.

6.2. Local counts by maximum index

For $I\in\mathcal C_\infty$, define \[ r(I):=\max I, \qquad m_I:=\mathrm{lcm}(n_i:\ i\in I), \] and choose one residue class \[ b_I\pmod{m_I} \] solving all congruences $n\equiv a_i\pmod{n_i}$ for $i\in I$.

For each $r\ge 1$, define \begin{equation} \mathcal D_r:=\{I\in\mathcal C_\infty:\ r(I)=r\}, \qquad C_r:=|\mathcal D_r|. \tag{6.2} \end{equation} Thus $C_r$ counts the compatible subfamilies that are first introduced when index $r$ becomes active.

Also define the local density increment \begin{equation} \gamma_r:=\sum_{I\in\mathcal D_r}(-1)^{|I|}\frac1{m_I}. \tag{6.3} \end{equation} Since the stage-density inclusion--exclusion formula from Round 5 gives \[ \delta_k = 1+\sum_{\substack{I\in\mathcal C_\infty\\ r(I)\le k}}(-1)^{|I|}\frac1{m_I}, \] we have \begin{equation} \gamma_r=\delta_r-\delta_{r-1}\qquad(r\ge 1), \tag{6.4} \end{equation} with the convention $\delta_0:=1$.

6.3. Exact global inclusion--exclusion for $A$

Lemma 6.2 (global activated inclusion--exclusion).

For every integer $n\ge 1$, \begin{equation} \mathbf 1_A(n) = 1+ \sum_{\substack{I\in\mathcal C_\infty\\ n_{r(I)}\le n}} (-1)^{|I|}\,\mathbf 1_{\,n\equiv b_I\ (\mathrm{mod}\ m_I)}. \tag{6.5} \end{equation} Only finitely many terms are present in the sum.

Proof. Let \[ K(n):=\max\{k:\ n_k\le n\}, \] with the understanding that $K(n)=0$ if $n<n_1$. The active constraints at the integer $n$ are exactly those indexed by $\{1,\dots,K(n)\}$. Therefore, by ordinary finite inclusion--exclusion on this active set, \[ \mathbf 1_A(n)= \sum_{I\subseteq\{1,\dots,K(n)\}}(-1)^{|I|}\mathbf 1_{\cap_{i\in I}\{n\equiv a_i\ (\mathrm{mod}\ n_i)\}}. \] The empty subset contributes $1$. A nonempty subset contributes $0$ unless it is compatible, in which case its intersection is exactly the single residue class $n\equiv b_I\pmod{m_I}$. Since $I\subseteq\{1,\dots,K(n)\}$ is equivalent to $n_{r(I)}\le n$, this is precisely (6.5). $\square$

6.4. A sharper general criterion

Let \[ S(x):=\sum_{\substack{1\le n\le x\\ n\in A}}\frac1n. \] For $x\ge n_1$, let $K(x)$ be the unique index with \[ n_{K(x)}\le x<n_{K(x)+1}. \]

Theorem 6.3 (local compatible-subfamily criterion).

Assume $n_1\ge 2$ and define $C_r$ by (6.2). If \begin{equation} F_K:=\sum_{r=1}^{K}\frac{C_r}{n_r}=o(\log n_K)\qquad(K\to\infty), \tag{6.6} \end{equation} then the logarithmic density $\delta_{\log}(A)$ exists and equals \begin{equation} \delta:=\lim_{k\to\infty}\delta_k. \tag{6.7} \end{equation}

Proof. Fix $x\ge n_1$ and write $K:=K(x)$. Summing (6.5) over $1\le n\le x$ with weight $1/n$ gives \begin{equation} S(x) = \sum_{n\le x}\frac1n + \sum_{r=1}^{K}\ \sum_{I\in\mathcal D_r} (-1)^{|I|} \sum_{\substack{n_r\le n\le x\\ n\equiv b_I\ (\mathrm{mod}\ m_I)}}\frac1n. \tag{6.8} \end{equation} For each fixed $I\in\mathcal D_r$, apply the Round 5 single-progression harmonic estimate with $X=n_r$ and $Y=\lfloor x\rfloor+1$: \[ \sum_{\substack{n_r\le n\le x\\ n\equiv b_I\ (\mathrm{mod}\ m_I)}}\frac1n = \frac1{m_I}\log\frac{x}{n_r}+O\!\left(\frac1{n_r}\right), \] uniformly in $I$ and $x$. Therefore (6.8) becomes \begin{equation} S(x) = \log x+O(1) + \sum_{r=1}^{K}\gamma_r\log\frac{x}{n_r} + O\!\left(\sum_{r=1}^{K}\frac{C_r}{n_r}\right). \tag{6.9} \end{equation} Using (6.4), this is \begin{equation} S(x) = \log x+O(1) + \sum_{r=1}^{K}(\delta_r-\delta_{r-1})\log\frac{x}{n_r} + O(F_K). \tag{6.10} \end{equation} Now summation by parts gives the exact identity \begin{equation} \log x+ \sum_{r=1}^{K}(\delta_r-\delta_{r-1})\log\frac{x}{n_r} = \log n_1 + \sum_{r=1}^{K-1}\delta_r\log\frac{n_{r+1}}{n_r} + \delta_K\log\frac{x}{n_K}. \tag{6.11} \end{equation} Insert this into (6.10) and divide by $\log x$: \begin{equation} \frac{S(x)}{\log x} = \frac{\log n_1}{\log x} + \sum_{r=1}^{K-1}w_r(x)\,\delta_r + \nu_K(x)\,\delta_K + o(1), \tag{6.12} \end{equation} where \begin{equation} w_r(x):=\frac{\log(n_{r+1}/n_r)}{\log x}, \qquad \nu_K(x):=\frac{\log(x/n_K)}{\log x}. \tag{6.13} \end{equation} Because of (6.6), \[ \frac{F_K}{\log x}\le \frac{F_K}{\log n_K}\to 0. \] Also the weights are nonnegative and satisfy \begin{equation} \sum_{r=1}^{K-1}w_r(x)+\nu_K(x)=\frac{\log(x/n_1)}{\log x}\longrightarrow 1. \tag{6.14} \end{equation} It remains to show that the weighted average of $\delta_r$ tends to $\delta$. Fix $\varepsilon>0$ and choose $R$ so large that \[ \delta\le \delta_r\le \delta+\varepsilon\qquad(r\ge R) \] (using the Round 2 monotone convergence $\delta_r\downarrow\delta$). Let \[ \alpha_R(x):=\sum_{r=1}^{R-1}w_r(x). \] Then \[ 0\le \alpha_R(x)\le \frac{\log(n_R/n_1)}{\log x}\to 0. \] For large $x$ (so that $K\ge R$), the tail contribution in (6.12) satisfies \[ \delta\left(\sum_{r=R}^{K-1}w_r(x)+\nu_K(x)\right) \le \sum_{r=R}^{K-1}w_r(x)\delta_r+\nu_K(x)\delta_K \le (\delta+\varepsilon)\left(\sum_{r=R}^{K-1}w_r(x)+\nu_K(x)\right). \] The initial contribution is bounded between $0$ and $\alpha_R(x)$. Since the total weight tends to $1$ by (6.14), it follows that \[ \liminf_{x\to\infty}\left(\sum_{r=1}^{K-1}w_r(x)\delta_r+\nu_K(x)\delta_K\right)\ge \delta, \] and \[ \limsup_{x\to\infty}\left(\sum_{r=1}^{K-1}w_r(x)\delta_r+\nu_K(x)\delta_K\right)\le \delta+\varepsilon. \] As $\varepsilon>0$ is arbitrary, the weighted average converges to $\delta$. Equation (6.12) therefore yields \[ \frac{S(x)}{\log x}\longrightarrow \delta. \] This is exactly $\delta_{\log}(A)=\delta$. $\square$

Comparison with Round 5.

Round 5 used the cumulative counts \[ M_k=\#\{I\subseteq\{1,\dots,k\}: I\text{ compatible}\}. \] The new theorem replaces the stagewise error condition \[ \sum_{k\le K}\frac{M_k}{n_k}=o(\log n_K) \] by the sharper local condition (6.6). The conceptual gain is that a compatible subfamily $I$ is now charged once, at its activation index $r(I)=\max I$, instead of being re-charged on every later interval.

6.5. Graph-theoretic reformulation

For each $r\ge 1$, define the backward neighborhood of $r$ in the compatibility graph by \[ N^{-}(r):=\{i<r:\ i\sim r\}, \qquad d_r^{-}:=|N^{-}(r)|. \]

Lemma 6.4 (local counts are clique counts).

For each $r\ge 1$, the number $C_r$ equals the number of cliques in the induced subgraph $G[N^{-}(r)]$, with the empty clique included. In particular, \begin{equation} C_r\le 2^{d_r^{-}}. \tag{6.15} \end{equation}

Proof. A set $I\in\mathcal D_r$ has the form \[ I=J\cup\{r\},\qquad J\subseteq\{1,\dots,r-1\}. \] By Lemma 6.1, $I$ is compatible if and only if it is a clique in $G$. Since $r\in I$, this is equivalent to saying that every vertex of $J$ lies in $N^{-}(r)$ and that $J$ itself is a clique in the induced graph on $N^{-}(r)$. Thus compatible sets with maximum index $r$ are in bijection with cliques of $G[N^{-}(r)]$, including the empty clique $J=\varnothing$. The bound (6.15) is immediate. $\square$

Corollary 6.5 (degree-based criterion).

If \begin{equation} \sum_{r=1}^{K}\frac{2^{d_r^{-}}}{n_r}=o(\log n_K)\qquad(K\to\infty), \tag{6.16} \end{equation} then $\delta_{\log}(A)$ exists.

Proof. Combine (6.15) with Theorem 6.3. $\square$

Corollary 6.6 (bounded backward degree).

If the compatibility graph satisfies \[ \sup_{r\ge 1} d_r^{-}\le D<\infty \] and \begin{equation} \sum_{r=1}^{K}\frac1{n_r}=o(\log n_K), \tag{6.17} \end{equation} then $\delta_{\log}(A)$ exists. In particular, this holds whenever $d_r^{-}\le D$ uniformly and \[ \sum_{r\ge 1}\frac1{n_r}<\infty \] (for example, if $n_r\gg r^{1+\eta}$ for some $\eta>0$).

Proof. Under the bounded-degree hypothesis, (6.15) gives $C_r\le 2^D$ for every $r$. Hence \[ \sum_{r=1}^{K}\frac{C_r}{n_r} \le 2^D\sum_{r=1}^{K}\frac1{n_r}, \] so Theorem 6.3 applies. $\square$

Adversarial Verification

We stress-test the new argument.

• No illegal infinite inclusion--exclusion. Formula (6.5) is not an infinite alternating series at a fixed integer $n$. Only indices with $n_i\le n$ are active, so the sum is finite and is exactly ordinary inclusion--exclusion on the active constraints.
• Correct use of the generalized CRT. Lemma 6.1 is proved, not assumed. The key induction step checks compatibility with the combined modulus $L_J$ through the exact identity \[ \gcd(L_J,n_r)=\mathrm{lcm}\left(\gcd(n_j,n_r):\ j\in J\right). \]
• The new error term really is local. In (6.8), each compatible family $I$ contributes one activated progression beginning at $n_{r(I)}$, so its harmonic approximation error is paid once as $O(1/n_{r(I)})$. This is the central improvement over the Round 5 stagewise bookkeeping.
• Main-term algebra. Identity (6.11) is an exact summation-by-parts identity. It recovers the same weighted-average structure as in Rounds 2--4, so the monotone convergence of $\delta_k$ can be reused safely.
• Intermediate $x$, not just endpoints. The proof works for arbitrary $x$, because the final active progression sums are taken directly up to $x$ in (6.8); no endpoint reduction is needed.
• Graph reformulation. The identification of compatible sets with cliques is exact by Lemma 6.1. Hence $C_r$ really is the local clique count in the backward neighborhood of $r$.
• Edge case $n_1=1$. As in earlier rounds, if $n_1=1$ then every integer is congruent to $a_1\pmod 1$, so $A=\varnothing$ and $\delta_{\log}(A)=0$. Theorem 6.3 is therefore stated for $n_1\ge 2$.

Final

UNRESOLVED (BUT STRICTLY ADVANCED).

The main new advance is Theorem 6.3, which sharpens the Round 5 inclusion--exclusion criterion from the cumulative stage counts $M_k$ to the local activation counts $C_r$: \[ \sum_{r\le K}\frac{C_r}{n_r}=o(\log n_K) \quad\Longrightarrow\quad \delta_{\log}(A)\ \text{exists}. \] This is a real improvement in mechanism: each compatible subfamily is now charged exactly once, at the moment it first becomes active.

The graph-theoretic reformulation in Lemma 6.4 and Corollaries 6.5--6.6 isolates a sharper obstruction than any previous round:

A counterexample must have large local clique counts in the compatibility graph, on a harmonic scale, \[ \sum_{r\le K}\frac{C_r}{n_r}\not=o(\log n_K). \] Equivalently, sparse backward neighborhoods (or any hypothesis forcing small clique counts in those neighborhoods) cannot support a counterexample.

So the remaining open core is now narrower than after Round 5: one must combine large activated compatible-subfamily complexity with failure of all previously proved lcm/discrepancy criteria.

Completion Estimate

COMPLETION: 88%.

References

• No external references were used in this round.
• The argument builds directly on the vetted Round 1--Round 5 results supplied in the conversation.