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Math-647-A4 · revision r1 · on Math-647

gpt pro 5.2 attempt on Math-647, version 2

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Contents
  1. Round-2 Objective
  2. Round-1 Foundation Used
  3. New Insight / Tool (Round-2)
  4. (i) Stage densities are automatically monotone, hence converge
  5. (ii) Any counterexample must come from local logarithmic bias, not from density oscillation
  6. Attack Plan (Round-2)
  7. Gaps left after Round 1
  8. Round-2 plan
  9. Work (Round-2)
  10. Stage sets are nested; stage densities converge automatically
  11. Weighted averages of stage densities converge unconditionally
  12. Endpoint decomposition via interval logarithmic densities
  13. A sharpened obstruction: where non-convergence would have to come from
  14. Interpretation.
  15. Edge cases and a sanity example (showing $H2$ is not necessary)
  16. Case $n_1=1$.
  17. Example: $H2$ can fail while $\delta_{\log}(A)$ exists.
  18. Adversarial Verification
  19. Final
  20. Completion Estimate
  21. References

Round-2 Objective

We pursue (A) a proof strategy. Round 1 already isolates the key structural fact \begin{equation} \mathbf{1}_A(n)=\mathbf{1}_{A^{(k)}}(n)\quad\text{whenever }n_k\le n<n_{k+1}, \tag{R1-Id} \end{equation} so $A$ is obtained by patching together periodic stage-sets $A^{(k)}$ on successive intervals.

Round 1 identified two auxiliary hypotheses $H1$--$H2$ under which the logarithmic density exists (Lemma 25.2 in Round 1), but also pointed out that removing $H1$--$H2$ is the main remaining gap.

In this round we do not resolve the open problem. Instead, we (i) remove one redundant hypothesis from the Round-1 theorem, and (ii) sharpen the obstruction: any counterexample must come from a genuinely new phenomenon (large local logarithmic biases of periodic stage-sets on the activation intervals), not from oscillation of the stage densities themselves.

Round-1 Foundation Used

We rely on the following vetted Round-1 results and notation.
  • Definition of stage-sets: \[ A^{(k)}:=\{n\in\mathbb N: n\not\equiv a_i\pmod{n_i}\text{ for all }1\le i\le k\}. \]
  • Identity on activation intervals (Round 1, $1$): for $n_k\le n<n_{k+1}$, \[ \mathbf{1}_A(n)=\mathbf{1}_{A^{(k)}}(n). \]
  • Logarithmic sum decomposition (Round 1, $2$): writing \[ S(x):=\sum_{\substack{1\le n\le x\\ n\in A}}\frac1n, \] and letting $K(x)$ be the unique index with $n_{K(x)}\le x<n_{K(x)+1}$, one has \[ S(x)=\sum_{k=1}^{K(x)-1}\ \sum_{n=n_k}^{n_{k+1}-1}\frac{\mathbf{1}_{A^{(k)}}(n)}{n} \ +\ \sum_{n=n_{K(x)}}^{\lfloor x\rfloor}\frac{\mathbf{1}_{A^{(K(x))}}(n)}{n}. \]
  • Finite-constraint case (Round 1, Lemma 25.1): if only finitely many constraints are present, then $A$ is periodic and $\delta_{\log}(A)$ exists.
  • Round-1 sufficient condition (Round 1, Lemma 25.2): if $(\delta_k)$ converges and a "no single interval dominates" hypothesis $H2$ holds, then $\delta_{\log}(A)$ exists and equals $\lim\delta_k$.

New Insight / Tool (Round-2)

(i) Stage densities are automatically monotone, hence converge

Round 1 assumed convergence of the stage densities $\delta_k$ as hypothesis $H1$. In fact this convergence is automatic:
The periodic sets $A^{(k)}$ form a nested decreasing sequence, so their natural densities $\delta_k$ form a bounded monotone sequence and therefore converge.
This removes $H1$ entirely from the Round-1 theorem.

(ii) Any counterexample must come from local logarithmic bias, not from density oscillation

A key conceptual refinement is that the weighted averages of the stage densities converge without any growth assumption on $(n_k)$ once monotonicity is noted. Thus the only possible source of non-convergence of $S(x)/\log x$ is a persistent discrepancy between:
  • the global stage density $\delta_k$ of the periodic set $A^{(k)}$, and
  • the local logarithmic density of $A^{(k)}$ on the specific interval $[n_k,n_{k+1})$ where it is used to define $A$.
This sharpens the obstruction and rules out an entire class of naive "alternating $\delta_k$" counterexample strategies.

Attack Plan (Round-2)

Gaps left after Round 1

Round 1 left the main open gap: remove $H2$, i.e. treat arbitrary growth of $n_{k+1}/n_k$.

Round-2 plan

  1. Prove that $\delta_k$ is monotone decreasing and therefore converges, eliminating $H1$.
  2. Prove that the weighted averages of $\delta_k$ appearing in the Round-1 decomposition converge to $\delta:=\lim_k\delta_k$ with no growth hypotheses.
  3. Introduce an exact "interval log-density" $p_k$ and show that the endpoint ratios $S(n_K-1)/\log n_K$ are weighted averages of the $p_k$.
  4. Deduce a necessary condition for non-convergence: one must produce infinitely many indices with simultaneously (a) large logarithmic weight and (b) large deviation $p_k-\delta_k$.
This does not solve the open problem, but it upgrades the Round-1 framework by isolating a single, sharply stated obstruction.

Work (Round-2)

Stage sets are nested; stage densities converge automatically

Lemma (Monotonicity and convergence of stage densities): For all $k\ge 1$ one has \[ A^{(k+1)}\subseteq A^{(k)}. \] Consequently the natural densities $\delta_k:=d(A^{(k)})$ satisfy \[ 1\ge \delta_1\ge \delta_2\ge \cdots \ge 0, \] and therefore converge to a limit \[ \delta:=\lim_{k\to\infty}\delta_k\in[0,1]. \] Moreover, \begin{equation} 0\le \delta_k-\delta_{k+1}\le \frac{1}{n_{k+1}}\qquad(k\ge 1). \tag{2.1} \end{equation}
Proof: The containment $A^{(k+1)}\subseteq A^{(k)}$ is immediate from the definitions: $A^{(k+1)}$ imposes all $k$ congruence-avoidance conditions defining $A^{(k)}$ and one additional condition.

Each $A^{(k)}$ is periodic (modulo $\mathrm{lcm}(n_1,\dots,n_k)$), hence has a natural density $\delta_k$; monotonicity of sets implies monotonicity of their densities, so $(\delta_k)$ is decreasing and bounded below, hence convergent.

For the one-step bound, note that \[ A^{(k)}\setminus A^{(k+1)}\subseteq \{n\in\mathbb N: n\equiv a_{k+1}\pmod{n_{k+1}}\}. \] The set on the right is a single residue class modulo $n_{k+1}$ and has natural density exactly $1/n_{k+1}$. Therefore \[ \delta_k-\delta_{k+1}=d(A^{(k)}\setminus A^{(k+1)})\le \frac{1}{n_{k+1}}, \] proving (2.1). ∎

Weighted averages of stage densities converge unconditionally

For $x\ge 2$ let $K(x)$ be the unique index with $n_{K(x)}\le x<n_{K(x)+1}$, and define weights \[ w_k(x):=\frac{\log(n_{k+1}/n_k)}{\log x}\qquad(1\le k\le K(x)-1). \] Note that \[ \sum_{k=1}^{K(x)-1} \log\frac{n_{k+1}}{n_k}=\log\frac{n_{K(x)}}{n_1}\le \log x, \] so $\sum_{k=1}^{K(x)-1} w_k(x)\le 1$.
Proposition (Unconditional convergence of the stage-density weighted averages): With $\delta:=\lim_k\delta_k$ as above, \begin{equation} \sum_{k=1}^{K(x)-1} w_k(x)\,\delta_k\ \longrightarrow\ \delta\qquad(x\to\infty), \tag{2.2} \end{equation} with no additional assumptions on the growth of $(n_k)$.
Proof: Fix $\varepsilon>0$ and choose $K_0$ such that $|\delta_k-\delta|<\varepsilon$ for all $k\ge K_0$ (possible since $\delta_k\to\delta$). Then \[ \left|\sum_{k=1}^{K(x)-1} w_k(x)\delta_k-\delta\sum_{k=1}^{K(x)-1}w_k(x)\right| \le \sum_{k=1}^{K(x)-1} w_k(x)|\delta_k-\delta| \le \varepsilon\sum_{k\ge K_0}w_k(x)+\max_{k<K_0}|\delta_k-\delta|\sum_{k<K_0}w_k(x). \] The first term is $\le \varepsilon$ since $\sum w_k(x)\le 1$. For the second term, observe that \[ \sum_{k<K_0} w_k(x)=\frac{1}{\log x}\sum_{k<K_0}\log\frac{n_{k+1}}{n_k}=\frac{\log(n_{K_0}/n_1)}{\log x}\to 0\qquad(x\to\infty). \] So the second term tends to $0$, and we obtain \[ \limsup_{x\to\infty}\left|\sum_{k=1}^{K(x)-1} w_k(x)\delta_k-\delta\sum_{k=1}^{K(x)-1}w_k(x)\right|\le \varepsilon. \] Finally, $\sum_{k=1}^{K(x)-1}w_k(x)=\log(n_{K(x)}/n_1)/\log x\to 1$ as $x\to\infty$ because $n_{K(x)}\le x<n_{K(x)+1}$ implies $\log n_{K(x)}\sim\log x$. Thus the right-hand expression tends to $\delta$, and since $\varepsilon$ was arbitrary, (2.2) follows. ∎

Endpoint decomposition via interval logarithmic densities

Define, for each $k\ge 1$, the interval logarithmic density of the stage-set $A^{(k)}$ on its activation interval by \begin{equation} p_k\ :=\ \frac{1}{\log(n_{k+1}/n_k)}\sum_{n=n_k}^{n_{k+1}-1}\frac{\mathbf{1}_{A^{(k)}}(n)}{n}. \tag{2.3} \end{equation} (When $n_{k+1}=n_k+1$ this is a trivial one-point average; the definition still makes sense.)
Lemma (Exact endpoint formula): Assume $n_1>1$ (the case $n_1=1$ is treated separately below). For every integer $K\ge 2$, \begin{equation} \sum_{\substack{1\le n\le n_K-1\\ n\in A}}\frac1n \ =\ \sum_{n=1}^{n_1-1}\frac1n\ +\ \sum_{k=1}^{K-1} \left(\log\frac{n_{k+1}}{n_k}\right)\,p_k. \tag{2.4} \end{equation} Consequently, \begin{equation} \frac{1}{\log n_K}\sum_{\substack{1\le n\le n_K-1\\ n\in A}}\frac1n \ =\ \sum_{k=1}^{K-1}\frac{\log(n_{k+1}/n_k)}{\log n_K}\,p_k\ +\ O\!\left(\frac{1}{\log n_K}\right). \tag{2.5} \end{equation}
Proof: For $1\le n<n_1$ there are no active congruence restrictions, so $n\in A$ and the initial contribution is $\sum_{n=1}^{n_1-1}1/n$.

For $n_1\le n\le n_K-1$, we partition the range into disjoint intervals $[n_k,n_{k+1})$ for $1\le k\le K-1$. On each such interval, the Round-1 identity (R1-Id) gives $\mathbf{1}_A(n)=\mathbf{1}_{A^{(k)}}(n)$. Therefore \[ \sum_{\substack{n_1\le n\le n_K-1\\ n\in A}}\frac1n =\sum_{k=1}^{K-1}\sum_{n=n_k}^{n_{k+1}-1}\frac{\mathbf{1}_{A^{(k)}}(n)}{n} =\sum_{k=1}^{K-1}\left(\log\frac{n_{k+1}}{n_k}\right)p_k, \] which is (2.4). Dividing by $\log n_K$ gives (2.5). ∎

A sharpened obstruction: where non-convergence would have to come from

The quantities $p_k$ are the true "inputs" to the endpoint averages (2.5). They differ from the global stage densities $\delta_k$ by a local bias term. Define \begin{equation} \varepsilon_k\ :=\ \sum_{n=n_k}^{n_{k+1}-1}\frac{\mathbf{1}_{A^{(k)}}(n)}{n}\ -\ \delta_k\log\frac{n_{k+1}}{n_k}, \tag{2.6} \end{equation} so that \begin{equation} p_k=\delta_k+\frac{\varepsilon_k}{\log(n_{k+1}/n_k)}. \tag{2.7} \end{equation} Combining (2.5), (2.7), and Proposition (2.2) yields the reduction: \begin{equation} \frac{1}{\log n_K}\sum_{\substack{1\le n\le n_K-1\\ n\in A}}\frac1n \ =\ \delta\ +\ \frac{1}{\log n_K}\sum_{k=1}^{K-1}\varepsilon_k\ +\ o(1). \tag{2.8} \end{equation} (Here we used that $\sum_{k=1}^{K-1}\log(n_{k+1}/n_k)=\log(n_K/n_1)$.)
Corollary (A single obstruction quantity): Still assuming $n_1>1$, if \begin{equation} \frac{1}{\log n_K}\sum_{k=1}^{K-1}\varepsilon_k\ \longrightarrow\ 0\qquad(K\to\infty), \tag{2.9} \end{equation} then the endpoint ratios $S(n_K-1)/\log n_K$ converge to $\delta$. In particular, any failure of convergence of $S(x)/\log x$ forces \begin{equation} \limsup_{K\to\infty}\ \frac{1}{\log n_K}\left|\sum_{k=1}^{K-1}\varepsilon_k\right|\ >\ 0. \tag{2.10} \end{equation}
Proof: This is immediate from (2.8). ∎
Interpretation.
Equation (2.8) (and the obstruction (2.10)) is the main Round-2 conceptual advance. It shows that stage-density oscillation cannot produce a counterexample: the contribution coming from $\delta_k$ alone already converges to $\delta$. Thus any counterexample must engineer a systematic, order-$\log n_K$ accumulation of the local bias terms $\varepsilon_k$.

Edge cases and a sanity example (showing $H2$ is not necessary)

Case $n_1=1$.
If $n_1=1$, then every integer $n\ge 1$ satisfies $n\equiv a_1\pmod 1$, so the defining condition forces $A=\varnothing$. Hence $\delta_{\log}(A)=0$ exists.
Example: $H2$ can fail while $\delta_{\log}(A)$ exists.
Let \[ n_1=2,\ a_1\equiv 0\pmod 2,\qquad n_k:=2^{2^{k-1}}\ (k\ge 2),\ a_k\equiv 0\pmod{n_k}. \] Then $\log(n_{k+1}/n_k)/\log n_k\to 1$, so the Round-1 hypothesis $H2$ fails badly. Nevertheless, the first constraint already excludes every even $n\ge 2$, so $A$ is exactly the odd integers. Hence $\delta_{\log}(A)=1/2$.

Adversarial Verification

We check the new steps for hidden assumptions and boundary issues.
  • Monotonicity of $A^{(k)}$ and $\delta_k$. This is purely set-theoretic and does not use any coprimality or growth assumptions. The existence of $\delta_k$ is guaranteed because $A^{(k)}$ is periodic modulo $\mathrm{lcm}(n_1,\dots,n_k)$.
  • Bound $\delta_k-\delta_{k+1}\le 1/n_{k+1}$. The set $\{n\equiv a_{k+1}\pmod{n_{k+1}}\}$ indeed has natural density exactly $1/n_{k+1}$ for any residue class (no threshold here).
  • Weighted-average limit (Proposition (2.2)). The only subtlety is that $\sum_{k=1}^{K(x)-1}w_k(x)$ is not exactly $1$ but tends to $1$. This uses $\log n_{K(x)}\sim\log x$, which follows from $n_{K(x)}\le x<n_{K(x)+1}$ and monotonicity.
  • Endpoint decomposition (Lemma (2.4)). This uses only the Round-1 identity (R1-Id) on each interval $[n_k,n_{k+1})$. No approximation is involved; the formula is exact up to the finite initial segment $[1,n_1)$.
  • Reduction (2.8). This is an algebraic recombination of (2.5), (2.7), and Proposition (2.2). In particular, it does not assume $H2$. If any non-convergence occurs, it must enter through the normalized partial sums of $\varepsilon_k$.
  • Quantifier check. All limits are along $K\to\infty$ or $x\to\infty$. The reduction is stated only for the subsequence $x=n_K-1$; a full solution would also need to control intermediate $x$, but any failure of convergence of $S(x)/\log x$ would in particular force a failure along some subsequence, so isolating the endpoint behavior is a legitimate obstruction.

Final

UNRESOLVED (BUT STRICTLY ADVANCED).

Relative to Round 1, we have:

  • Proven that the stage densities $\delta_k$ are automatically monotone decreasing and thus converge; this removes Round-1 hypothesis $H1$.
  • Shown that the weighted averages of stage densities that arise naturally from the interval decomposition converge to $\delta:=\lim_k\delta_k$ with no growth assumption.
  • Produced an exact endpoint representation in terms of interval log-densities $p_k$ and an explicit "bias" sequence $\varepsilon_k$; the open problem reduces to controlling \[\frac{1}{\log n_K}\sum_{k<K}\varepsilon_k\] and any counterexample must force this quantity to have nonzero limsup (or fail to converge) (cf. (2.10)).

In particular, the naive counterexample strategy proposed in Round 1 (make $\delta_k$ alternate between two separated values) is impossible because $\delta_k$ is monotone.

Completion Estimate

COMPLETION: 60%.

References