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Math-647-A5 · revision r1 · on Math-647

gpt pro 5.2 attempt on Math-647, version 3

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Manuscript

Contents
  1. Round-4 Objective
  2. Round-3 Foundation Used
  3. New Insight / Tool (Round-4)
  4. (i) A bounded-lcm structural hypothesis that preserves the Round 3 mechanism
  5. (ii) Harmonic sampling from an arbitrary starting point
  6. (iii) (LC) forces geometric growth
  7. Attack Plan (Round-4)
  8. Work (Round-4)
  9. Setup and residue sets modulo $L_k$
  10. Discrepancy and its recursion under (LC)
  11. (LC) forces geometric growth
  12. Shifted harmonic averaging for periodic sets
  13. Existence of logarithmic density under bounded lcm-factor
  14. Examples and scope
  15. (LC) is strictly more general than the nested case.
  16. Adversarial Verification
  17. Final
  18. Completion Estimate
  19. References

Round-4 Objective

We pursue (A) a proof strategy by extending the Round 3 "nested moduli" theorem to a strictly larger structural class. Round 3 established existence of $\delta_{\log}(A)$ when $n_1\mid n_2\mid n_3\mid\cdots$. The mechanism was that, at stage $k$, the relevant period equals $n_k$, so the new constraint removes exactly one residue class inside one period, yielding a linear discrepancy bound and summable harmonic errors.

In this round we generalize this mechanism to cases where the stage period $\mathrm{lcm}(n_1,\dots,n_k)$ may exceed $n_k$, but only by a uniformly bounded factor. This rules out a broad class of potential counterexamples (including those with huge gaps $n_{k+1}/n_k$ so that the Round 1/2 "no single interval dominates" hypothesis $H2$ can fail).

Round-3 Foundation Used

We rely on the following Round 3 (and earlier) ingredients.
  • Stage sets and activation identity (Round 1): $A^{(k)}:=\{n: n\not\equiv a_i\!\!\pmod{n_i}\ \forall i\le k\}$ and $\mathbf 1_A(n)=\mathbf 1_{A^{(k)}}(n)$ for $n_k\le n<n_{k+1}$.
  • Monotone convergence of stage densities (Round 2): $A^{(k+1)}\subseteq A^{(k)}$, hence $\delta_k:=d(A^{(k)})$ is decreasing and converges to $\delta:=\lim_k\delta_k$.
  • Discrepancy method blueprint (Round 3): define a discrepancy parameter for the residue set of a periodic stage-set, prove a linear-in-$k$ discrepancy recursion under a "delete $O(1)$ points per period" structure, and use Abel summation to convert discrepancy bounds into harmonic-sum error bounds.

New Insight / Tool (Round-4)

(i) A bounded-lcm structural hypothesis that preserves the Round 3 mechanism

Let \begin{equation} L_k := \mathrm{lcm}(n_1,\dots,n_k). \tag{4.1} \end{equation} We introduce the hypothesis: \begin{equation} \exists C\ge 1\ \forall k\ge 1,\qquad \frac{L_k}{n_k}\le C. \tag{LC} \end{equation} Since $n_k\mid L_k$, this is equivalent to writing $L_k=d_k n_k$ with integers $1\le d_k\le C$. Under (LC), the $k$-th excluded residue class $a_k\pmod{n_k}$ occupies exactly $d_k\le C$ residues in one period $\bmod L_k$. Thus each new stage deletes only $O_C(1)$ residues per period (not necessarily just one), and we can re-run the Round 3 discrepancy recursion.

(ii) Harmonic sampling from an arbitrary starting point

Round 3 used a harmonic averaging lemma starting at a period boundary. Here the activation interval starts at $n_k$, which need not be a multiple of $L_k$. We prove a shifted version: for any periodic set with discrepancy $\Delta$, the harmonic sum over $[X,Y)$ differs from $\delta\log(Y/X)$ by $O(\Delta/X)$.

(iii) (LC) forces geometric growth

A crucial bonus: (LC) implies a uniform growth factor $n_{k+1}\ge (1+1/C)n_k$. Hence $\sum_k k/n_k$ converges, so the accumulated harmonic errors are bounded.

Attack Plan (Round-4)

The remaining obstacle after Round 3 was that, in general, the stage period is $L_k$, which may be much larger than $n_k$, preventing the "one deletion per period" discrepancy recursion. Under (LC) we recover this deletion structure with $\le C$ deletions per period.

Concretely we will:

  1. Work modulo $L_k$, defining the allowed residue set $R_k\subset \{0,\dots,L_k-1\}$.
  2. Prove a recursion: $R_k$ is obtained from the lifted set for stage $k-1$ by deleting at most $d_k\le C$ residues.
  3. Deduce a discrepancy recursion $\Delta_k\le \Delta_{k-1}+O(C)$, hence $\Delta_k=O_C(k)$.
  4. Prove the shifted harmonic averaging lemma with error $O(\Delta_k/n_k)$.
  5. Sum the errors across stages (using geometric growth) and conclude $S(x)/\log x\to \delta$, i.e. $\delta_{\log}(A)$ exists.

Work (Round-4)

Setup and residue sets modulo $L_k$

Assume $n_1\ge 1$ and $n_1<n_2<\cdots$, with residue classes $a_i\pmod{n_i}$, and define $A$ as in Round 1. If $n_1=1$ then $A=\varnothing$ (since every $n\ge 1$ satisfies $n\equiv a_1\pmod 1$), hence $\delta_{\log}(A)=0$. So we henceforth assume \begin{equation} n_1\ge 2. \tag{4.2} \end{equation}

For each $k\ge 1$, define $L_k$ by (4.1) and the residue set \begin{equation} R_k := \left\{r\in\{0,1,\dots,L_k-1\}: \forall i\le k,\ r\not\equiv a_i\pmod{n_i}\right\}. \tag{4.3} \end{equation} Then $A^{(k)}$ is periodic modulo $L_k$ with \begin{equation} \mathbf 1_{A^{(k)}}(n)=\mathbf 1_{R_k}(n\bmod L_k). \tag{4.4} \end{equation} Let \begin{equation} \delta_k := \frac{|R_k|}{L_k}=d(A^{(k)}). \tag{4.5} \end{equation} By Round 2, $(\delta_k)$ decreases and converges to $\delta\in[0,1]$.

Discrepancy and its recursion under (LC)

For a set $R\subseteq\{0,\dots,q-1\}$ of density $\delta=|R|/q$, define its (prefix) discrepancy \begin{equation} \Delta(R;q) := \max_{0\le t\le q}\left|\,|R\cap\{0,1,\dots,t-1\}|-\delta t\,\right|. \tag{4.6} \end{equation} For stage $k$ we write \begin{equation} \Delta_k := \Delta(R_k;L_k). \tag{4.7} \end{equation}
Lemma (Lift-and-delete structure under (LC)): Assume (LC), so $L_k=d_k n_k$ with $1\le d_k\le C$. Let \begin{equation} \widetilde R_{k-1} := \{r\in\{0,\dots,L_k-1\}: r\bmod L_{k-1}\in R_{k-1}\} \tag{4.8} \end{equation} be the lift of $R_{k-1}$ to modulus $L_k$. Then \begin{equation} R_k = \widetilde R_{k-1}\setminus E_k, \tag{4.9} \end{equation} where \begin{equation} E_k \subseteq \{r\in\{0,\dots,L_k-1\}: r\equiv a_k\pmod{n_k}\} \quad\text{and}\quad |E_k|\le d_k\le C. \tag{4.10} \end{equation} Moreover, $\widetilde R_{k-1}$ is a disjoint union of $L_k/L_{k-1}$ translates of $R_{k-1}$ by multiples of $L_{k-1}$, so \begin{equation} \Delta(\widetilde R_{k-1};L_k)=\Delta(R_{k-1};L_{k-1})=\Delta_{k-1}. \tag{4.11} \end{equation}
Proof: Because $L_{k-1}\mid L_k$, membership in the first $k-1$ congruence conditions depends only on $r\bmod L_{k-1}$, which gives (4.8). Imposing the additional condition $r\not\equiv a_k\pmod{n_k}$ removes from $\widetilde R_{k-1}$ those residues congruent to $a_k\pmod{n_k}$; this yields (4.9)--(4.10). Since $n_k\mid L_k$, the congruence $r\equiv a_k\pmod{n_k}$ has exactly $L_k/n_k=d_k$ solutions modulo $L_k$, so $|E_k|\le d_k\le C$.

Finally, $\widetilde R_{k-1}$ is obtained by repeating the pattern of $R_{k-1}$ in consecutive blocks of length $L_{k-1}$, hence its prefix discrepancy equals that of $R_{k-1}$, giving (4.11). ∎

Lemma (Linear discrepancy bound under (LC)): Assume (LC). Then \begin{equation} \Delta_k \le \Delta_{k-1}+2C\qquad(k\ge 2). \tag{4.12} \end{equation} In particular, $\Delta_k\ll_C k$.
Proof: Work in modulus $L_k$. By Lemma (4.9), $R_k$ is obtained from $\widetilde R_{k-1}$ by deleting at most $C$ points. Hence for every prefix length $t$ we have \begin{equation} \left|\,|R_k\cap[0,t)|-|\widetilde R_{k-1}\cap[0,t)|\,\right|\le C. \tag{4.13} \end{equation} Also, deleting at most $C$ points changes the density by at most $C/L_k$: \begin{equation} |\delta_k-\delta(\widetilde R_{k-1})|\le \frac{C}{L_k}. \tag{4.14} \end{equation} Therefore, for every $t\in[0,L_k]$, \begin{align} \Bigl|\,|R_k\cap[0,t)|-\delta_k t\,\Bigr| &\le \Bigl|\,|\widetilde R_{k-1}\cap[0,t)|-\delta(\widetilde R_{k-1})t\,\Bigr| \\ &\quad + \bigl|\,|R_k\cap[0,t)|-|\widetilde R_{k-1}\cap[0,t)|\,\bigr| \\ &\quad + |\delta_k-\delta(\widetilde R_{k-1})|\,t \\ &\le \Delta(\widetilde R_{k-1};L_k)+C+\frac{C}{L_k}\,L_k. \end{align} Using (4.11) gives $\Delta_k\le \Delta_{k-1}+2C$, which is (4.12). Iterating yields $\Delta_k\ll_C k$. ∎

(LC) forces geometric growth

Lemma (Uniform growth factor): Assume (LC). Then for every $k\ge 1$, \begin{equation} \frac{n_{k+1}}{n_k}\ge 1+\frac{1}{C}. \tag{4.15} \end{equation} Consequently $n_k\gg_C (1+1/C)^k$ and \begin{equation} \sum_{k\ge 1}\frac{k}{n_k}<\infty. \tag{4.16} \end{equation}
Proof: Since $L_{k+1}\ge \mathrm{lcm}(n_k,n_{k+1})=\dfrac{n_k n_{k+1}}{\gcd(n_k,n_{k+1})}$ and (LC) gives $L_{k+1}\le C n_{k+1}$, we obtain \begin{equation} \frac{n_k}{\gcd(n_k,n_{k+1})}\le C, \qquad\text{so}\qquad \gcd(n_k,n_{k+1})\ge \frac{n_k}{C}. \tag{4.17} \end{equation} Write $n_k=d u$ and $n_{k+1}=d v$ with $d=\gcd(n_k,n_{k+1})$ and $\gcd(u,v)=1$. Then (4.17) gives $u\le C$. Since $n_{k+1}>n_k$, we have $v>u$, hence $v\ge u+1$. Therefore \[\frac{n_{k+1}}{n_k}=\frac{v}{u}\ge \frac{u+1}{u}\ge 1+\frac{1}{C},\] proving (4.15). Geometric growth follows by iteration. Finally, (4.16) holds because $\sum k r^{-k}$ converges for every $r>1$. ∎

Shifted harmonic averaging for periodic sets

Lemma (Shifted harmonic averaging via discrepancy): Let $B\subseteq\mathbb N$ be periodic with period $q\ge 1$, and let $\delta$ be its (natural) density. Let $R\subseteq\{0,\dots,q-1\}$ be the corresponding residue set (so $n\in B\iff n\bmod q\in R$), and let $\Delta:=\Delta(R;q)$. Then for all real $1\le X<Y$, \begin{equation} \sum_{X\le n<Y}\frac{\mathbf 1_B(n)}{n} =\delta\sum_{X\le n<Y}\frac{1}{n}+O\!\left(\frac{\Delta}{X}\right). \tag{4.18} \end{equation} Equivalently, \begin{equation} \sum_{X\le n<Y}\frac{\mathbf 1_B(n)}{n} =\delta\log\frac{Y}{X}+O\!\left(\frac{\Delta}{X}\right)+O\!\left(\frac{1}{X}\right). \tag{4.19} \end{equation} The implied constants are absolute.
Proof: Define the centered periodic function $g(n):=\mathbf 1_B(n)-\delta$. Let $G(t):=\sum_{0\le n<t} g(n)$. By definition of discrepancy, for $0\le t\le q$ we have $|G(t)|\le \Delta$. Also $G(t+q)=G(t)$ because $\sum_{0\le n<q}g(n)=|R|-\delta q=0$. Thus $|G(t)|\le \Delta$ for all integers $t\ge 0$.

For integers $m\ge X$, define partial sums from $X$: $H(m):=\sum_{X\le n\le m} g(n)$. Then $H(m)=G(m+1)-G(X)$, so \begin{equation} |H(m)|\le 2\Delta\qquad(m\ge X). \tag{4.20} \end{equation}

Apply Abel summation: \begin{align} \sum_{X\le n<Y}\frac{g(n)}{n} &=\frac{H(\lfloor Y\rfloor-1)}{\lfloor Y\rfloor} +\sum_{m=\lceil X\rceil}^{\lfloor Y\rfloor-2} H(m)\left(\frac{1}{m}-\frac{1}{m+1}\right). \end{align} Using (4.20) and $\sum_{m\ge X}(1/m-1/(m+1))=1/X$ gives \[\left|\sum_{X\le n<Y}\frac{g(n)}{n}\right|\ll \frac{\Delta}{X}. \] Since $\mathbf 1_B=\delta+g$, this is (4.18). The conversion to (4.19) uses $\sum_{X\le n<Y}1/n=\log(Y/X)+O(1/X)$. ∎

Existence of logarithmic density under bounded lcm-factor

Theorem (Bounded lcm-factor case): Assume $n_1\ge 2$ and that (LC) holds, i.e. $\mathrm{lcm}(n_1,\dots,n_k)\le C n_k$ for all $k$. Then the logarithmic density $\delta_{\log}(A)$ exists, and \begin{equation} \delta_{\log}(A)=\delta:=\lim_{k\to\infty}\delta_k, \tag{4.21} \end{equation} where $\delta_k=d(A^{(k)})$.
Proof: Let $S(x):=\sum_{\substack{1\le n\le x\\ n\in A}}\frac{1}{n}$. Fix large $x$ and let $K=K(x)$ be the unique index with $n_K\le x<n_{K+1}$. By the activation identity (Round 1), for $n_k\le n<n_{k+1}$ we have $\mathbf 1_A(n)=\mathbf 1_{A^{(k)}}(n)$. Hence \begin{equation} S(x)=\sum_{n< n_1}\frac{1}{n} +\sum_{k=1}^{K-1}\sum_{n=n_k}^{n_{k+1}-1}\frac{\mathbf 1_{A^{(k)}}(n)}{n} +\sum_{n=n_K}^{\lfloor x\rfloor}\frac{\mathbf 1_{A^{(K)}}(n)}{n}. \tag{4.22} \end{equation}

We approximate each inner sum by $\delta_k\log(n_{k+1}/n_k)$. Apply Lemma (4.18) to the periodic set $A^{(k)}$, whose period is $L_k$ and whose residue set is $R_k$. With $X=n_k$ and $Y=n_{k+1}$ we obtain \begin{equation} \sum_{n=n_k}^{n_{k+1}-1}\frac{\mathbf 1_{A^{(k)}}(n)}{n} =\delta_k\sum_{n=n_k}^{n_{k+1}-1}\frac{1}{n} +O\!\left(\frac{\Delta_k}{n_k}\right). \tag{4.23} \end{equation} Similarly, for the final partial block, \begin{equation} \sum_{n=n_K}^{\lfloor x\rfloor}\frac{\mathbf 1_{A^{(K)}}(n)}{n} =\delta_K\sum_{n=n_K}^{\lfloor x\rfloor}\frac{1}{n} +O\!\left(\frac{\Delta_K}{n_K}\right). \tag{4.24} \end{equation} By Lemma (4.12), $\Delta_k\ll_C k$. By Lemma (4.16), $\sum k/n_k<\infty$. Therefore \begin{equation} \sum_{k\ge 1}\frac{\Delta_k}{n_k}<\infty, \tag{4.25} \end{equation} so the total contribution of the error terms in (4.23)--(4.24) is bounded uniformly in $x$. Using $\sum_{n=u}^{v}1/n=\log(v/u)+O(1/u)$ and absorbing the bounded error, (4.22) becomes \begin{equation} S(x)=O(1) +\sum_{k=1}^{K-1}\delta_k\log\frac{n_{k+1}}{n_k} +\delta_K\log\frac{x}{n_K}. \tag{4.26} \end{equation} Divide by $\log x$: \begin{equation} \frac{S(x)}{\log x} =\sum_{k=1}^{K-1}\frac{\log(n_{k+1}/n_k)}{\log x}\,\delta_k +\frac{\log(x/n_K)}{\log x}\,\delta_K +o(1). \tag{4.27} \end{equation} The weights in (4.27) are nonnegative. Their total mass is \[ W(x):=\sum_{k=1}^{K-1}\frac{\log(n_{k+1}/n_k)}{\log x}+\frac{\log(x/n_K)}{\log x} =\frac{\log(x/n_1)}{\log x} =1+O\!\left(\frac{1}{\log x}\right). \] Define normalized weights $\widetilde w_k:=w_k/W(x)$, so that $\sum_{k=1}^{K}\widetilde w_k=1$. Then (4.27) may be rewritten as \[ \frac{S(x)}{\log x}=W(x)\sum_{k=1}^{K}\widetilde w_k\,\delta_k+o(1). \] Since $W(x)\to 1$, it suffices to show $\sum_{k=1}^{K}\widetilde w_k\delta_k\to\delta$. Fix $\varepsilon>0$ and choose $K_0$ with $\delta_{K_0}-\delta<\varepsilon$. Let $\alpha(x):=\sum_{k< K_0}\widetilde w_k$. Then \[ \alpha(x)\le \frac{1}{W(x)}\,\frac{\log(n_{K_0}/n_1)}{\log x}\longrightarrow 0\qquad(x\to\infty). \] For $k\ge K_0$ we have $\delta\le \delta_k\le \delta_{K_0}$, hence \[ \delta\,(1-\alpha(x))\le \sum_{k=1}^{K}\widetilde w_k\delta_k \le \alpha(x)+\delta_{K_0}(1-\alpha(x))=\delta_{K_0}+\alpha(x)(1-\delta_{K_0}). \] Taking $x\to\infty$ and then $\varepsilon\to 0$ gives $\sum_{k=1}^{K}\widetilde w_k\delta_k\to\delta$, and therefore $S(x)/\log x\to\delta$, proving (4.21).

∎

Examples and scope

(LC) is strictly more general than the nested case.
Round 3 corresponds to $C=1$ (since then $L_k=n_k$ for all $k$). For $C>1$ there are many non-nested examples. For instance, fix an odd integer sequence $M_j\uparrow\infty$ with $M_{j+1}\gg M_j$ (e.g. $M_j=10^{10^j}$) and define \[n_{2j-1}:=3M_j,\qquad n_{2j}:=6M_j.\] Then $n_{2j-1}\nmid n_{2j-2}$ (so the moduli are not nested), while $\mathrm{lcm}(n_1,\dots,n_{2j-1})=6M_j=2n_{2j-1}$ and $\mathrm{lcm}(n_1,\dots,n_{2j})=6M_j=n_{2j}$, so (LC) holds with $C=2$. The ratios $n_{2j-1}/n_{2j-2}$ can be made enormous, so the Round 1/2 hypothesis $H2$ can fail badly even though $\delta_{\log}(A)$ exists by Theorem (4.21).

Adversarial Verification

We stress-test the new argument.
  • Edge case $n_1=1$. Handled: then $A=\varnothing$ and $\delta_{\log}(A)=0$.
  • Correctness of the lift step. The lift $\widetilde R_{k-1}$ in (4.8) uses only that $L_{k-1}\mid L_k$, which always holds by definition of lcm.
  • Size of the deleted set. Because $n_k\mid L_k$, the congruence class $a_k\pmod{n_k}$ has exactly $L_k/n_k=d_k$ representatives in $[0,L_k)$. Under (LC), $d_k\le C$, so deletion is indeed $O_C(1)$.
  • Discrepancy recursion. The only place where "deleting $\le C$ points" is used is (4.13) and the density-shift bound (4.14). Both are sharp and unconditional.
  • Shifted harmonic averaging lemma. The periodicity of the centered prefix sum $G(t)$ is checked explicitly; no hidden assumption about starting at a period boundary is used.
  • Summability of errors. Theorem (4.21) uses $\sum \Delta_k/n_k<\infty$. This is guaranteed because $\Delta_k\ll_C k$ and (LC) forces geometric growth of $n_k$. The growth lemma (4.15) is derived from a clean gcd argument.
  • Quantifier check. The conclusion is a full limit as $x\to\infty$, not only along endpoints $x=n_K$. This is ensured because we control the final partial interval $[n_K,x]$ via (4.24).

Final

UNRESOLVED (BUT STRICTLY ADVANCED).

Beyond Round 3 (nested moduli), we have proved a strictly stronger unconditional existence theorem:

If the stage lcm $\mathrm{lcm}(n_1,\dots,n_k)$ is always within a bounded factor of the current modulus $n_k$, then $\delta_{\log}(A)$ exists.
Equivalently, we rule out counterexamples in the regime where each new constraint deletes only $O(1)$ residues in one stage period.

The general open case still hinges on controlling the accumulated bias terms when $L_k/n_k$ is unbounded.

Completion Estimate

COMPLETION: 78%.

References