gpt pro 5.4 attempt on Math-647, version 4

Round-9 Objective

We pursue (C) obstruction/correction. Round 8 proved that if the finite-stage densities \[ \delta_k:=d(A^{(k)}) \] tend to $0$, then $A$ has natural density $0$, hence logarithmic density $0$. The exact unresolved gap after Round 8 is therefore the branch \[ \lim_{k\to\infty}\delta_k>0. \]

The new goal of this round is to determine whether that branch is genuinely simpler. We show that it is not: for every fixed $c<1$, the original problem is equivalent to its restriction to instances whose finite-stage densities are all at least $c$. In particular, any hypothetical counterexample can be transformed into one with \[ \inf_k d(A^{(k)})\ge 1-\varepsilon \] for any prescribed $\varepsilon>0$.

Thus the zero-density branch isolated in Round 8, while correct, is not a decisive reduction of the open problem. This rules out an entire class of proof strategies based on eventual smallness of the stage densities.

Round-8 Foundation Used

We rely on the following vetted earlier results.

• Stage sets. For any instance $(n_i,a_i)$, the finite-stage set \[ A^{(k)}:=\{n\in\mathbb N:\ n\not\equiv a_i \pmod{n_i}\ \forall\ 1\le i\le k\} \] is periodic, so its natural density \[ \delta_k:=d(A^{(k)}) \] exists.
• Round 2 monotonicity. One has $A^{(k+1)}\subseteq A^{(k)}$, hence \[ 1\ge \delta_1\ge \delta_2\ge \cdots \ge 0, \] so $\delta_k$ always converges.
• Periodic logarithmic asymptotics. If $P\subseteq\mathbb N$ is periodic with natural density $d(P)$, then \[ \sum_{\substack{n\le x\\ n\in P}}\frac1n = d(P)\log x+O(1). \] In particular, each residue class modulo $M$ contributes $(1/M)\log x+O(1)$ in harmonic weight.
• Round 8 zero-density theorem. If $\delta_k\to 0$, then $A$ has natural density $0$, hence logarithmic density $0$.

New Insight / Tool (Round-9)

The new tool is an affine padding transformation. Start from any instance $(m_j,b_j)$ and choose an integer $M\ge 3$. We build a new instance by \[ n_1=M,\quad a_1\equiv 0\pmod M, \qquad n_{j+1}:=Mm_j,\quad a_{j+1}:=Mb_j+1 \pmod{Mm_j}. \] This has three effects.

1. The residue class $0\pmod M$ is deleted immediately.
2. The residue classes $2,3,\dots,M-1\pmod M$ are permanently protected: no later congruence can ever hit them.
3. The residue class $1\pmod M$ carries a perfect affine copy of the original problem under the map $t\mapsto Mt+1$.

Hence the transformed set is exactly \[ \{1\}\ \cup\ \{n\equiv 2,3,\dots,M-1\pmod M\}\ \cup\ \{Mt+1:\ t\in B\}, \] where $B$ is the original set. The harmonic sum therefore satisfies \[ S_{A_M}(x)=\frac{M-2}{M}\log x+\frac1M S_B(x/M)+O(1), \] so upper/lower logarithmic densities are affinely related. At the same time, every finite-stage density of the transformed instance is bounded below by $(M-2)/M$.

This shows that the whole problem can be embedded into an arbitrarily high-density branch.

Attack Plan (Round-9)

The exact gap left after Round 8 is whether the remaining branch $\lim\delta_k>0$ is easier than the general problem.

We prove the following claims.

1. Given any nontrivial instance $(m_j,b_j)$, construct a transformed instance $(n_i,a_i)$ depending on a parameter $M\ge 3$.
2. Prove an exact description of the transformed set $A_M$ and of its finite-stage sets $A_M^{(k)}$.
3. Deduce an exact affine relation between the upper/lower logarithmic densities of $A_M$ and those of the original set $B$.
4. Deduce an exact affine relation between the stage densities of $A_M$ and those of $B$.
5. Conclude that, for every fixed $c<1$, the original problem is equivalent to its restriction to instances with \[ \inf_k d(A^{(k)})\ge c. \]

This overcomes the Round 8 obstacle by proving that the positive-density branch is not a genuine simplification of the original problem.

Work (Round-9)

Let \[ 1\le m_1<m_2<\cdots \] be an arbitrary increasing sequence of integers, with residues $b_j\pmod{m_j}$, and let \[ B:=\{t\in\mathbb N:\ \forall j\ge 1,\ (t<m_j)\ \text{or}\ (t\not\equiv b_j\!\!\pmod{m_j})\} \] be the associated set. For each $k\ge 1$, define the finite-stage periodic set \[ B^{(k)}:=\{t\in\mathbb N:\ t\not\equiv b_j\pmod{m_j}\ \forall\ 1\le j\le k\} \] and let \[ \beta_k:=d(B^{(k)}). \] By the Round 2 monotonicity result, $(\beta_k)$ decreases to a limit \[ \beta:=\lim_{k\to\infty}\beta_k\in[0,1]. \]

If $m_1=1$, then $B=\varnothing$, so the original instance is trivial and cannot be a counterexample to existence. Thus for the reduction below we may and do assume \begin{equation} m_1\ge 2. \tag{9.1} \end{equation}

Fix an integer $M\ge 3$. Define a new instance $(n_i,a_i)$ by \begin{equation} n_1:=M,\qquad a_1:=0, \tag{9.2} \end{equation} and for $j\ge 1$, \begin{equation} n_{j+1}:=Mm_j, \qquad a_{j+1}:=Mb_j+1. \tag{9.3} \end{equation} Because of (9.1), we have $n_2=Mm_1>M=n_1$, and strict increase follows from strict increase of $(m_j)$.

Let $A_M$ be the set associated with the transformed instance, and for each $k\ge 1$ let \[ A_M^{(k)}:=\{n\in\mathbb N:\ n\not\equiv a_i\pmod{n_i}\ \forall\ 1\le i\le k\} \] be its $k$-stage periodic set.

9.1. Exact description of the transformed set

Lemma 9.1.

For the transformed instance (9.2)--(9.3), one has the exact identity \begin{equation} A_M = \{1\} \cup \{n\in\mathbb N:\ n\equiv r\pmod M\ \text{for some }2\le r\le M-1\} \cup \{Mt+1:\ t\in B\}. \tag{9.4} \end{equation}

Proof. Let $n\in\mathbb N$.

If $n\equiv r\pmod M$ with $2\le r\le M-1$, then $n$ does not satisfy the first congruence $n\equiv 0\pmod M$, and no later congruence can hold either, because every later residue class $a_{j+1}=Mb_j+1$ is congruent to $1\pmod M$. Hence every such $n$ lies in $A_M$.

If $n\equiv 0\pmod M$, then $n\ge M=n_1$ and $n\equiv a_1\pmod{n_1}$, so $n\notin A_M$.

It remains to analyze the numbers $n\equiv 1\pmod M$. Write \[ n=Mt+1 \] with $t\ge 0$ an integer. The special case $t=0$ gives $n=1$, and clearly $1\in A_M$ since $1<n_i$ for every $i$. Now assume $t\ge 1$. The first congruence does not apply, because $Mt+1\not\equiv 0\pmod M$. For $j\ge 1$, the $(j+1)$-st condition is \[ (Mt+1<Mm_j)\ \text{or}\ (Mt+1\not\equiv Mb_j+1\pmod{Mm_j}). \] Since $t$ and $m_j$ are integers, this is equivalent to \[ (t<m_j)\ \text{or}\ (t\not\equiv b_j\pmod{m_j}). \] Thus $Mt+1\in A_M$ if and only if $t\in B$. Combining the three residue classes proves (9.4). $\square$

9.2. Exact description of the finite stages

Lemma 9.2.

For every $k\ge 1$ one has \begin{equation} A_M^{(k+1)} = \{1\} \cup \{n\in\mathbb N:\ n\equiv r\pmod M\ \text{for some }2\le r\le M-1\} \cup \{Mt+1:\ t\in B^{(k)}\}. \tag{9.5} \end{equation} Consequently, \begin{equation} d(A_M^{(k+1)})=\frac{M-2}{M}+\frac1M\beta_k. \tag{9.6} \end{equation} In particular, \begin{equation} \inf_{\ell\ge 1} d(A_M^{(\ell)})\ge \frac{M-2}{M}. \tag{9.7} \end{equation}

Proof. The proof is identical to that of Lemma 9.1, except that one now keeps only the first $k$ embedded congruences. Numbers congruent to $2,3,\dots,M-1\pmod M$ survive automatically, numbers divisible by $M$ fail the first stage, and for numbers of the form $Mt+1$ with $t\ge 1$, avoiding the first $k$ embedded congruences is equivalent to requiring \[ t\not\equiv b_j\pmod{m_j}\qquad (1\le j\le k), \] i.e. $t\in B^{(k)}$. This proves (9.5).

The singleton $\{1\}$ has density $0$. The protected residue classes $2,\dots,M-1\pmod M$ contribute density $(M-2)/M$, and the affine copy of $B^{(k)}$ inside the class $1\pmod M$ contributes density $(1/M)\beta_k$. Hence (9.6). Since $\beta_k\ge 0$, inequality (9.7) follows. $\square$

9.3. Harmonic relation and transfer of logarithmic nonconvergence

Define upper and lower logarithmic densities by \[ \overline{\delta}_{\log}(E):= \limsup_{x\to\infty}\frac{1}{\log x}\sum_{\substack{n\le x\\ n\in E}}\frac1n, \qquad \underline{\delta}_{\log}(E):= \liminf_{x\to\infty}\frac{1}{\log x}\sum_{\substack{n\le x\\ n\in E}}\frac1n. \] For $E=B$ and $E=A_M$, write \[ S_B(y):=\sum_{\substack{t\le y\\ t\in B}}\frac1t, \qquad S_{A_M}(x):=\sum_{\substack{n\le x\\ n\in A_M}}\frac1n. \]

Proposition 9.3.

For the transformed instance (9.2)--(9.3), \begin{equation} S_{A_M}(x)=\frac{M-2}{M}\log x+\frac1M S_B(x/M)+O(1) \qquad (x\to\infty), \tag{9.8} \end{equation} where the implied constant may depend on $M$ and on the original instance, but not on $x$. Consequently, \begin{align} \underline{\delta}_{\log}(A_M) &= \frac{M-2}{M}+\frac1M\underline{\delta}_{\log}(B), \tag{9.9}\\ \overline{\delta}_{\log}(A_M) &= \frac{M-2}{M}+\frac1M\overline{\delta}_{\log}(B). \tag{9.10} \end{align} In particular, $A_M$ has a logarithmic density if and only if $B$ does.

Proof. From the exact decomposition (9.4), \begin{align} S_{A_M}(x) &= \sum_{r=2}^{M-1}\sum_{\substack{n\le x\\ n\equiv r\, (\mathrm{mod}\ M)}}\frac1n +\mathbf 1_{x\ge 1} +\sum_{\substack{1\le t\le (x-1)/M\\ t\in B}}\frac1{Mt+1}. \tag{9.11} \end{align} For each fixed residue class $r\pmod M$, periodic logarithmic asymptotics give \[ \sum_{\substack{n\le x\\ n\equiv r\, (\mathrm{mod}\ M)}}\frac1n = \frac1M\log x+O(1). \] Summing over $r=2,\dots,M-1$ yields \begin{equation} \sum_{r=2}^{M-1}\sum_{\substack{n\le x\\ n\equiv r\, (\mathrm{mod}\ M)}}\frac1n = \frac{M-2}{M}\log x+O(1). \tag{9.12} \end{equation}

For the embedded copy of $B$, write $y:=\lfloor (x-1)/M\rfloor$. Then \begin{align} \sum_{\substack{1\le t\le y\\ t\in B}}\frac1{Mt+1} &= \frac1M\sum_{\substack{1\le t\le y\\ t\in B}}\frac1t +\sum_{\substack{1\le t\le y\\ t\in B}}\left(\frac1{Mt+1}-\frac1{Mt}\right). \tag{9.13} \end{align} Now \[ \left|\frac1{Mt+1}-\frac1{Mt}\right|=\frac{1}{Mt(Mt+1)}\ll_M \frac1{t^2}, \] and $\sum_{t\ge 1}t^{-2}<\infty$, so the second sum in (9.13) is $O(1)$. Hence \begin{equation} \sum_{\substack{1\le t\le y\\ t\in B}}\frac1{Mt+1} = \frac1M S_B(y)+O(1) = \frac1M S_B(x/M)+O(1), \tag{9.14} \end{equation} since changing the cutoff from $y$ to $x/M$ affects at most one term.

Substituting (9.12) and (9.14) into (9.11) proves (9.8). Dividing by $\log x$ and using \[ \frac{\log(x/M)}{\log x}\to 1 \qquad (x\to\infty) \] immediately gives (9.9) and (9.10). The final claim follows because upper and lower logarithmic densities coincide for $A_M$ if and only if they coincide for $B$. $\square$

9.4. Equivalence of the full problem with high-density restrictions

Theorem 9.4.

Fix any real number $c<1$. The following are equivalent.

1. (i) Every instance of Erdős Problem 25 has a logarithmic density.
2. (ii) Every instance of Erdős Problem 25 satisfying \begin{equation} \inf_{k\ge 1} d(A^{(k)})\ge c \tag{9.15} \end{equation} has a logarithmic density.

Proof. The implication (i)$\Rightarrow$(ii) is immediate.

Assume (ii). Let $(m_j,b_j)$ be an arbitrary instance. If $m_1=1$, then the associated set $B$ is empty and there is nothing to prove. So assume $m_1\ge 2$. Choose an integer $M\ge 3$ so large that \begin{equation} \frac{M-2}{M}\ge c. \tag{9.16} \end{equation} Construct the transformed instance $(n_i,a_i)$ by (9.2)--(9.3). By Lemma 9.2, \[ \inf_{\ell\ge 1} d(A_M^{(\ell)})\ge \frac{M-2}{M}\ge c. \] Hence, by assumption (ii), the transformed set $A_M$ has a logarithmic density. Proposition 9.3 then implies that the original set $B$ also has a logarithmic density. Since the original instance was arbitrary, (i) follows. $\square$

Corollary 9.5.

For every $\varepsilon>0$, the following are equivalent.

1. (a) There exists a counterexample to Erdős Problem 25.
2. (b) There exists a counterexample to Erdős Problem 25 for which \begin{equation} \inf_{k\ge 1} d(A^{(k)})\ge 1-\varepsilon. \tag{9.17} \end{equation}

More quantitatively, if the original instance has upper and lower logarithmic densities differing by \begin{equation} \overline{\delta}_{\log}(B)-\underline{\delta}_{\log}(B)=\Delta>0, \tag{9.18} \end{equation} then the transformed instance with parameter $M$ satisfies \begin{equation} \overline{\delta}_{\log}(A_M)-\underline{\delta}_{\log}(A_M)=\frac{\Delta}{M}, \tag{9.19} \end{equation} while still obeying \begin{equation} \inf_{k\ge 1} d(A_M^{(k)})\ge \frac{M-2}{M}. \tag{9.20} \end{equation}

Proof. This is immediate from Proposition 9.3 and Lemma 9.2 after choosing $M\ge 2/\varepsilon$. $\square$

Interpretation.

Round 8 proved that the branch $\delta_k\to 0$ is always benign. Theorem 9.4 shows that the complementary branch is not merely "what remains": it is actually equivalent to the full problem. In fact, one may restrict attention to instances whose finite-stage densities stay above any fixed level $c<1$, or even above $1-\varepsilon$ for arbitrarily small $\varepsilon>0$.

So any proof strategy that hopes to settle the full problem by using only eventual smallness of the stage densities cannot succeed on its own.

Adversarial Verification

We attempt to break the new argument.

• Strict increase of the transformed moduli. The construction uses $n_1=M$ and $n_{j+1}=Mm_j$. This is strictly increasing because $m_1\ge 2$ (the case $m_1=1$ is trivial) and $(m_j)$ is strictly increasing.
• Thresholds map exactly. For numbers of the form $Mt+1$, the transformed condition \[ Mt+1<Mm_j \] is equivalent to $t<m_j$ because $t,m_j\in\mathbb Z$. Also \[ Mt+1\equiv Mb_j+1\pmod{Mm_j} \iff t\equiv b_j\pmod{m_j}. \] Thus the embedded copy really is exact, not approximate.
• Protected residue classes are truly protected. Every later forbidden residue is congruent to $1\pmod M$, so residue classes $2,\dots,M-1\pmod M$ can never be touched after the first step.
• The singleton $\{1\}$ causes no issue. It appears in both the actual set and the finite-stage sets, but contributes zero natural density and only $O(1)$ harmonic weight.
• Transfer of nonconvergence. Proposition 9.3 gives an additive affine relation between the upper/lower logarithmic densities, so nonexistence of the limit transfers exactly.
• Compatibility with Round 8. There is no contradiction. Round 8 correctly solved the branch $\delta_k\to 0$. The new theorem shows only that this branch is not a genuine reduction of the full problem, because arbitrary instances can be padded into high-density ones.
• Sharpness of the obstruction. The restriction $c<1$ is essential: no nontrivial instance can satisfy $d(A^{(k)})=1$ for every $k$, since the first stage already removes a residue class.

Final

UNRESOLVED (BUT STRICTLY ADVANCED).

The main new advance is Theorem 9.4:

For every fixed $c<1$, Erdős Problem 25 is equivalent to its restriction to instances satisfying \[ \inf_{k\ge 1} d(A^{(k)})\ge c. \]

Equivalently, any counterexample can be padded into one whose finite-stage densities stay above $1-\varepsilon$ for any prescribed $\varepsilon>0$.

This is a genuine obstruction result. It shows that the Round 8 reduction to the branch $\delta_k\to 0$ --- while correct --- is not a meaningful simplification of the full problem. The unresolved core is therefore not merely the positive-density branch; rather, the arbitrarily high-density branch already contains the full difficulty of the original question.

Completion Estimate

COMPLETION: 95%.

References

• T. F. Bloom, Erdős Problem #25, official problem page, https://www.erdosproblems.com/25, accessed 2026-03-08.
• No external mathematical references were used in the proofs of this round.