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gpt pro 5.4 attempt on Math-784

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Contents
  1. Erdős Problem 371
  2. Problem statement.
  3. Tools in this round.
  4. 1) ROUND-2 OBJECTIVE
  5. 2) Round-1 FOUNDATION USED
  6. 3) NEW INSIGHT / TOOL (ROUND-2)
  7. 4) ATTACK PLAN (ROUND-2)
  8. 5) WORK (ROUND-2)
  9. Step 1: a fully proved new theorem -- both sides have positive lower density.
  10. Step 2: logarithmic density is exactly $1/2$.
  11. Step 3: at almost all scales, the ordinary density is also $1/2$.
  12. Step 4: the best currently published explicit lower bound.
  13. 6) ADVERSARIAL VERIFICATION
  14. 7) FINAL
  15. 8) COMPLETION ESTIMATE (MANDATORY)
  16. 9) REFERENCES
  17. 10) OUTPUT FORMAT

Erdős Problem 371

Problem statement.
Let $P^{+}(n)$ denote the largest prime factor of $n$. Show that the set of $n$ with $P^{+}(n)<P^{+}(n+1)$ has density $1/2$.
Tools in this round.
Web browsing: YES. Computation: YES.

1) ROUND-2 OBJECTIVE

I am not pursuing path (A), because the full natural-density statement is still open. The Round-2 goal is therefore strict advancement: replace Round 1's mere infinitude results by bona fide density theorems, and isolate the exact remaining obstruction between what is known and the desired limit $1/2$.

2) Round-1 FOUNDATION USED

I rely on two Round-1 facts.
  • For every $n\geq 1$, one has $P^{+}(n)\neq P^{+}(n+1)$, since consecutive integers are coprime.
  • Round 1 gave infinite families in both directions: $P^{+}(2^{m})<P^{+}(2^{m}+1)$ and $P^{+}(2^{m}-1)>P^{+}(2^{m})$.
These facts will now be strengthened to genuine density statements.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredients are three published theorems.
  • Ter"av"ainen's window theorem: for any exponents $0<a<b<1$ and $0<c<d<1$, the set of $n$ for which $P^{+}(n)$ and $P^{+}(n+1)$ lie simultaneously in the windows $[n^{a},n^{b}]$ and $[n^{c},n^{d}]$ has positive asymptotic lower density.
  • Ter"av"ainen's logarithmic-density theorem: the set $\{n:P^{+}(n)<P^{+}(n+1)\}$ has logarithmic density $1/2$.
  • Tao--Ter"av"ainen's almost-all-scales theorem, together with the explicit lower-density bound $0.2017$ of L"u--Wang.
The first item lets me prove positive lower density for both inequalities directly, which is already a substantial advance over Round 1.

4) ATTACK PLAN (ROUND-2)

The exact gap left after Round 1 was the absence of any mean-cancellation theorem for the sign sequence \[ f(n)=\begin{cases} +1,& P^{+}(n)<P^{+}(n+1),\\ -1,& P^{+}(n)>P^{+}(n+1). \end{cases} \] To prove density $1/2$ one would need \[ \sum_{n\leq X} f(n)=o(X). \] I do not have that. Instead I will do the next best rigorous thing.
  1. Use Ter"av"ainen's window theorem to obtain positive lower density for both sides.
  2. Record the stronger known theorems: logarithmic density $1/2$, density $1/2$ at almost all scales, and explicit lower density at least $0.2017$.
  3. Explain precisely why these do not yet imply the full natural density statement.

5) WORK (ROUND-2)

Let \[ S=\{n\geq 1: P^{+}(n)<P^{+}(n+1)\}. \] Write \[ \underline d(A)=\liminf_{X\to\infty} \frac{1}{X}\#(A\cap [1,X]) \] for asymptotic lower density.
Step 1: a fully proved new theorem -- both sides have positive lower density.
Ter"av"ainen proved the following.

If $a,b,c,d\in (0,1)$ with $a<b$ and $c<d$, then the set \[ E_{a,b;c,d}=\{n\geq 1: n^{a}\leq P^{+}(n)\leq n^{b},\; n^{c}\leq P^{+}(n+1)\leq n^{d}\} \] has positive asymptotic lower density.

Choose, for example, \[ a=\frac{1}{10},\qquad b=\frac{1}{5},\qquad c=\frac{4}{5},\qquad d=\frac{9}{10}. \] Then $a<b$ and $c<d$, so $E_{a,b;c,d}$ has positive lower density. For every $n\in E_{a,b;c,d}$ we have \[ P^{+}(n)\leq n^{1/5}<n^{4/5}\leq P^{+}(n+1), \] whence $n\in S$. Therefore \[ \underline d(S)>0. \]

Now reverse the windows. Let \[ F=E_{4/5,9/10;1/10,1/5}. \] Again Ter"av"ainen's theorem gives $\underline d(F)>0$, and for every $n\in F$ one has \[ P^{+}(n)\geq n^{4/5}>n^{1/5}\geq P^{+}(n+1), \] so $n\notin S$. Hence \[ \underline d(\mathbf N\setminus S)>0. \] This strictly strengthens the Round-1 infinitude result.

Step 2: logarithmic density is exactly $1/2$.
Ter"av"ainen also proved that the logarithmic density \[ \delta_{\log}(A)=\lim_{X\to\infty}\frac{1}{\log X}\sum_{\substack{n\leq X\\ n\in A}}\frac{1}{n} \] exists for $A=S$ and satisfies \[ \delta_{\log}(S)=\frac{1}{2}. \] So the conjectured symmetry is already known on logarithmic average.
Step 3: at almost all scales, the ordinary density is also $1/2$.
Tao and Ter"av"ainen proved that there exists an exceptional set $\mathcal X_{0}$ of logarithmic density $0$ such that \[ \lim_{\substack{X\to\infty\\ X\notin \mathcal X_{0}}} \frac{1}{X}\#\{n\leq X: n\in S\}=\frac{1}{2}. \] Thus the natural-density statement is true for all scales outside a logarithmically negligible exceptional set.
Step 4: the best currently published explicit lower bound.
L"u and Wang proved \[ \#\{n<x:P^{+}(n)<P^{+}(n+1)\}>(0.2017-o(1))x, \] and the same lower bound for the complementary inequality. Therefore \[ \underline d(S)\geq 0.2017, \qquad \underline d(\mathbf N\setminus S)\geq 0.2017. \] Since $S$ and its complement partition the positive integers by the Round-1 no-ties observation, one also gets the corresponding upper bounds \[ \overline d(S)\leq 0.7983, \qquad \overline d(\mathbf N\setminus S)\leq 0.7983. \]

6) ADVERSARIAL VERIFICATION

The points most likely to conceal a mistake are the following.
  • Window inclusion. In Step 1 the inequalities are strict for $n>1$ because $1/5<4/5$. So every $n$ in the first window set really belongs to $S$, and every $n$ in the reversed window set really belongs to the complement.
  • Density notions. Positive lower density, logarithmic density, and "density at almost all scales" are three different notions. None of Steps 1--4 silently identifies them.
  • Why don't Steps 2 and 3 already prove the conjecture? Because a set of scales of logarithmic density $0$ can still contain subsequences along which the ordinary Cesàro averages oscillate. So the Tao--Ter"av"ainen theorem is stronger than anything in Round 1, but weaker than the existence of an ordinary limit along all $X$.
  • Compatibility with Round 1. The new results strengthen, rather than contradict, Round 1. Infinitude follows immediately from positive lower density.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED)

Round 2 upgrades the Round-1 output substantially:

  • both inequalities now have rigorously proved positive lower density;
  • the set $S$ is known to have logarithmic density $1/2$;
  • the ordinary density equals $1/2$ at almost all scales;
  • the best explicit published lower bound is $0.2017$ for both $S$ and its complement.
The exact missing step is now sharply isolated: one must upgrade these partial density results to an ordinary asymptotic density along all scales.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 70%

9) REFERENCES

  • J. Ter"av"ainen, On binary correlations of multiplicative functions, Forum Math. Sigma 6 $2018$, e10.
  • T. Tao and J. Ter"av"ainen, The structure of correlations of multiplicative functions at almost all scales, with applications to the Chowla and Elliott conjectures, Algebra & Number Theory 13 $2019$, no. 9.
  • X. L"u and Z. Wang, On the largest prime factors of consecutive integers, Monatsh. Math. 206 $2025$, 403--418.
  • T. F. Bloom, Erdős Problem #371, accessed 2026-03-07.

10) OUTPUT FORMAT

Erdős problem number: $x=371$. This section is written as a drop-in LaTeX fragment and is included in the combined file 369_371_373_round2_solutions.tex.