gpt pro 5.4 attempt on Math-791

Erdős Problem #385

1) ROUND-2 OBJECTIVE

I pursue (C) obstruction/partial reduction. Round-1 already localized every possible bad value of $n$ to the even prime-shift case $n=p+1$, so the most promising next step is not to re-run the original argument but to extract sharp consequences from the Round-1 bounds. The goal in this round is to prove two genuinely stronger statements: first, that the set of bad integers has density $0$; second, that the trivial upper bound $F(n)-n\leq \sqrt n$ is attained on the correct scale infinitely often, in the precise sense that \[ \limsup_{n\to\infty}\frac{F(n)-n}{\sqrt n}=1. \]

2) Round-1 FOUNDATION USED

I use the following Round-1 results.

1. Round-1 Lemma 1: if $n\geq 5$ is odd, then $F(n)\geq n+1$.
2. Round-1 Lemma 2: if $n\geq 6$ is even, then $F(n)\geq n$.
3. Round-1 Lemma 3: if $n\geq 6$ is even and $n-1$ is composite, then $F(n)>n$.
4. Round-1 Corollary 4: if $F(n)\leq n$, then $n$ is even, $n-1$ is prime, and in fact $F(n)=n$.
5. Round-1 Lemma 5: for $n=q^2+1$ with $q$ prime, one has $F(n)-n\geq q-1$.
6. Round-1 Lemma 6: for all $n\geq 5$, one has $F(n)\leq n+\sqrt n$.

3) NEW INSIGHT / TOOL (ROUND-2)

The new idea is to combine the Round-1 square-sequence lower bound with the Round-1 trivial upper bound at the level of the normalized ratio $(F(n)-n)/\sqrt n$. This turns the earlier weak statement \[ \limsup_{n\to\infty}(F(n)-n)=\infty \] into the sharp asymptotic statement \[ \limsup_{n\to\infty}\frac{F(n)-n}{\sqrt n}=1. \] A second new observation is that Round-1 Corollary 4 immediately forces the bad set into a subset of the shifted primes $\{p+1\}$, giving an unconditional zero-density theorem.

4) ATTACK PLAN (ROUND-2)

The exact gaps left after Round-1 were:

1. The main questions $F(n)>n$ eventually and $F(n)-n\to\infty$ were still open.
2. The earlier lower bound only showed that $F(n)-n$ is unbounded on a subsequence, without identifying the true maximal order.
3. The earlier reduction to the case $n-1$ prime had not yet been converted into a density statement.

The plan is therefore:

1. Use Round-1 Lemmas 5 and 6 together to pin down the exact value of the limsup ratio.
2. Use Round-1 Corollary 4 together with the prime number theorem to show that bad integers have density $0$.

This does not solve the original problem, but it does close two concrete gaps left by Round-1.

5) WORK (ROUND-2)

Theorem 385.1. One has \[ \limsup_{n\to\infty}\frac{F(n)-n}{\sqrt n}=1. \]

Proof. Round-1 Lemma 6 gives the uniform upper bound \[ \frac{F(n)-n}{\sqrt n}\leq 1 \qquad (n\geq 5), \] so \[ \limsup_{n\to\infty}\frac{F(n)-n}{\sqrt n}\leq 1. \] On the other hand, if $q$ is prime and $n=q^2+1$, then Round-1 Lemma 5 gives \[ F(n)-n\geq q-1. \] Hence along this subsequence \[ \frac{F(n)-n}{\sqrt n} \geq \frac{q-1}{\sqrt{q^2+1}}. \] As $q\to\infty$, the right-hand side tends to $1$. Therefore \[ \limsup_{n\to\infty}\frac{F(n)-n}{\sqrt n}\geq 1. \] Combining the upper and lower bounds gives the claim. $\square$

Corollary 385.2. The bad set \[ \mathcal B:=\{n\geq 5: F(n)\leq n\} \] has natural density $0$. More precisely, \[ \#\{n\leq x: F(n)\leq n\}\leq \pi(x-1)+O(1)=o(x). \] In particular, almost all integers are good.

Proof. By Round-1 Corollary 4, every bad integer $n\geq 6$ is even and satisfies $n-1$ prime. Thus \[ \{n\leq x: F(n)\leq n\}\subseteq \{p+1\leq x: p\text{ prime}\}\cup\{5\}. \] Hence \[ \#\{n\leq x: F(n)\leq n\}\leq \pi(x-1)+1. \] Since $\pi(x)=o(x)$, the density is $0$. $\square$

Corollary 385.3. The trivial upper bound from Round-1 is asymptotically best possible on an infinite subsequence: for infinitely many $n$, \[ F(n)=n+(1-o(1))\sqrt n. \]

Proof. This is just a restatement of Theorem 385.1. $\square$

6) ADVERSARIAL VERIFICATION

The new work has three places where hidden issues could occur.

1. Could the subsequence $n=q^2+1$ fail to be admissible? No. For every prime $q$, the integer $m=q^2<n=q^2+1$ is composite, so Round-1 Lemma 5 applies exactly.
2. Could the upper bound be too weak to identify the limsup ratio? No. The bound $F(n)-n\leq \sqrt n$ is uniform in $n$, while the square subsequence gives ratios tending to $1$ from below, so the limsup is forced to be exactly $1$.
3. Could the density-zero claim miss some bad odd numbers? No. Round-1 Lemma 1 already ruled out all odd $n\geq 5$, and Round-1 Corollary 4 completely characterizes the only possible bad numbers.

No quantifier problem remains in the new claims.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

The strict advance beyond Round-1 is twofold:

1. a sharp normalized limsup theorem, \[ \limsup_{n\to\infty}\frac{F(n)-n}{\sqrt n}=1, \] which substantially strengthens the earlier statement that $F(n)-n$ is merely unbounded on a subsequence;
2. an unconditional density theorem: the bad set has density $0$, so almost all integers are good.

The original eventual-positivity question and the full limit $F(n)-n\to\infty$ remain open.

8) COMPLETION ESTIMATE

COMPLETION: 48%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #385, erdosproblems.com/385, accessed 2026-03-07.
2. T. Tao, Erdos problem #385, the parity problem, and Siegel zeroes, blog post, 2024.