gpt pro 5.4 attempt on Math-722

Problem 233 --- Round 2

1) ROUND-2 OBJECTIVE

Path (C): obstruction/correction. The conjectured bound \[ \sum_{n\le N} d_n^2 \ll N(\log N)^2 \] remains open. Round 1 had only the lower bound and a trivial upper bound through the maximal gap. In this round I do two strictly new things:

• I prove the exact equivalence between the $N$-formulation and the standard $x$-formulation \[ \sum_{p_n\le x} d_n^2 \ll x\log x; \]
• I import the current unconditional mean-square theorem of Stadlmann and translate it carefully into an explicit bound in the original $N$-variable.

2) Round-1 FOUNDATION USED

I use the Round 1 notation $d_n=p_{n+1}-p_n$ and the Round 1 lower bound \[ \sum_{n\le N} d_n^2\ge \frac{(p_{N+1}-2)^2}{N}\gg N(\log N)^2, \] which I do not reprove here.

3) NEW INSIGHT / TOOL (ROUND-2)

The new ingredients are:

• the exact identity \[ M(x):=\sum_{p_n\le x} d_n^2 = \sum_{n\le \pi(x)} d_n^2; \]
• Stadlmann's theorem that \[ \frac{1}{\pi(x)}\sum_{p_n\le x} d_n^2 = O_{\varepsilon}(x^{23/100+\varepsilon}) \qquad (\varepsilon>0), \] which implies \[ M(x)\ll_{\varepsilon} \frac{x^{123/100+\varepsilon}}{\log x}. \]

4) ATTACK PLAN (ROUND-2)

The exact Round 1 gap was the absence of any nontrivial upper bound in the original variable $N$. So I first prove that the conjecture is equivalent to the $x$-formulation. Then I plug the current best unconditional $x$-bound into this equivalence to obtain a genuine unconditional theorem about \[ S(N):=\sum_{n\le N} d_n^2. \]

5) WORK (ROUND-2)

Define \[ M(x):=\sum_{p_n\le x} d_n^2. \] If $\pi(x)=N$, then the condition $p_n\le x$ is equivalent to $n\le N$, so \[ M(x)=\sum_{n\le N} d_n^2. \] In particular, \[ M(p_N)=\sum_{n\le N} d_n^2 = S(N). \] This makes the two formulations equivalent.

Equivalence lemma. The following are equivalent:

• (i) $S(N)\ll N(\log N)^2$;
• (ii) $M(x)\ll x\log x$.

Proof. Assume (i). Let $N=\pi(x)$. Then $M(x)=S(N)\ll N(\log N)^2$. By the prime number theorem, \[ N\asymp \frac{x}{\log x} \qquad\mbox{and}\qquad \log N\asymp \log x, \] so $M(x)\ll x\log x$. Thus (ii) follows.

Conversely, assume (ii). Then \[ S(N)=M(p_N)\ll p_N\log p_N. \] Again by the prime number theorem, $p_N\asymp N\log N$, hence \[ S(N)\ll N(\log N)^2. \] So (i) follows. This proves the equivalence.

Now apply Stadlmann's theorem: \[ \frac{M(x)}{\pi(x)}=O_{\varepsilon}(x^{23/100+\varepsilon}). \] Multiplying by $\pi(x)$ and using $\pi(x)\ll x/\log x$, we get \[ M(x)\ll_{\varepsilon} \frac{x^{123/100+\varepsilon}}{\log x}. \] Set $x=p_N$. Then \[ S(N)=M(p_N)\ll_{\varepsilon} \frac{p_N^{123/100+\varepsilon}}{\log p_N}. \] Using $p_N\ll N\log N$ and $\log p_N\asymp \log N$, we conclude that \[ S(N)\ll_{\varepsilon} N^{123/100+\varepsilon}(\log N)^{23/100+\varepsilon}. \] Therefore Round 2 proves the following corrected theorem.

Corrected theorem proved in Round 2. For every $\varepsilon>0$, \[ \sum_{n\le N} d_n^2 \ll_{\varepsilon} N^{123/100+\varepsilon}(\log N)^{23/100+\varepsilon}. \] Combining this with the Round 1 lower bound gives the rigorous sandwich \[ N(\log N)^2 \ll \sum_{n\le N} d_n^2 \ll_{\varepsilon} N^{123/100+\varepsilon}(\log N)^{23/100+\varepsilon}. \]

6) ADVERSARIAL VERIFICATION

• Any indexing error in $M(p_N)$? No. Since $p_n\le p_N$ iff $n\le N$, we have exactly \[ M(p_N)=\sum_{n\le N} d_n^2. \]
• Did the argument use only valid asymptotic input? Yes: only the prime number theorem bounds $\pi(x)\asymp x/\log x$ and $p_N\asymp N\log N$.
• Could the Stadlmann theorem have used a different normalization? The cited source states explicitly that the average considered is \[ \frac{1}{\pi(x)}\sum_{p_n\le x}(p_{n+1}-p_n)^2, \] which is exactly the normalization used here.
• Does this settle the conjecture? No. The exponent of $N$ is still $1.23+\varepsilon$ rather than $1$, so the main barrier remains substantial.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED). The conjectured bound $\sum_{n\le N} d_n^2\ll N(\log N)^2$ is still open, but Round 2 replaces the Round 1 trivial upper bound by the current best unconditional type of estimate in the original variable: \[ \sum_{n\le N} d_n^2 \ll_{\varepsilon} N^{123/100+\varepsilon}(\log N)^{23/100+\varepsilon}. \] It also proves the exact equivalence between the $N$-formulation and the standard $x$-formulation.

8) COMPLETION ESTIMATE

COMPLETION: 35%

9) REFERENCES

• J. Stadlmann, On the mean square gap between primes, arXiv:2212.10867.
• J. Stadlmann, Topics in sieve theory, D.Phil. thesis, University of Oxford, 2024.