Erdős Problem #233

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.

Erdős Problem #233: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

For $d_n=p_{n+1}-p_n$, prove $\sum_{n\le N}d_n^2\ll N(\log N)^2$.

All supplied versions considered: 233.tex. The source changes between prime-index and prime-size formulations and imports a mean-square upper bound. I isolate precisely the tail information needed by the conjecture; no improvement in the arithmetic exponent is claimed.

2) WORK

For $N\ge3$ put $u_n=d_n/\log N$ and $A_j=\#\{n\le N:u_n>2^j\}$ for $j\ge0$. Pointwise geometric summation gives \[ \sum_{j\ge0}4^j\mathbf1_{u>2^j}\le\tfrac43u^2, \qquad u^2\le1+4\sum_{j\ge0}4^j\mathbf1_{u>2^j}\quad(u\ge0). \] For the second inequality, if $u>1$ let $j$ be the last threshold strictly below $u$; then $u\le2^{j+1}$, including powers of two by taking the preceding index. Hence \[ \sum_{n\le N}d_n^2\ll N(\log N)^2 \quad\Longleftrightarrow\quad \sum_{j\ge0}4^j A_j\ll N. \] This is an exact dyadic-tail reformulation up to absolute constants.

For example, the uniform estimates \[ A_j\le {CN\over4^j(1+j)^{1+\epsilon}}\quad(j\ge0) \] would suffice. The superficially similar bound $A_j\le CN/4^j$ does not suffice through this argument, since each occupied scale contributes order $N$. Thus even the correct quadratic tail exponent must be accompanied by summability across scales. This locates a separate issue from merely bounding the largest prime gap.

3) ADVERSARIAL VERIFICATION

The sums over $j$ are finite at every fixed $N$, and strict threshold conventions handle equality correctly. The sufficient tail bound is not proved for prime gaps. A nonsummable majorant is not a counterexample to the conjecture.

4) FINAL

UNRESOLVED. The required arithmetic control of the aggregate tail mass remains absent. The dyadic identity and sufficient condition are verified reductions, not a solution.

5) SOURCES CHECKED

Read 233.tex; checked Stadlmann, https://arxiv.org/abs/2212.10867, on 2026-09-23. Its abstract confirms the $0.23+\epsilon$ mean-square exponent; no assertion that this is the latest best bound is needed here.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 8%

Problem #233 (second moment of prime gaps)

1. Formal restatement

Let $p_n$ be the $n$th prime and $d_n:=p_{n+1}-p_n$. The conjectured bound is \[ \sum_{n\le N} d_n^2 \ll N(\log N)^2\qquad (N\to\infty), \] where $\log$ denotes the natural logarithm and the implied constant is absolute.

Equivalently, writing $x\approx p_{N+1}$ (so $\pi(x)\approx N$), this asks for a bound of order $\sum_{p_n\le x}(p_{n+1}-p_n)^2 \ll x\log x$ up to log factors, consistent with Cramér-type heuristics.

2. Quick literature/context check (browsing available)

The Erdős Problems page (OPEN) records:(https://www.erdosproblems.com/233)

• Cramér $1936$ proved $\sum_{n\le N} d_n^2 = O(N(\log N)^4)$ assuming the Riemann Hypothesis (RH).
• Selberg $1943$, still assuming RH, improved this slightly to $\sum_{n\le N} \frac{d_n^2}{n}\ll (\log N)^4$.
• The prime number theorem gives the lower bound $\sum_{n\le N} d_n^2\gg N(\log N)^2$.

For unconditional progress toward bounding the mean square gap, Stadlmann $2022$ proves that the average squared gap among primes $\le x$ is $O(x^{0.23+\varepsilon})$, i.e. \[ \frac{1}{\pi(x)}\sum_{p_n\le x}(p_{n+1}-p_n)^2 = O(x^{0.23+\varepsilon}). \] (J. Stadlmann, On the mean square gap between primes, arXiv:2212.10867.) This implies the (much weaker than conjectured) bound $\sum_{p_n\le x}(p_{n+1}-p_n)^2 = O(x^{1.23+\varepsilon})$.

A convenient summary of the historical exponents (Heath-Brown, Peck, Maynard, Stadlmann) for bounds of the form $\sum_{p_n\le x}(p_{n+1}-p_n)^2\ll x^{\theta+\varepsilon}$ appears in the paper "The sequence of prime gaps is graphic".(https://arxiv.org/pdf/2205.00580 (see Remark after Theorem 2.4).)

3. Attack plan

A natural approach to the conjectured $N(\log N)^2$ upper bound would be to combine:

1. strong bounds on primes in short intervals on average (second moment of $\pi(x+h)-\pi(x)$ for $h$ in ranges comparable to typical gaps);
2. a way to convert those average short-interval estimates into a second-moment estimate for consecutive gaps;
3. ideally a "near-Poisson" heuristic for prime counts in disjoint intervals, or deep zero-density estimates / pair-correlation input.

Conditionally, RH gives $\sum d_n^2\ll N(\log N)^4$ (Cramér), so the main quantitative barrier is saving two logarithms.

4. Work

4.1. Fast reality check (small $N$)

For the first few gaps: \[(d_1,d_2,d_3,d_4,d_5,\dots)=(1,2,2,4,2,\dots),\] so $\sum_{n\le 5} d_n^2 = 1^2+2^2+2^2+4^2+2^2=29$. The asymptotic comparison with $N(\log N)^2$ is not meaningful for such tiny $N$ (since $\log 1=0$), but it confirms the definitions.

4.2. A rigorous lower bound via Cauchy--Schwarz (assuming PNT-scale growth)

4.3. A trivial upper bound (too weak)

Let $D_N:=\max_{1\le n\le N} d_n$. Then \[ \sum_{n\le N} d_n^2 \le D_N\sum_{n\le N} d_n = D_N\,(p_{N+1}-2). \] Without strong control of $D_N$ this is far from the conjectured $N(\log N)^2$ scale.

4.4. Numerical experiment (finite data)

Computing the first $N$ gaps and the sum $S(N):=\sum_{n\le N} d_n^2$ for $N\le 10^6$ gives: This supports the heuristic that $S(N)$ is of order $N(\log N)^2$ with a moderate constant, but of course does not prove any asymptotic bound.

5. Verification

• The Cauchy--Schwarz lower bound is exact and uses only telescoping of gaps.
• The numerical ratios are stable in the tested range, consistent with the conjectured order.

6. Final

7. Completion estimate

5%. I can supply the standard lower bound and numerical evidence, and can quote known conditional/unconditional partial results, but the conjectured upper bound seems to require significantly deeper analytic number theory.