Reused writeup on Math-5943 (gpt pro 5.2)

Erdős Problem 852 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/852

Saved collaborative-database status: open. Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/852.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/852.tex.

FORMAL RESTATEMENT

Let $p_1=2<p_2=3<p_3=5<\cdots$ be the increasing sequence of primes and define the prime gaps \[ d_n := p_{n+1}-p_n \qquad (n\ge 1). \] Ambiguity note. The problem text says: "$h(x)$ be maximal such that for some $n<x$ the numbers $d_n,\ldots,d_{n+h(x)-1}$ are all distinct." As written, the endpoint index $n+h(x)-1$ is not constrained to be $\le x$.

Minimal corrected convention used below. For an integer $x\ge 1$, define \[ h(x) := \max\left\{h\in\mathbb{N}: \exists n\in\{1,2,\dots,x-h+1\}\ \text{s.t.}\ d_n,d_{n+1},\dots,d_{n+h-1}\ \text{are pairwise distinct}\right\}. \] Equivalently, $h(x)$ is the maximum length of a contiguous block of pairwise distinct values inside the finite sequence $(d_1,\dots,d_x)$.

The task is to estimate $h(x)$ as $x\to\infty$, and in particular to decide whether: \[ (\exists c>0)(\forall x\ \text{large})\quad h(x)>(\log x)^c,\qquad\text{and whether }\quad h(x)=o(\log x). \]

QUICK LITERATURE/CONTEXT CHECK

I do not import external results beyond what is explicitly stated in the problem file. The statement notes that Brun's sieve implies $h(x)\to\infty$ as $x\to\infty$.

ATTACK PLAN

• Proof-track (lower bounds): Try to construct, via sieve ideas, intervals containing primes whose consecutive gaps realize many distinct even values.
• Proof-track (upper bounds): Relate "distinct gaps" to the size of the largest gap available in the block, or to the growth of $p_n$ via $\sum d_i = p_{n+h}-p_n$.
• Disproof-track: Search computationally for evidence that $h(x)$ grows too slowly/too fast compared with $(\log x)^c$ or $\log x$.

I can only execute the elementary upper-bound track and a reality-check computation here.

WORK

Lemma 852.1 (Even gaps past $p_2$). For every $n\ge 2$, $d_n$ is an even positive integer.

Proof. For $n\ge 2$, both primes $p_n$ and $p_{n+1}$ are odd (since $p_1=2$ is the only even prime). The difference of two odd integers is even, and $p_{n+1}>p_n$ so $d_n>0$. $\square$

Lemma 852.2 (Distinct even gaps force quadratic total length). Fix integers $n\ge 2$ and $h\ge 1$. If the $h$ consecutive gaps $d_n,d_{n+1},\dots,d_{n+h-1}$ are pairwise distinct, then \[ p_{n+h}-p_n = \sum_{i=0}^{h-1} d_{n+i} \ \ge\ h(h+1). \]

Proof. By Lemma 852.1, each $d_{n+i}$ is an even positive integer. Write $d_{n+i}=2e_i$ with $e_i\in\mathbb{N}$. Pairwise distinctness of the $d_{n+i}$ implies pairwise distinctness of the $e_i$.

Among $h$ distinct positive integers, the minimum possible sum is $1+2+\cdots+h = h(h+1)/2$ (achieved by the set $\{1,2,\dots,h\}$). Therefore \[ \sum_{i=0}^{h-1} e_i \ge \frac{h(h+1)}{2} \quad\Rightarrow\quad \sum_{i=0}^{h-1} d_{n+i} = 2\sum_{i=0}^{h-1} e_i \ge h(h+1). \] Finally, telescoping gives $\sum_{i=0}^{h-1} d_{n+i} = (p_{n+1}-p_n)+\cdots+(p_{n+h}-p_{n+h-1})=p_{n+h}-p_n$. $\square$

Lemma 852.3 (A basic max-gap upper bound). For every integer $x\ge 1$, \[ h(x) \le \max\{d_1,d_2,\dots,d_x\}. \]

Proof. Take any block $d_n,\dots,d_{n+h-1}$ of length $h$ consisting of pairwise distinct positive integers. Then the maximum of these $h$ integers is at least $h$ (because $\{1,2,\dots,h\}$ is the smallest possible set of $h$ distinct positives in the pointwise order). Therefore $h\le \max(d_n,\dots,d_{n+h-1})$. Maximizing over all such blocks inside $\{1,\dots,x\}$ yields $h(x)\le \max_{1\le i\le x} d_i$. $\square$

Proposition 852.4 (A $p_{x+1}$-based upper bound). For every integer $x\ge 1$, \[ h(x)\left(h(x)-1\right) \le p_{x+1}-3. \] In particular, $h(x) < \sqrt{p_{x+1}}+1$.

Proof. Let $h:=h(x)$ and choose $n\in\{1,\dots,x-h+1\}$ such that $d_n,\dots,d_{n+h-1}$ are pairwise distinct.

Case 1: $n\ge 2$. Then Lemma 852.2 applies to the length-$h$ block and gives $p_{n+h}-p_n=\sum_{i=0}^{h-1} d_{n+i}\ge h(h+1)\ge h(h-1)$. Since $n+h\le x+1$ we have $p_{n+h}\le p_{x+1}$, and also $p_n\ge p_2=3$, so $h(h-1)\le p_{n+h}-p_n \le p_{x+1}-3$.

Case 2: $n=1$. If $h=1$ then the claimed inequality $h(h-1)\le p_{x+1}-3$ is $0\le p_{x+1}-3$, which holds because $p_{x+1}\ge p_2=3$. If $h\ge 2$, then the sub-block $d_2,\dots,d_h$ has length $h-1$, consists of even gaps, and is still pairwise distinct. Applying Lemma 852.2 with start index $2$ and length $h-1$ gives $p_{h+1}-p_2=\sum_{i=2}^{h} d_i\ge (h-1)h$. Because $h\le x$ we have $p_{h+1}\le p_{x+1}$, hence $(h-1)h\le p_{x+1}-3$.

In both cases, $h(h-1)\le p_{x+1}-3$. Finally, $h(h-1)<p_{x+1}$ implies $h^2<p_{x+1}+h$, hence $h<\sqrt{p_{x+1}}+1$. $\square$

FAST REALITY CHECK (computed small cases). I computed $h(x)$ for various $x$ by generating the first $x+1$ primes and applying a standard sliding-window "longest subarray with distinct entries" scan to the gap sequence $(d_1,\dots,d_x)$.

Results:

First 20 gaps:
[1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2]

h(x) for x=1..20:
1:1, 2:2, 3:2, 4:2, 5:2, 6:2, 7:2, 8:2, 9:3, 10:3,
11:3, 12:3, 13:3, 14:3, 15:3, 16:3, 17:3, 18:3, 19:3, 20:3.

Selected larger x:
 h(10)=3
 h(100)=6
 h(1,000)=9
 h(10,000)=13
 h(100,000)=15
 h(200,000)=16
 h(1,000,000)=18

For x=1,000,000 one witness block of length 18 occurs at gap indices
849,875..849,892 with gaps
[4, 12, 32, 6, 16, 80, 24, 22, 72, 8, 28, 30, 84, 2, 46, 14, 10, 26]
(all distinct), spanning primes 13,010,143 to 13,010,659.

VERIFICATION

• Quantifiers/indices: Under the corrected convention, a "block inside the first $x$ gaps" means $n+h-1\le x$. All lemmas/proofs use only such blocks.
• Edge cases: $d_1=1$ is the only odd prime gap; Lemma 852.1 explicitly starts at $n\ge 2$. Lemma 852.2 also starts at $n\ge 2$.
• Lemma 852.2 minimal-sum step: The reduction $d_{n+i}=2e_i$ is valid because all these gaps are even; distinctness is preserved under division by $2$.
• Computation: The reported $h(x)$ values were obtained by an exact scan of the finite list of gaps (no randomness). The witness block for $x=10^6$ was explicitly extracted and checked to have 18 distinct entries.

FINAL

UNRESOLVED

1. (i) Strongest proved partial result. Lemma 852.3 gives $h(x)\le\max_{1\le n\le x} d_n$. Proposition 852.4 gives the unconditional bound $h(x)(h(x)-1)\le p_{x+1}-3$, hence $h(x)<\sqrt{p_{x+1}}+1$.
2. (ii) First gap (crisp). Prove a nontrivial lower bound of the form $h(x)>(\log x)^c$ for some absolute $c>0$ and all sufficiently large $x$ (or else construct infinitely many $x$ violating it).
3. (iii) Top 3 next moves.
   1. Prove (via sieve) existence of a block of consecutive primes $p_n,\dots,p_{n+m}$ with pairwise distinct gaps $d_n,\dots,d_{n+m-1}$ of length $m\gg (\log x)^c$ when $n\asymp x$.
   2. Strengthen the "$h\le\max d$" upper bound by proving that among $\asymp\log x$ consecutive gaps some value must repeat (e.g. by showing only $o(\log x)$ many distinct gap values can occur in such a window).
   3. Extend computations to much larger $x$ (e.g. $x\ge 10^8$) to test whether $h(x)$ grows like $\log\log x$, a small power of $\log x$, or something else.
4. (iv) Minimal counterexample structure. A counterexample to "$h(x)>(\log x)^c$ eventually" would be a sequence $x_j\to\infty$ such that every block of length $\lceil(\log x_j)^c\rceil$ inside $(d_1,\dots,d_{x_j})$ contains a repeated gap value.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.