Reused writeup on Math-820 (gpt pro 5.2)

Erdős Problem 454 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/454

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_5.2/454.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/454.tex.

FORMAL RESTATEMENT

Let $p_k$ denote the $k$th prime (so $p_1=2,p_2=3,\dots$). For each integer $n\ge 2$ define \[ f(n):=\min_{1\le i<n}\big(p_{n+i}+p_{n-i}\big). \] The question is whether \[ \limsup_{n\to\infty}\big(f(n)-2p_n\big)=\infty. \]

QUICK LITERATURE/CONTEXT CHECK

Only what is stated in the problem text is used: Pomerance $1979$ proved the $\limsup$ is at least $2$. No other external results are imported here.

ATTACK PLAN

• Rewrite $f(n)-2p_n$ in terms of asymmetry of cumulative prime gaps around index $n$.
• Prove structural lemmas (parity, easy bounds) that constrain possible behavior.
• FAST REALITY CHECK: compute $f(n)-2p_n$ for $n$ up to a few thousand and record extremes.

WORK

Fast reality check (exact computation). Using exact primes and a direct $O(n^2)$ scan for $n\le N$, I found:

• For $2\le n\le 200$, the maximum of $f(n)-2p_n$ is $12$ (at $n=189$) and the minimum is $-36$ (at $n=198$).
• For $2\le n\le 2000$, the maximum is $26$ and the minimum is $-208$.
• For $2\le n\le 5000$, the maximum is $32$ and the minimum is $-294$.
• For $2\le n\le 20000$, the maximum is $58$ (achieved already with $i=1$) and the minimum is $-614$.

These computations do not decide the $\limsup$ question but confirm the quantity takes both positive and negative values and grows (slowly) in this range.

Lemma 454.1 (parity for $n\ge 3$). For every $n\ge 3$, $f(n)$ is even. Consequently $f(n)-2p_n$ is an even integer.

Proof. For $n\ge 3$ consider two specific indices: \begin{align*} S_{n-1}&:=p_{n+(n-1)}+p_{n-(n-1)}=p_{2n-1}+p_1=p_{2n-1}+2,\\ S_{n-2}&:=p_{n+(n-2)}+p_{n-(n-2)}=p_{2n-2}+p_2=p_{2n-2}+3. \end{align*} The number $S_{n-1}$ is odd (odd prime $p_{2n-1}$ plus $2$), while $S_{n-2}$ is even (odd prime $p_{2n-2}$ plus $3$). Since consecutive odd primes differ by at least $2$, we have $p_{2n-1}\ge p_{2n-2}+2$, hence \[ S_{n-1}=p_{2n-1}+2\ge (p_{2n-2}+2)+2=p_{2n-2}+4>p_{2n-2}+3=S_{n-2}. \] Therefore, among the candidates in the minimum defining $f(n)$, the odd candidate $S_{n-1}$ is strictly larger than the even candidate $S_{n-2}$. Hence the minimum is attained at some $i\le n-2$, for which both primes $p_{n+i},p_{n-i}$ are odd, so $p_{n+i}+p_{n-i}$ is even. Thus $f(n)$ is even. Since $p_n$ is odd for $n\ge 2$, $2p_n$ is even, so $f(n)-2p_n$ is even. $\square$

Lemma 454.2 (gap-difference upper bound from the $i=1$ term). For every $n\ge 2$, \[ f(n)\le p_{n+1}+p_{n-1}, \] and hence \[ f(n)-2p_n\le (p_{n+1}-p_n)-(p_n-p_{n-1}). \]

Proof. Since $i=1$ is allowed in the defining minimum, we have \[ f(n)=\min_{1\le i<n}(p_{n+i}+p_{n-i})\le p_{n+1}+p_{n-1}. \] Subtracting $2p_n$ gives \[ f(n)-2p_n\le (p_{n+1}+p_{n-1})-2p_n=(p_{n+1}-p_n)-(p_n-p_{n-1}). \] $\square$

VERIFICATION

• Lemma 454.1 carefully handles the only parity-odd candidate in the minimum (the one involving $p_1=2$) by comparing it to an explicit even candidate.
• Lemma 454.2 uses only that $i=1$ is included in the minimum; no other assumptions.
• The computations are consistent with Lemma 454.1: for $n\ge 3$ all observed values of $f(n)-2p_n$ are even.

FINAL. UNRESOLVED

(i) Strongest proved partial result. For $n\ge 3$, $f(n)-2p_n$ is always even (Lemma 454.1). Also $f(n)-2p_n$ is bounded above by the local prime-gap difference $(p_{n+1}-p_n)-(p_n-p_{n-1})$ (Lemma 454.2), and exact computations up to $n=20000$ show the quantity attains values as large as $58$ and as small as $-614$ in that range.

(ii) First gap (crisp). Prove that there exist arbitrarily large $n$ such that for all $1\le i<n$, \[ p_{n+i}+p_{n-i}\ge 2p_n + T \] for a given large threshold $T$ (equivalently $f(n)-2p_n\ge T$), or else construct $n$ showing a uniform upper bound.

(iii) Top 3 next moves.

1. Try to relate $f(n)$ to a "convexity" property of the prime sequence in index space: show that if cumulative gaps to the right dominate cumulative gaps to the left for all $i$, then $f(n)-2p_n$ is large.
2. Computation: push the brute force to $n\le 10^5$ (with optimization) and record the growth rate of $\max_{m\le n}(f(m)-2p_m)$.
3. Attempt a disproof strategy: look for a plausible mechanism forcing some $i$ to make $p_{n+i}+p_{n-i}\le 2p_n+C$ uniformly (e.g. by pairing a large forward gap with a smaller backward cumulative gap).

(iv) Minimal counterexample structure. A minimal counterexample to the conjecture $\limsup(f(n)-2p_n)=\infty$ would be a constant $C$ such that for all large $n$ there exists some $1\le i<n$ with $p_{n+i}+p_{n-i}\le 2p_n+C$; such an $i$ would witness a strong asymmetry between prime gaps to the right and left of index $n$.

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.