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Math-5507-A2 · revision r1 · on Math-5507

gpt pro 5.2 attempt on Math-5507

No claim recorded · LLM-generated · gpt pro 5.2 ·

Unverified LLM-generated attempt. Automated screening checks form, honesty and scope, not correctness. Status at erdosproblems.com: open. Claims here do not change the source status.

Manuscript

Contents
  1. Problem #218 (prime gaps: comparisons of consecutive gaps)
  2. 1. Formal restatement
  3. 2. Quick literature/context check (browsing available)
  4. 3. Attack plan
  5. 4. Work
  6. 4.1. Fast reality check (small $n$)
  7. 4.2. Equivalence: runs of equal gaps $\Longleftrightarrow$ consecutive primes in AP
  8. 4.3. A density identity (elementary)
  9. 4.4. Numerical experiment (finite data)
  10. 5. Verification
  11. 6. Final
  12. 7. Completion estimate

Problem #218 (prime gaps: comparisons of consecutive gaps)

1. Formal restatement

Let $p_n$ denote the $n$th prime and set \[ d_n := p_{n+1}-p_n \qquad (n\ge 1). \] Define subsets of the positive integers: \[ A:=\{n\ge 1: d_{n+1}\ge d_n\},\quad B:=\{n\ge 1: d_{n+1}\le d_n\},\quad E:=\{n\ge 1: d_{n+1}=d_n\}. \] For a set $S\subseteq\mathbb N$ the (natural) density is \[ \delta(S):=\lim_{N\to\infty}\frac{|S\cap\{1,2,\dots,N\}|}{N} \] if the limit exists.

The problem statement (as given) asserts:

  1. $\delta(A)=\tfrac12$ and $\delta(B)=\tfrac12$;
  2. $E$ is infinite (infinitely many $n$ with $d_{n+1}=d_n$);
  3. (Erdős, 1985) for every $k\ge 1$ there exists $n$ such that \[d_n=d_{n+1}=\cdots=d_{n+k}. \]

Minimal clarifications / equivalences.

  • The sets $A$ and $B$ always satisfy $A\cup B=\mathbb N$ and $A\cap B=E$.
  • If both $\delta(A)$ and $\delta(B)$ exist and equal $\tfrac12$, then necessarily $\delta(E)=0$ (because $\delta(A)+\delta(B)=1+\delta(E)$).
  • The condition $d_n=d_{n+1}=\cdots=d_{n+k}$ is equivalent to saying that $p_n,p_{n+1},\dots,p_{n+k+1}$ form an arithmetic progression (and in fact are consecutive primes). A short proof is given below.

2. Quick literature/context check (browsing available)

The Erdős Problems site lists this as OPEN and records the formulation above, including the remark that the system $d_n=\cdots=d_{n+k}$ is equivalent to consecutive primes in arithmetic progression.(https://www.erdosproblems.com/218)

In the associated discussion thread (comments on the same page), T. Tao notes that it "should be possible" (via sieve methods) to show the set $E$ has density $0$; this is not presented there as a completed refereed proof.(https://www.erdosproblems.com/218?forum=1)

No published proof of the full density-$\tfrac12$ claims is supplied on that page, and I did not locate (in a quick browse) a definitive resolution; so I treat the main statements as open.

3. Attack plan

A plausible route to the density claims would be:
  1. Prove $\delta(E)=0$ (density zero of equal consecutive gaps). A natural idea is to bound the count of indices $n\le N$ with $d_{n+1}=d_n$ by an upper-bound sieve for prime constellations of type $\{0,d,2d\}$ (plus a treatment of large $d$ via a Markov/averaging argument).
  2. Show that the "strict" sets $\{n: d_{n+1}>d_n\}$ and $\{n: d_{n+1}<d_n\}$ have equal density (or at least both have density $\tfrac12$ after removing $E$). This would likely require some distributional symmetry of pairs $(d_n,d_{n+1})$ which is far beyond current unconditional technology.
  3. For the long-run conjecture $d_n=\cdots=d_{n+k}$ for all $k$, relate it to prime $k$-tuple / consecutive-primes-in-AP phenomena (stronger than Green--Tao, which gives APs of primes but not consecutiveness).

4. Work

4.1. Fast reality check (small $n$)

The first primes are $2,3,5,7,11,13,17,19,\dots$ giving gaps \[ (d_1,d_2,d_3,d_4,d_5,d_6,\dots)=(1,2,2,4,2,4,2,\dots). \] Then $d_2\ge d_1$ (true), $d_3\ge d_2$ (equality), and $d_4\ge d_3$ (strict). So $A$ is nonempty and $E$ already occurs at $n=2$.

4.2. Equivalence: runs of equal gaps $\Longleftrightarrow$ consecutive primes in AP

Lemma: Fix $k\ge 1$. For an index $n\ge 1$, the following are equivalent:
  1. $d_n=d_{n+1}=\cdots=d_{n+k}=d$ for some integer $d\ge 1$.
  2. The primes $p_n,p_{n+1},\dots,p_{n+k+1}$ satisfy $p_{n+j}=p_n+j d$ for all $0\le j\le k+1$; i.e. they form an arithmetic progression of length $k+2$.
In particular, they are consecutive primes lying in that progression.
Proof: (a)$\Rightarrow$(b): By definition $p_{n+1}=p_n+d$, and if $p_{n+j+1}-p_{n+j}=d$ for $0\le j\le k$, then summing gives \[ p_{n+m}=p_n+\sum_{j=0}^{m-1}(p_{n+j+1}-p_{n+j})=p_n+md \] for every $1\le m\le k+1$.

(b)$\Rightarrow$(a): If $p_{n+j}=p_n+j d$, then \[ d_{n+j}=p_{n+j+1}-p_{n+j}=(p_n+(j+1)d)-(p_n+j d)=d \] for $0\le j\le k$. ∎

4.3. A density identity (elementary)

Proposition: Assume the natural densities $\delta(A)$ and $\delta(B)$ exist. Then $\delta(E)$ exists and \[ \delta(A)+\delta(B)=1+\delta(E). \] In particular, if $\delta(A)=\delta(B)=\tfrac12$, then $\delta(E)=0$.
Proof: For every $n$, at least one of $d_{n+1}\ge d_n$ or $d_{n+1}\le d_n$ holds, so $A\cup B=\mathbb N$. Moreover, both hold iff $d_{n+1}=d_n$, so $A\cap B=E$. Hence for each $N$, \[ |A\cap[1,N]|+|B\cap[1,N]|=|[1,N]|+|E\cap[1,N]|=N+|E\cap[1,N]|. \] Divide by $N$ and let $N\to\infty$. ∎

4.4. Numerical experiment (finite data)

I computed the first $10^6$ prime gaps and measured the frequencies of $d_{n+1}>d_n$, $d_{n+1}<d_n$, and $d_{n+1}=d_n$. The results (for comparisons among the first $N$ gaps) were:
$N$$\frac{\#\{d_{n+1}>d_n\}}{N}$$\frac{\#\{d_{n+1}<d_n\}}{N}$$\frac{\#\{d_{n+1}=d_n\}}{N}$
$10^3$$0.480$$0.461$$0.059$
$10^4$$0.489$$0.462$$0.049$
$10^5$$0.482$$0.481$$0.038$
$5\cdot 10^5$$0.4837$$0.4826$$0.0337$
$10^6$$0.484183$$0.483839$$0.031978$

Equivalently, including equality, one finds $\#\{d_{n+1}\ge d_n\}/N\approx 0.516$ and $\#\{d_{n+1}\le d_n\}/N\approx 0.516$ at $N=10^6$, with equality occurring about $3.2\%$ of the time in this range. These computations are only heuristic evidence and do not address existence of densities.

5. Verification

  • The equivalence lemma is a direct telescoping-sum check and has no hidden cases.
  • The density identity uses only set algebra $A\cup B=\mathbb N$ and $A\cap B=E$.
  • The numeric table is consistent with the identity $\#\{\ge\}+\#\{\le\}=N+\#\{=\}$.

6. Final

UNRESOLVED.
I did not obtain a complete proof or counterexample to the density-$\tfrac12$ statements or to the infinitude of $E$, and I did not resolve the stronger "arbitrarily long runs" conjecture.
(i) Strongest fully proved partial result: The identity $A\cup B=\mathbb N$, $A\cap B=E$ implies $\delta(A)+\delta(B)=1+\delta(E)$ whenever these densities exist; in particular, the simultaneous claims $\delta(A)=\delta(B)=\tfrac12$ force $\delta(E)=0$. Also, $d_n=\cdots=d_{n+k}$ is equivalent to $k+2$ consecutive primes in an arithmetic progression.
(ii) First gap that blocks completion: Showing $\delta(E)=0$ (or even proving $E$ is infinite) appears to require nontrivial upper bounds for the count of consecutive 3-term prime arithmetic progressions (or related prime-constellation counts) in a range, beyond what I derived here.
(iii) Top 3 next moves: $1$ Try to implement the sieve/averaging strategy sketched in the forum discussion to prove $\delta(E)=0$; $2$ look for any known results on sign changes/oscillation of consecutive prime gaps and whether they yield lower bounds for $\#\{d_{n+1}>d_n\}$ and $\#\{d_{n+1}<d_n\}$; $3$ study "chains of consecutive prime gaps" literature for any distributional symmetry of $(d_n,d_{n+1})$.
(iv) What a minimal counterexample would look like: To refute $\delta(A)=\tfrac12$ one would need to show a persistent bias in comparisons $d_{n+1}\ge d_n$ versus $d_{n+1}\le d_n$ that does not vanish as $n\to\infty$ (equivalently, a positive-density bias in strict inequalities after removing the equality set).

7. Completion estimate

10%. I can prove basic identities and equivalences and compute evidence, but the core density assertions seem to require deep, currently unavailable distributional control of consecutive prime gaps.