we-solve (beta)

Math-723-A2 · revision r1 · on Math-723

gpt pro 5.2 attempt on Math-723

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. Erdős Problem #234
  2. Lemma 234.1 (Mean of normalized gaps).
  3. Corollary 234.2 (A Markov-type bound on large normalized gaps).

Erdős Problem #234

  1. FORMAL RESTATEMENT.

    Let $p_n$ denote the $n$th prime ($p_1=2,p_2=3,\dots$) and let \[d_n := p_{n+1}-p_n \qquad (n\ge 1)\] be the $n$th prime gap. Throughout, $\log$ denotes the natural logarithm.

    For $c\ge 0$, define the set \[A_c := \left\{n\in\mathbb{N} : n\ge 2\ \text{and}\ \frac{d_n}{\log n} < c\right\}. \] (We restrict to $n\ge 2$ so that $\log n$ is defined.)

    Define the natural density of $A_c$ (if it exists) by \[ f(c) := \lim_{N\to\infty} \frac{1}{N}\,\left|A_c\cap\{1,2,\dots,N\}\right|. \]

    Claim/conjecture. For every $c\ge 0$, the limit defining $f(c)$ exists, and the resulting function $f:[0,\infty)\to[0,1]$ is continuous.

    Edge cases. For $c=0$, $A_0=\varnothing$ (since $d_n>0$), so $f(0)=0$ if the above definition is used.

  2. QUICK LITERATURE/CONTEXT CHECK.

    This is an "existence of limiting distribution" problem for the normalized gaps $a_n:=d_n/\log n$. Heuristically, in Cramér/Hardy--Littlewood models one expects a limiting distribution (often Poisson/exponential for suitably normalized gaps), which would in particular imply the existence of $f(c)$ and its continuity.

    Unconditionally, one has very strong information on extremes of prime gaps (liminf $=0$ in the $d_n/\log p_n$ normalization, and arbitrarily large multiples of the average gap), but no unconditional theorem gives the full limiting distribution of the normalized gaps.

  3. ATTACK PLAN.

    Proof strategies.

    • $P1$ Try to show convergence of the empirical distribution functions $F_N(c):=\frac1N|\{2\le n\le N: d_n/\log n<c\}|$ to a limit using deep distributional results on primes in short intervals / correlations of primes.
    • $P2$ Prove existence of the density for all $c$ by deriving asymptotics for counts of gaps $\le c\log n$ from (a strong form of) the Hardy--Littlewood prime tuples conjecture.

    Disproof/construction strategies.

    • $D1$ Attempt to use known irregularity phenomena for primes in short intervals (à la Maier) to construct a $c$ for which $F_N(c)$ has distinct limit points.
    • $D2$ Attempt to show $f$ cannot be continuous by forcing a nontrivial point mass in any putative limiting distribution.

    Best path chosen here. In this write-up, the proof track $P1$--$P2$ and the disproof track $D1$--$D2$ both stall at the point where one would need substantially finer distributional control of prime gaps than is established here. Accordingly, I record rigorous partial results that any full solution must be consistent with: (i) the normalized gaps have mean $1+o(1)$; (ii) this yields quantitative bounds on upper/lower densities for large $c$.

  4. WORK.

    Lemma 234.1 (Mean of normalized gaps).
    Let $a_n:=\dfrac{p_{n+1}-p_n}{\log n}$ for $n\ge 2$. Assuming the Prime Number Theorem, one has \[ \lim_{N\to\infty}\frac{1}{N}\sum_{n=2}^{N} a_n = 1. \]
    Proof: Write $d_n:=p_{n+1}-p_n$. For $N\ge 2$ consider \[ S(N):=\sum_{n=2}^{N}\frac{d_n}{\log n}. \] For each $n\ge 2$ we have \[ \frac{d_n}{\log n}=\int_{p_n}^{p_{n+1}} \frac{dt}{\log n}. \] On the interval $[p_n,p_{n+1})$ the prime-counting function satisfies $\pi(t)=n$, hence $\log n = \log \pi(t)$ there. Summing over $n=2,\dots,N$ and concatenating the integrals gives the identity \[ S(N)=\int_{p_2}^{p_{N+1}}\frac{dt}{\log\pi(t)}. \] By the Prime Number Theorem, $\pi(t)\sim t/\log t$, so for sufficiently large $t$ we have \[ \log\pi(t)=\log t + O(\log\log t). \] For large $t$, the error term satisfies $|O(\log\log t)|\le \tfrac12\log t$, so the function $u\mapsto 1/(\log t+u)$ can be expanded to first order, yielding \[ \frac{1}{\log\pi(t)} = \frac{1}{\log t} + O\!\left(\frac{\log\log t}{(\log t)^2}\right). \] Therefore, with $X:=p_{N+1}$, \[ S(N)=\int_{p_2}^{X}\frac{dt}{\log t} \;+ O\!\left(\int_{p_2}^{X}\frac{\log\log t}{(\log t)^2}\,dt\right). \] The main integral is the logarithmic integral $\operatorname{li}(X)$ up to an additive constant. By integration by parts, \[ \int_{2}^{X}\frac{dt}{\log t}=\frac{X}{\log X}+O\!\left(\frac{X}{(\log X)^2}\right). \] For the error integral, since the integrand is eventually increasing in $t$ and bounded above by its value at $X$, one has \[ \int_{2}^{X}\frac{\log\log t}{(\log t)^2}\,dt \ll \frac{X\log\log X}{(\log X)^2}. \] Combining these estimates gives \[ S(N)=\frac{X}{\log X}+O\!\left(\frac{X\log\log X}{(\log X)^2}\right). \] Finally, by the Prime Number Theorem in the equivalent form $p_{N+1}\sim (N+1)\log(N+1)$, we have $X\sim N\log N$ and $\log X = \log N + O(\log\log N)$. Hence \[ \frac{X}{\log X}=N\,(1+o(1)), \qquad \frac{X\log\log X}{(\log X)^2}=N\,o(1). \] Therefore $S(N)=N+o(N)$, and dividing by $N$ yields \[ \frac{1}{N}\sum_{n=2}^{N}a_n = \frac{S(N)}{N} \to 1. \] ∎
    Corollary 234.2 (A Markov-type bound on large normalized gaps).
    Assume Lemma 234.1. Fix $c>0$. Then \[ \limsup_{N\to\infty}\frac{1}{N}\left|\{2\le n\le N: a_n\ge c\}\right|\le \frac{1}{c}. \] In particular, for $c>1$, \[ \liminf_{N\to\infty}\frac{1}{N}\left|\{2\le n\le N: a_n< c\}\right|\ge 1-\frac{1}{c}. \]
    Proof: For each $N$ and $c>0$, \[ \sum_{n=2}^{N}a_n \ge \sum_{\substack{2\le n\le N\\ a_n\ge c}} a_n \ge c\,\left|\{2\le n\le N: a_n\ge c\}\right|. \] Divide by $N$ and take $\limsup$; Lemma 234.1 gives $\lim_{N\to\infty}\frac1N\sum_{n=2}^N a_n =1$, hence the displayed bound. The second inequality is the complement. ∎

    What this does not give. These arguments control only the first moment of $a_n$ and hence give only coarse bounds on the distribution of $a_n$; they do not address existence of the density $f(c)$ for a fixed $c$, nor continuity of $f$.

  5. VERIFICATION.

    • Quantifiers. Lemma 234.1 uses only $N\to\infty$ and PNT; the conversion $S(N)=\int dt/\log\pi(t)$ is exact.
    • Edge cases. The definition starts at $n\ge 2$ so $\log n$ is defined. Any finite number of initial terms does not affect densities.
    • Error control. The step $1/\log\pi(t)=1/\log t + O(\log\log t/(\log t)^2)$ requires $\log\pi(t)=\log t + O(\log\log t)$ and that $\log t$ dominates $\log\log t$; this holds for all sufficiently large $t$.

  6. FINAL.

    UNRESOLVED.

    • (i) Strongest fully proved partial result obtained here. Assuming PNT, the mean of the normalized gaps $a_n=(p_{n+1}-p_n)/\log n$ tends to $1$, and hence for every fixed $c>0$, $\limsup_{N\to\infty}\frac1N|\{n\le N: a_n\ge c\}|\le 1/c$.
    • (ii) First gap. No unconditional (or even PNT-level) method in this write-up proves that the limit defining $f(c)$ exists for an arbitrary fixed $c$, nor continuity of the putative limit function.
    • (iii) Top 3 next moves.
      • Target an "asymptotic for the count of gaps $\le c\log n$" from a strong Hardy--Littlewood prime tuples input.
      • Try to convert known irregularity results for primes in short intervals into oscillations of $F_N(c)$ to attack existence of the density.
      • Study whether any currently known moment information on prime gaps (even conditional) implies continuity/no-atom properties of a limiting law.
    • (iv) What a minimal counterexample would likely look like. A value $c_0$ and two sequences $N_j\to\infty$, $M_j\to\infty$ such that $F_{N_j}(c_0)$ and $F_{M_j}(c_0)$ converge to different limits, i.e. a persistent oscillation in the frequency of "$d_n< c_0\log n$".

  7. COMPLETION ESTIMATE (MANDATORY).
    COMPLETION: 35%.