gpt pro 5.2 attempt on Math-690
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Contents
Erdős Problem #145 (Moments of gaps between squarefree numbers)
1) FORMAL RESTATEMENT
Let $1=s_1<s_2<s_3<\cdots$ be the increasing sequence of squarefree positive integers. For $x\ge 1$, let \[ N(x):=|\{n:\ s_n\le x\}|. \] For $\alpha\ge 0$, define \[ M_{\alpha}(x):=\frac{1}{x}\sum_{n\le N(x)}(s_{n+1}-s_n)^{\alpha}. \] (The sum is finite because $N(x)<\infty$ for finite $x$, and $s_{N(x)+1}$ is defined because the squarefree numbers are infinite.)The question is: for every fixed $\alpha\ge 0$, does the limit \[ \lim_{x\to\infty} M_{\alpha}(x) \] exist?
2) QUICK LITERATURE / CONTEXT CHECK
- Under the $abc$ conjecture, Granville $1998$ derived the asymptotic predicted by Erdős for all moments of the gaps between squarefree numbers.
- Unconditionally, Chan $2023$ proved an asymptotic $\sum_{s_{k+1}\le x}(s_{k+1}-s_k)^{\gamma}\sim B(\gamma)x$ for $0\le\gamma<3.75$, which in particular implies existence of the limit for $0\le\alpha<3.75$.
- For the special exponents $\alpha=0$ and $\alpha=1$, the limit exists and can be proved elementarily: $\alpha=0$ is the natural density of squarefree numbers ($6/\pi^2$), and $\alpha=1$ telescopes to $x$ up to a negligible boundary term.
3) ATTACK PLAN
Proof track.- Prove the cases $\alpha=0$ and $\alpha=1$ directly (baseline, fully rigorous).
- For general $\alpha$, expand moments in terms of local configurations of squarefree/non-squarefree patterns and estimate via inclusion--exclusion. This rapidly leads to difficult correlation sums of $\mu^2(n)$.
- Use known deep results as black boxes (Chan 2023; conditional Granville 1998) to describe the state of the art.
Disproof track.
- A counterexample would require that $M_{\alpha}(x)$ oscillates without settling. This would likely come from persistent, scale-dependent irregularities in the frequency of very large gaps.
- Heuristically, squarefree numbers behave like a weakly dependent Bernoulli process with density $6/\pi^2$, suggesting $M_{\alpha}(x)$ should converge for all $\alpha$, but converting this to proof is hard.
4) WORK
Phase 1 sanity check by computation.
A quick sieve computation up to $10^6$ (not a proof) suggests stabilization of $M_{\alpha}(x)$ for a few sample $\alpha$: \[ \begin{array}{c|c} \alpha & M_{\alpha}(10^6)\\ \hline 0 & 0.607929\\ 1 & 1.000004\\ 2 & 2.041249\\ 3 & 6.624972 \end{array} \] These are consistent with the existence of limits (and with $M_0(\infty)=6/\pi^2\approx0.607927$).Case $\alpha=0$.
Observe that $(s_{n+1}-s_n)^0=1$ for all $n$, hence \[ M_0(x)=\frac{1}{x}\sum_{n\le N(x)}1=\frac{N(x)}{x}. \] Thus we need $N(x)/x\to 6/\pi^2$.Lemma (Counting squarefree numbers): Let $Q(x):=|\{n\le x: n\ \text{squarefree}\}|$. Then \[
Q(x)=\frac{x}{\zeta(2)}+O(\sqrt{x})
\qquad (x\to\infty).
\] In particular $Q(x)/x\to 1/\zeta(2)=6/\pi^2$.
Proof: Use the identity $\mu^2(n)=\sum_{d^2\mid n}\mu(d)$ (since $\mu^2$ is the indicator of squarefree integers) to write \[
Q(x)=\sum_{n\le x}\mu^2(n)=\sum_{n\le x}\sum_{d^2\mid n}\mu(d)=\sum_{d\le \sqrt{x}}\mu(d)\Big\lfloor \frac{x}{d^2}\Big\rfloor.
\] Split \[
\Big\lfloor \frac{x}{d^2}\Big\rfloor=\frac{x}{d^2}+O(1)
\] and obtain \[
Q(x)=x\sum_{d\le\sqrt{x}}\frac{\mu(d)}{d^2}+O\Big(\sum_{d\le\sqrt{x}}1\Big).
\] The error term is $O(\sqrt{x})$. For the main term, note that $\sum_{d\ge1}\frac{\mu(d)}{d^2}=1/\zeta(2)$ and the tail satisfies $\sum_{d>\sqrt{x}}\frac{|\mu(d)|}{d^2}\le\sum_{d>\sqrt{x}}\frac1{d^2}=O(1/\sqrt{x})$. Therefore \[
\sum_{d\le\sqrt{x}}\frac{\mu(d)}{d^2}=\frac1{\zeta(2)}+O\Big(\frac1{\sqrt{x}}\Big),
\] so $Q(x)=\frac{x}{\zeta(2)}+O(\sqrt{x})$. Finally, $N(x)=Q(x)$ because $N(x)$ counts squarefree integers $\le x$. ∎
Corollary (Limit for $\alpha=0$): $\displaystyle \lim_{x\to\infty} M_0(x)=\frac{1}{\zeta(2)}=\frac{6}{\pi^2}$.
Proof: Immediate from Lemma Lemma (Counting squarefree numbers) and $M_0(x)=N(x)/x$. ∎
Case $\alpha=1$.
Let $N=N(x)$. Then the gap sum telescopes: \[ \sum_{n\le N}(s_{n+1}-s_n)=s_{N+1}-s_1. \] Hence \begin{equation} \label{wsref-0-65-71-3a-31-34-35-2d-61-6c-70-68-61-31-2d-74-65-6c-65-73-63-6f-70-65} M_1(x)=\frac{s_{N(x)+1}-s_1}{x}. \end{equation} Since $s_1=1$, it suffices to prove $s_{N(x)+1}/x\to 1$.Lemma (Relative next-squarefree gap is $o(x)$): Let $t(x):=s_{N(x)+1}$ be the least squarefree integer strictly larger than $x$. Then \[
\lim_{x\to\infty}\frac{t(x)}{x}=1.
\]
Proof: We already know from Lemma Lemma (Counting squarefree numbers) that $N(u)/u\to 1/\zeta(2)$ as $u\to\infty$. Fix $\varepsilon>0$. Suppose for contradiction that there exist arbitrarily large $x$ with $t(x)>(1+\varepsilon)x$. For such an $x$, there are no squarefree integers in $(x,(1+\varepsilon)x]$, hence $N((1+\varepsilon)x)=N(x)$. Then \[
\frac{N((1+\varepsilon)x)}{(1+\varepsilon)x}=\frac{N(x)}{(1+\varepsilon)x}.
\] Taking $x\to\infty$ along this subsequence gives \[
\lim_{x\to\infty}\frac{N((1+\varepsilon)x)}{(1+\varepsilon)x}=\frac{1}{1+\varepsilon}\cdot \lim_{x\to\infty}\frac{N(x)}{x}=\frac{1}{(1+\varepsilon)\zeta(2)}.
\] But the left-hand limit must equal $1/\zeta(2)$ because $(1+\varepsilon)x\to\infty$ as well. This contradiction shows that for all sufficiently large $x$, $t(x)\le (1+\varepsilon)x$. Since always $t(x)>x$, we have $1< t(x)/x\le 1+\varepsilon$ for large $x$, hence $\limsup t(x)/x\le 1+\varepsilon$. As $\varepsilon>0$ is arbitrary, $\limsup t(x)/x\le 1$, and therefore $t(x)/x\to 1$. ∎
Corollary (Limit for $\alpha=1$): $\displaystyle \lim_{x\to\infty} M_1(x)=1$.
Proof: Combine (eq:145-alpha1-telescope) with Lemma lem:145-nextgap and $s_1=1$. ∎
Beyond $\alpha=1$.
For $\alpha\ge 2$, proving existence of $\lim_{x\to\infty}M_{\alpha}(x)$ appears to require nontrivial control of higher correlations of $\mu^2(n)$ (equivalently, counts of squarefree patterns), and unconditional results currently only cover a finite range of exponents (e.g. $\alpha<3.75$ by Chan).5) VERIFICATION
- The definitions match the problem statement: we sum over $s_n\le x$ and use the corresponding gap to $s_{n+1}$.
- The proofs for $\alpha=0$ and $\alpha=1$ are complete and unconditional.
- Lemma lem:145-nextgap uses only the existence of the density limit for squarefrees (proved in Lemma Lemma (Counting squarefree numbers)).
6) FINAL
UNRESOLVED. The full statement "the limit exists for every $\alpha\ge 0$" remains open unconditionally.- (i) Strongest fully proved partial result obtained:
- The limit exists for $\alpha=0$ and equals $6/\pi^2$ (Corollary cor:145-alpha0).
- The limit exists for $\alpha=1$ and equals $1$ (Corollary cor:145-alpha1).
- (From the literature) the limit exists for all $0\le\alpha<3.75$ (Chan 2023) and, assuming $abc$, for all $\alpha\ge 0$ (Granville 1998).
- (ii) Exact first gap: I do not have an unconditional argument for existence of $\lim_{x\to\infty}M_{\alpha}(x)$ for all $\alpha\ge 0$ (e.g. for $\alpha\ge 4$), beyond citing the known partial ranges.
- (iii) Top 3 next moves:
- Strengthen unconditional estimates for correlation sums of $\mu^2$ in short patterns to push the known range of $\alpha$ upward.
- Seek a probabilistic/ergodic formulation of squarefree indicator that gives a renewal-type limit theorem for the gap process, from which all moments would follow.
- Explore whether modern sieve technology (e.g. adaptations of methods used for prime gaps) can control the tail of the gap distribution strongly enough to imply convergence of all moment averages.
- (iv) What a minimal counterexample would likely look like: persistent oscillation in the frequency of long non-squarefree runs at scales comparable to $x$, causing $M_{\alpha}(x)$ to have different subsequential limits. Any such construction seems incompatible with the strong "random-like" heuristics for squarefrees, but no impossibility proof is known to me.