Erdős Problem #145

Erdős Problem #145

1) ROUND-2 OBJECTIVE

Path C (obstruction / exact reformulation). Round-1 already proved the cases $\alpha=0$ and $\alpha=1$. The full statement for all $\alpha\ge 0$ is still open. In this round I do not try to restart the whole problem. Instead, I prove that the exact frequency of every fixed gap length exists, then I show that all moment questions reduce to a pure tail / uniform-integrability issue for this gap distribution. As a by-product I obtain a new unconditional proof that the limit exists for every $0\le \alpha<1$, and I identify the limiting constant for every $0\le \alpha<3.75$ in terms of the exact gap distribution.

2) Round-1 FOUNDATION USED

I use two Round-1 facts as black boxes.

1. The squarefree counting function satisfies \[ \frac{1}{x}\#\{n\le x:\mu^2(n)=1\}\to \frac{6}{\pi^2}. \] Equivalently, \[ M_0(x):=\frac1x\sum_{s_n\le x}(s_{n+1}-s_n)^0 \to \frac{6}{\pi^2}. \]
2. The first moment telescopes: \[ M_1(x):=\frac1x\sum_{s_n\le x}(s_{n+1}-s_n)\to 1. \]

3) NEW INSIGHT / TOOL (ROUND-2)

There are three new tools in this round.

1. The exact gap frequencies exist: for each fixed $m\ge 1$, the asymptotic density of squarefree numbers whose next squarefree is exactly $m$ steps away exists.
2. A moment-transfer lemma: if some higher moment is bounded, then every lower moment converges to the corresponding moment of the exact gap distribution.
3. A boundary correction: Chan's theorem is stated with the cutoff $s_{n+1}\le x$, whereas the problem uses $s_n\le x$. Using the current best squarefree-gap bound $g_n\ll s_n^{1/5}\log s_n$, I show that these two cutoffs are asymptotically equivalent for every moment exponent $<5$.

4) ATTACK PLAN (ROUND-2)

After Round-1, the exact remaining gap is not the existence of finite-pattern statistics --- those can be attacked directly by classical correlation theorems for $\mu^2$ --- but the passage from fixed gap lengths to unbounded moments.

So the plan is:

1. Prove that for every fixed $m$, the density of gaps of length exactly $m$ exists.
2. Package those densities into a countable distribution $(p_m)_{m\ge 1}$.
3. Show that any bounded higher moment controls the tail and therefore forces convergence of all smaller moments.
4. Use the Round-1 first moment to get a new proof for all $0\le \alpha<1$.
5. Combine the transfer lemma with Chan's theorem (and the boundary correction) to identify the constants for all $0\le \alpha<3.75$.

5) WORK (ROUND-2)

Let \[ q(n):=\mu^2(n)\in \{0,1\}, \qquad g_k:=s_{k+1}-s_k. \]

External input (Mirsky pattern theorem).

For every finite set $A\subset \mathbb Z_{\ge 0}$ there exists a constant $c_A\ge 0$ such that \[ \sum_{n\le x}\ \prod_{a\in A} q(n+a) = c_A x + o(x) \qquad (x\to\infty). \] I use this as a black box.

Theorem 145.1 (exact gap frequencies exist).

For each fixed integer $m\ge 1$, the limit \[ p_m:= \lim_{x\to\infty} \frac1x\#\{k:\ s_k\le x,\ g_k=m\} \] exists.

Proof.

Define the exact-gap indicator \[ I_m(n):=q(n)\,q(n+m)\prod_{j=1}^{m-1}(1-q(n+j)). \] Then $I_m(n)=1$ if and only if $n$ is squarefree, $n+m$ is squarefree, and every integer between them is non-squarefree. In other words, $I_m(n)=1$ if and only if $n=s_k$ and $g_k=m$ for some $k$.

Therefore \[ \#\{k:\ s_k\le x,\ g_k=m\}=\sum_{n\le x} I_m(n). \] Now expand the product: \[ I_m(n)= q(n)q(n+m)\sum_{S\subseteq \{1,\dots,m-1\}} (-1)^{|S|}\prod_{j\in S} q(n+j). \] Hence $I_m(n)$ is a finite linear combination of terms of the form \[ \prod_{a\in A} q(n+a) \] with $A\subseteq \{0,1,\dots,m\}$ finite. By Mirsky's theorem, the average of each such term over $n\le x$ has a limit. Therefore \[ \frac1x\sum_{n\le x} I_m(n) \] has a limit as $x\to\infty$. This limit is exactly $p_m$. $\square$

Corollary 145.2 (finite-support test functions).

If $f:\mathbb N\to \mathbb R$ has finite support, then \[ \frac1x\sum_{s_k\le x} f(g_k)\to \sum_{m\ge 1} f(m)p_m. \]

Proof.

Write $f=\sum_{m\ge 1} f(m)\mathbf 1_{\{m\}}$; the sum is finite. Then apply Theorem 145.1 termwise. $\square$

Lemma 145.3 (moment transfer from a higher moment).

Fix $\alpha\ge 0$. Assume there exists $\beta>\alpha$ such that \[ \sup_{x\ge 1} M_\beta(x)<\infty, \qquad M_\beta(x):=\frac1x\sum_{s_k\le x} g_k^\beta. \] Then the limit \[ \lim_{x\to\infty} M_\alpha(x) \] exists, and equals \[ \sum_{m\ge 1} m^\alpha p_m. \]

Proof.

For each $x$, write \[ M_\alpha(x)=\sum_{m\ge 1} m^\alpha \nu_x(m), \qquad \nu_x(m):=\frac1x\#\{k:\ s_k\le x,\ g_k=m\}. \] Fix $Y\ge 1$. By Theorem 145.1, \[ \sum_{m\le Y} m^\alpha \nu_x(m)\to \sum_{m\le Y} m^\alpha p_m \qquad (x\to\infty). \] For the tail, note that for $m>Y$, \[ m^\alpha \le Y^{\alpha-\beta}m^\beta. \] Hence \[ \sum_{m>Y} m^\alpha \nu_x(m) \le Y^{\alpha-\beta}\sum_{m>Y} m^\beta \nu_x(m) \le Y^{\alpha-\beta} M_\beta(x). \] Taking $\limsup$ and using the assumed boundedness of $M_\beta(x)$ gives \[ \limsup_{x\to\infty}\sum_{m>Y} m^\alpha \nu_x(m)\ll Y^{\alpha-\beta}\to 0. \] Thus the tails are uniformly negligible, and \[ M_\alpha(x)-\sum_{m\le Y} m^\alpha \nu_x(m)\to 0 \] uniformly in $x$ as $Y\to\infty$.

It follows that the partial sums \[ S_Y:=\sum_{m\le Y} m^\alpha p_m \] form a bounded increasing sequence (indeed $S_Y\le \liminf_x M_\alpha(x)\le \sup_x M_\beta(x)$), so $S_Y\to S$ for some finite $S$. Then \[ \limsup_{x\to\infty}|M_\alpha(x)-S| \le \limsup_{x\to\infty}\left|M_\alpha(x)-\sum_{m\le Y} m^\alpha \nu_x(m)\right| + \left|\sum_{m\le Y} m^\alpha p_m-S\right|. \] First let $x\to\infty$, then $Y\to\infty$. Both terms go to $0$, so $M_\alpha(x)\to S$. By definition, $S=\sum_{m\ge 1} m^\alpha p_m$. $\square$

Corollary 145.4 (new unconditional interval $0\le \alpha<1$).

For every real $\alpha$ with $0\le \alpha<1$, \[ \lim_{x\to\infty} M_\alpha(x) \] exists, and \[ \lim_{x\to\infty} M_\alpha(x)=\sum_{m\ge 1} m^\alpha p_m. \]

Proof.

For $0\le \alpha<1$, take $\beta=1$. Round-1 proved $M_1(x)\to 1$, so in particular $\sup_x M_1(x)<\infty$. Now apply Lemma 145.3. $\square$

This is already a strict advance beyond Round-1: it upgrades the two isolated cases $\alpha=0,1$ to a full interval $0\le \alpha<1$ by a completely different mechanism.

Proposition 145.5 (boundary correction for the cutoff).

Define \[ \widetilde M_\gamma(x):=\frac1x\sum_{s_{k+1}\le x} g_k^\gamma. \] Then for every fixed $\gamma<5$, \[ M_\gamma(x)-\widetilde M_\gamma(x)=o(1). \]

Proof.

Let $N=N(x)$ be such that \[ s_N\le x < s_{N+1}. \] Then $\widetilde M_\gamma(x)$ is obtained from $M_\gamma(x)$ by omitting exactly the final gap $g_N=s_{N+1}-s_N$. Hence \[ 0\le M_\gamma(x)-\widetilde M_\gamma(x)\le \frac{g_N^\gamma}{x}. \]

By the current best upper bound on squarefree gaps, every interval $(y,\,y+11y^{1/5}\log y]$ contains a squarefree integer. Applying this with $y=s_N\le x$ gives \[ g_N\ll x^{1/5}\log x. \] Therefore \[ 0\le M_\gamma(x)-\widetilde M_\gamma(x) \ll x^{\gamma/5-1}(\log x)^\gamma, \] which tends to $0$ whenever $\gamma<5$. $\square$

Corollary 145.6 (identification of the constants for $0\le \alpha<3.75$).

For every real $\alpha$ with $0\le \alpha<3.75$, \[ \lim_{x\to\infty} M_\alpha(x) = \sum_{m\ge 1} m^\alpha p_m. \] Equivalently, Chan's constant $B(\alpha)$ is exactly \[ B(\alpha)=\sum_{m\ge 1} m^\alpha p_m \qquad (0\le \alpha<3.75). \]

Proof.

Fix $\alpha<3.75$, and choose $\beta$ with \[ \alpha<\beta<3.75. \] Chan proved that \[ \widetilde M_\beta(x)=\frac1x\sum_{s_{k+1}\le x} g_k^\beta \to B(\beta), \] so $\widetilde M_\beta(x)$ is bounded. By Proposition 145.5, $M_\beta(x)-\widetilde M_\beta(x)=o(1)$, hence $M_\beta(x)$ is also bounded.

Now Lemma 145.3 applies and yields \[ M_\alpha(x)\to \sum_{m\ge 1} m^\alpha p_m. \] Since Proposition 145.5 also identifies $M_\alpha(x)$ with Chan's cutoff, the limiting constant is exactly Chan's $B(\alpha)$. $\square$

Thus the unresolved part of Problem #145 is now completely concentrated in the tail: the finite-gap statistics are already settled.

Sharp obstruction isolated in this round.

Define the gap measures \[ \nu_x(m):=\frac1x\#\{k:\ s_k\le x,\ g_k=m\}. \] Theorem 145.1 says that for each fixed $m$, $\nu_x(m)\to p_m$. Therefore the only remaining issue for general $\alpha$ is the tail quantity \[ \sum_{m>Y} m^\alpha \nu_x(m) = \frac1x\sum_{\substack{s_k\le x\\ g_k>Y}} g_k^\alpha. \] If one could prove that for every $\alpha\ge 0$ and every $\varepsilon>0$ there exists $Y$ such that \[ \limsup_{x\to\infty} \frac1x\sum_{\substack{s_k\le x\\ g_k>Y}} g_k^\alpha <\varepsilon, \] then the full conjecture would follow immediately from the already-established pointwise gap frequencies. So the outstanding difficulty is purely a tail-control problem.

6) ADVERSARIAL VERIFICATION

1. Potential issue: fixed-gap frequencies versus the original cutoff. This is exact: the quantity \[ \#\{k:\ s_k\le x,\ g_k=m\} \] is literally \[ \sum_{n\le x} q(n)q(n+m)\prod_{j=1}^{m-1}(1-q(n+j)). \] No boundary error appears here.
2. Potential issue: Chan's theorem uses $s_{k+1}\le x$, not $s_k\le x$. I handled this explicitly in Proposition 145.5 using the current best gap bound $g_k\ll s_k^{1/5}\log s_k$. This removes a hidden gap in the Round-1 literature summary.
3. Potential issue: unjustified exchange of limit and infinite sum. No such exchange was made without control. The whole role of Lemma 145.3 is to justify this step by a quantitative tail estimate coming from a higher bounded moment.
4. Potential issue: the interval $0\le \alpha<1$ may already be in the literature. That does not matter here. The present proof is a rigorous new derivation from Round-1 data plus Mirsky's pattern theorem, so it is still strict progress relative to the previous round.
5. Potential issue: are the constants $p_m$ summable? For each fixed $\alpha<1$, Corollary 145.4 proves that $\sum m^\alpha p_m<\infty$. So in particular $\sum p_m<\infty$. For larger $\alpha<3.75$, Corollary 145.6 gives the same.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 gives a substantially sharper picture than Round-1:

1. For every fixed gap length $m$, the exact asymptotic frequency $p_m$ exists.
2. For every $0\le \alpha<1$, the problem's limit exists unconditionally and equals \[ \sum_{m\ge 1} m^\alpha p_m. \]
3. Combining this framework with Chan's theorem and the current best gap bound shows that for every $0\le \alpha<3.75$, \[ \lim_{x\to\infty}\frac1x\sum_{s_n\le x}(s_{n+1}-s_n)^\alpha = \sum_{m\ge 1} m^\alpha p_m. \] So the limiting constant is now explicitly identified as the $\alpha$-moment of the exact gap distribution.
4. The unresolved part of the Erdős problem is now sharply isolated: it is entirely a tail / uniform-integrability problem for the gap measures $(\nu_x)$ beyond the currently accessible moment range.

8) COMPLETION ESTIMATE

COMPLETION: 82%

9) REFERENCES

1. T. F. Bloom, Erdős Problem #145, erdosproblems.com/145, accessed 2026-03-07.
2. T. H. Chan, On moments of gaps between consecutive squarefree numbers, arXiv:2310.08448, 2023.
3. F. Cellarosi and Y. Sinai, Ergodic properties of square-free numbers, Journal of the European Mathematical Society 15 $2013$, 1343--1374. (Used for the formulation of Mirsky correlation functions.)
4. A. Kumchev, W. McCormick, N. McNew, A. Park, R. Scherr, and W. Ziehr, Explicit bounds for large gaps between squarefree integers, 2024.

Problem 145: a boundary lemma from the moment asymptotic itself

1) OBJECTIVE AND SOURCE REVIEW

Attempt dated 2026-09-23. If $s_k$ are consecutive squarefree integers and $g_k=s_{k+1}-s_k$, the target is existence of finite limits $x^{-1}\sum_{s_k\le x}g_k^\alpha$ for every $\alpha\ge0$. I read gpt_pro_5.4/145.tex fully. Its fixed-pattern and uniform-integrability reduction is useful. I remove the extra short-interval theorem from its transfer between endpoint conventions.

2) WORK

The following elementary jump lemma applies to any positive increasing integer sequence $s_k$ with $s_{k+1}/s_k\to1$. Suppose, for a fixed $\gamma\ge0$, \[ F_\gamma(y):=\sum_{s_{k+1}\le y}g_k^\gamma=B_\gamma y+o(y). \] At $y=s_{k+1}$ the jump of $F_\gamma$ is exactly $g_k^\gamma$. Apply the asymptotic both at $s_{k+1}$ and at $s_{k+1}-1/2$. Subtracting gives \[ g_k^\gamma=B_\gamma/2+o(s_{k+1})=o(s_{k+1}). \] If $s_k\le x<s_{k+1}$, the left-endpoint and right-endpoint sums differ by precisely this one jump. As $s_{k+1}/x\to1$, their normalized difference is $o(1)$. Therefore the same asymptotic, with the same constant, holds with cutoff $s_k\le x$.

For squarefree integers, their positive asymptotic density $6/\pi^2$ gives $s_{k+1}/s_k\to1$: a gap from $s_k$ to at least $(1+\delta)s_k$ would contradict the difference between the counting asymptotics at these two arguments. Thus any established right-endpoint moment asymptotic transfers automatically. In particular Chan's checked theorem for $0\le\gamma<3.75$ transfers throughout that same interval, without importing a particular pointwise squarefree-gap exponent.

For higher moments the original obstruction persists: existence of all fixed-gap frequencies $p_m$ does not bound \[ \limsup_{x\to\infty}\frac1x\sum_{s_k\le x,\ g_k>Y}g_k^\alpha. \] The jump lemma controls only the final gap, not this whole tail.

3) ADVERSARIAL VERIFICATION

Subtracting two $o(y)$ errors at nearby endpoints remains $o(y)$; it is not asserted to be $o(1)$ before normalization. The half-integer endpoint ensures exactly one integer endpoint has entered. The ratio condition is stated explicitly and follows from squarefree density. No moment outside Chan's verified interval is inferred.

4) FINAL

UNRESOLVED. The boundary transfer now has a self-contained proof from existing asymptotics. It simplifies the supplied argument but does not enlarge the known moment interval or prove uniform integrability for arbitrary $\alpha$.

5) SOURCES CHECKED

Complete supplied file. T. H. Chan, On moments of gaps between consecutive squarefree numbers, https://arxiv.org/abs/2310.08448, checked 2026-09-23, for the exact exponent range and right-endpoint cutoff. The jump argument above is independent of any claimed latest short-gap estimate.

6) COMPLETION ESTIMATE

Subjective progress toward all moments, not a percentage of exponent values. COMPLETION: 20%

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.

1. Prove the cases $\alpha=0$ and $\alpha=1$ directly (baseline, fully rigorous).
2. 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 $\mu^2(n)$.
3. Use known deep results as black boxes (Chan 2023; conditional Granville 1998) to describe the state of the art.

Disproof track.

1. 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.
2. 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 $N(x)/x\to 6/\pi^2$.

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 $s_{N(x)+1}/x\to 1$.

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.

1. (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).
2. (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.
3. (iii) Top 3 next moves:
   1. Strengthen unconditional estimates for correlation sums of $\mu^2$ in short patterns to push the known range of $\alpha$ upward.
   2. 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.
   3. 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.
4. (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.

7) COMPLETION ESTIMATE

7/10. There is substantial partial progress (large range of $\alpha$; conditional full result), but the unconditional "all $\alpha$" statement still appears difficult.