Erdős Problem #464
Status Solved · erdosproblems.comLLM status Solved from source status
Erdős problems openaigpt 6 astra ultra 100% cataloguedSolvedSolvedStatement
Statement and interpretation.
Let $n_{j+1}\geq(1+\epsilon)n_j$, with fixed $\epsilon>0$. Must some irrational $\theta$ make the distances $\|\theta n_j\|$ non-dense? The displayed range $[0,1]$ on the problem page is automatically too large for distances to the nearest integer, which lie in $[0,1/2]$. We prove the substantive stronger conclusion \[ \inf_j\|\theta n_j\|>0 \tag{1} \] with $\theta$ irrational.Explicit external theorem.
Peres and Schlag, Theorem 3.1, prove the following bounded-block lacunarity result. There is a universal $c>0$ such that, if an increasing integer sequence $m_j$ satisfies \[ m_{j+M}>2m_j\quad(j\geq1),\qquad M\geq4, \] then some $\theta\in[0,1)$ has \[ \|m_j\theta\|\geq \frac{c}{M\log M}\quad(j\geq1). \tag{2} \] The normalization of $c$ absorbs the logarithm base. Their local-lemma proof is an external dependency. We give the additional reduction which ensures irrationality, rather than assuming that every parameter supplied by $2$ is irrational.Add a sparse sequence detecting every rational.
Let $S$ be the union of the given sequence with $\{r!:r\geq2\}$, and enumerate it increasingly without repetition as $m_1,m_2,\ldots$. In any interval $[x,2x]$, the original sequence contributes at most \[ \left\lfloor\frac{\log2}{\log(1+\epsilon)}\right\rfloor+1 \] members. The factorial sequence starting at $2!$ contributes at most one, since successive ratios are at least three. Thus, with \[ M=\left\lceil\frac{\log2}{\log(1+\epsilon)}\right\rceil+4, \] the interval $[m_j,2m_j]$ contains fewer than $M+1$ members of $S$. This proves $m_{j+M}>2m_j$. Apply $2$ to this union. It yields $1$, since every original $n_j$ belongs to $S$.If $\theta=a/b$ were rational, every factorial $r!$ with $r\geq b$ would make $r!\theta$ an integer, contradicting $2$. Hence $\theta$ is irrational. The same argument gives a uniform positive gap from zero, so the original distances are not dense even in their natural interval $[0,1/2]$.
Sources and scope.
Y. Peres and W. Schlag, Two Erdős problems on lacunary sequences: chromatic number and Diophantine approximation, Theorem 3.1, https://arxiv.org/abs/0706.0223. Current statement and earlier attribution to de Mathan and Pollington: https://www.erdosproblems.com/464, checked 24 September 2026. This is a complete deduction from the explicitly stated theorem, including the irrationality requirement.Completion estimate.
The stronger intended conclusion follows with irrationality proved explicitly.Restatement from Math-5667-A1 r1, with the source manuscript’s qualifications.
Original problem on erdosproblems.com
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Math-5667
Erdős Problem 464
Statement and interpretation.
Let $n_{j+1}\geq(1+\epsilon)n_j$, with fixed $\epsilon>0$. Must some irrational $\theta$ make the distances $\|\theta n_j\|$ non-dense? The displayed range $[0,1]$ on the problem page is automatically too large for distances to the nearest integer, which lie in $[0,1/2]$. We prove the substantive stronger conclusion \[ \inf_j\|\theta n_j\|>0 \tag{1} \] with $\theta$ irrational.Explicit external theorem.
Peres and Schlag, Theorem 3.1, prove the following bounded-block lacunarity result. There is a universal $c>0$ such that, if an increasing integer sequence $m_j$ satisfies \[ m_{j+M}>2m_j\quad(j\geq1),\qquad M\geq4, \] then some $\theta\in[0,1)$ has \[ \|m_j\theta\|\geq \frac{c}{M\log M}\quad(j\geq1). \tag{2} \] The normalization of $c$ absorbs the logarithm base. Their local-lemma proof is an external dependency. We give the additional reduction which ensures irrationality, rather than assuming that every parameter supplied by $2$ is irrational.Add a sparse sequence detecting every rational.
Let $S$ be the union of the given sequence with $\{r!:r\geq2\}$, and enumerate it increasingly without repetition as $m_1,m_2,\ldots$. In any interval $[x,2x]$, the original sequence contributes at most \[ \left\lfloor\frac{\log2}{\log(1+\epsilon)}\right\rfloor+1 \] members. The factorial sequence starting at $2!$ contributes at most one, since successive ratios are at least three. Thus, with \[ M=\left\lceil\frac{\log2}{\log(1+\epsilon)}\right\rceil+4, \] the interval $[m_j,2m_j]$ contains fewer than $M+1$ members of $S$. This proves $m_{j+M}>2m_j$. Apply $2$ to this union. It yields $1$, since every original $n_j$ belongs to $S$.If $\theta=a/b$ were rational, every factorial $r!$ with $r\geq b$ would make $r!\theta$ an integer, contradicting $2$. Hence $\theta$ is irrational. The same argument gives a uniform positive gap from zero, so the original distances are not dense even in their natural interval $[0,1/2]$.