Skip to main content
deleted 35 characters in body
Source Link
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

Not a complete answer: these sequences are called Beatty sequences. It's a nice exercise to show that $S(\alpha)$ and $S(\beta)$ partition the positive integers when $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, so the problem reduces to determining whenhence for $S(\alpha) \subseteq S(\beta)$. Certainly$\alpha, \beta$ satisfying this condition we can take $\alpha = n \beta$, and plausibly this is all we can do$S(n \alpha), S(m \beta)$ where $n, m \in \mathbb{N}$.

Not a complete answer: these sequences are called Beatty sequences. It's a nice exercise to show that $S(\alpha)$ and $S(\beta)$ partition the positive integers when $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, so the problem reduces to determining when $S(\alpha) \subseteq S(\beta)$. Certainly we can take $\alpha = n \beta$, and plausibly this is all we can do.

Not a complete answer: these sequences are called Beatty sequences. It's a nice exercise to show that $S(\alpha)$ and $S(\beta)$ partition the positive integers when $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, hence for $\alpha, \beta$ satisfying this condition we can take $S(n \alpha), S(m \beta)$ where $n, m \in \mathbb{N}$.

Source Link
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

Not a complete answer: these sequences are called Beatty sequences. It's a nice exercise to show that $S(\alpha)$ and $S(\beta)$ partition the positive integers when $\frac{1}{\alpha} + \frac{1}{\beta} = 1$, so the problem reduces to determining when $S(\alpha) \subseteq S(\beta)$. Certainly we can take $\alpha = n \beta$, and plausibly this is all we can do.