退出

Partitioning the positive integers into finite sets with sums in geometric progression

数论 Math StackExchange 1 票 1 回答 76 浏览 提问者: BomingY 2026-06-28 16:21
sequences-and-series combinatorics number-theory

问题内容

Here is a quite interesting problem I've come up with:

Let $\mathbb{N}^+ = \{1, 2, 3, \dots\}$.

Does there exist a sequence of sets $A_1, A_2, \dots$ such that:

$1$. $A_k \subset \mathbb{N}^+$ is nonempty and finite.

$2$. $A_i \cap A_j = \varnothing$ for $i \neq j$.

$3$. $\bigcup_{k=1}^{\infty} A_k = \mathbb{N}^+$.

$4$. The sums of the elements in each group form a geometric progression, i.e. $S_k = \sum_{x \in A_k} x = a r^{k-1}$ for some $a, r > 0$.

I'm curious whether such a partition exists. I would appreciate a proof or any ideas.

回答 (1)

French Man 4 票 已采纳 2026-06-28 17:32 原文

Yes, there's a simple construction. Let $$A_k=\left\{n\in\mathbb N^+:\frac{3^{k-1}+1}{2}\le n\le \frac{3^k-1}{2}\right\}$$ Because $A_k$ has $3^{k-1}$ terms and the average term is $3^{k-1}$ the sum of each $A_k$ is simply $9^{k-1}$