标题： 将集合$[n] = \{1, \dots, n\}$分成大小最多为$m$的子集，使得所有和都是$m$的幂 显示英文标题

标题： Partitioning set $[n] = \{1, \dots, n\}$ into subsets of size at most $m$ such that all sums are powers of $m$

摘要： 给定整数$n > 0$和$m > 1$，我们称集合$[n] = \{1, \dots, n\}$ {\em $m$ -好} 的一个划分是满足每个划分集合的大小至多为$m$且其中的数之和是一个$m$的幂，即$m^t$对于某个$t \geq 0$。 很容易看出，对于每个$n$都存在唯一的 2-good 分区，相反，对于每个固定的$m>3$，对于无限多个$n$，不存在$m$-good 分区。 Case$m=3$更加困难。 我们猜想对于每个$n$都存在 3-good 分区，并证明如果存在最小的反例，它至少为 101 并且必须属于集合 {\centering $N_o = \{n \equiv 2 \pmod 3\} \cap \{3^t+1 < n < (3^{t+1}+1)/2 \mid t \geq 4\}.$ } 对于这种情况，我们提供了一些部分结果。 我们还证明，如果$n \in N_u = \{1, 2, 3, 4, 3^t-4, 3^t-2, 3^t-1, 3^t, 3^t+1, 3^t+2, 3^t+3 \mid t > 1\}$，则 3-good 分区是唯一的，并且猜想其逆命题也成立。 显示英文摘要

摘要： Given integer $n > 0$ and $m > 1$, we call a partition of set $[n] = \{1, \dots, n\}$ {\em $m$-good} if each of the partitioning sets is of size at most $m$ and the sum of numbers in it is a power of $m$, that is, $m^t$ for some $t \geq 0$. It is easily seen that a unique 2-good partition exists for each $n$ and, in contrast, for each fixed $m>3$, for infinitely many $n$, no $m$-good partition exists. Case $m=3$ is more difficult. We conjecture that 3-good partitions exist for each $n$ and prove that a minimal counter-example, if any, is at least 101 and must belong to the set {\centering $N_o = \{n \equiv 2 \pmod 3\} \cap \{3^t+1 < n < (3^{t+1}+1)/2 \mid t \geq 4\}.$ } For this case we provide some partial results. We also show that a 3-good partition is unique if $n \in N_u = \{1, 2, 3, 4, 3^t-4, 3^t-2, 3^t-1, 3^t, 3^t+1, 3^t+2, 3^t+3 \mid t > 1\}$ and conjecture that the inverse holds too.