1996 Putnam B1

Define a **selfish** set to be a set which has its own

cardinality (number of elements) as an element. Find, with proof, the

number of subsets of $\{1, 2, \ldots, n\}$ which are *minimal*

selfish sets, that is, selfish sets none of whose proper subsets is selfish.

将**自私**集合定义为具有自己的集合

作为元素的基数(元素数量)。有证据地找出

$\{1, 2, \ldots, n\}$ 的*最小*子集数量

自私集，即其真子集都不自私的自私集。