I.e., the number of subsets S of $2^{[n]}$ such that no element of S is a subset of another? How to prove: $$A_1(n)=2^n$$ $$A_2(n)=\frac 12(4^n-2\cdot 3^n)+2^n$$ $$A_3(n)=\frac 16 (8^n-6\cdot 6^n +6\cdot 5^n +3\cdot 4^n - 6\cdot 3^n +2\cdot 2^n )$$ $$A_k(n)=\frac1{k!}\sum_{i=2}^{2^k}a_{k,i}i^n$$

  • $\begingroup$ What thoughts do you have? $\endgroup$ – Peter Taylor May 6 at 10:33

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