Erdős Problem #1108

declared status 'open'. Formalized: yes. Let$$A = \left\{ \sum_{n\in S}n! : S\subset \mathbb{N}\textrm{ finite}\right\}.$$If $k\geq 2$, then does $A$ contain only finitely many $k$th powers? Does it contain only finitely many powerful numbers? Current best: This was motivated in part by a problem of Mahler which he discussed with Erd\H{o}s a few days before his death in 1988: if $k\geq 5$ and$$A_k= \left\{ \sum_{n\in S}k^n : S\subset \mathbb{N}\textrm{ finite}\right\}$$then does $A_k$ contain only finitely many squares? Mahler showed that there are infinitely many squares in $A_k$ for $k\leq 4$, and found only one square for $k\geq 5$, namely$$1+7+7^2+7^3=400.$$Brindza and Erd\H{o}s [BrEr91] proved that, for any $r$, if $n_1!+\cdots+n_r!$ is powerful then $n_1\ll_r 1$. Prize: no. OEIS: A025494, A051761, A115645. Tags: factorials, number theory.

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