Are there infinitely many finite sets $S$ of primes where $\sum_{p\in S} {1/(p_i-1)}=1$? For example, $S = \{3,5,7,13\}$ gives $\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}=1$.
A few other such sets: $\{2\},\{3,5,7,19,37\},\{3,5,7,29,31,71\}$.
Are there infinitely many of these?
 A: I believe the answer to this is actually yes.
See Erdős–Graham problem.
This problem asked: if you partition all natural numbers larger than 1 into finitely many subsets, must the reciprocals of one of the subsets sum to 1? It was apparently proven in the affirmative in 2003.
By my reasoning, you could put all composite numbers in one subset, which already sums to more than 1 after taking the composites through 16, then divide up the remaining numbers (all primes) into any partitioning you like and be assured that at least one of those subsets will sum to 1. There are clearly infinitely many ways one could do that partitioning so as to obtain infinitely many different valid solution subsets.
Among other things, this would yield infinitely many distinct covering sets for Sierpinski numbers, which would therefore have infinitely many distinct possible coverage periods.

As pointed out below, I read this mistakenly so this isn't actually a solution. I'll leave this here as a potential jumping-off point for a solution, however; if one could assign the composites systematically into subsets in such a way that none of those subsets could contain a reciprocal sum to unity, then this would still work.
