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Hi, Thomas,
I have not given these problems a lot of thought. If there is a system of congruences that uses a finite set of primes P to show that k*2^n + 1 is always divisible by a prime in P (how coverings work for this problem), then it is true that k + 2^n will also always be divisible by a prime in P (and vice-versa). The dual Sierpinski problem (and similarly for the dual Riesel problem) boils down then to whether one can find such a system of congruences for which k*2^n + 1 is not in P for all n but k + 2^n is in P for some n (or vice-versa). It is reasonable to me to think such a system of congruences exist, but with too many other interesting mathematics I am working on, I don’t see spending time on trying to come up with such a system of congruences. I might mention this to a student I know who could be interested in the question though.
Kind regards,
Michael