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18
votes
Accepted
Why don't Zeilberger and Gosper's algorithms contradict Richardson's theorem?
(Comment turned into an answer:)
It's as simple as "the composition of hypergeometric terms is not hypergeometric". $f(n)=2^n$ is a hypergeometric term because $\frac{f(n+1)}{f(n)}=2$ is a rational te …
10
votes
Undecidable infinite analogs of NP-complete problems?
See, e.g., the discussion at Decidability of chess on an infinite board and the paper at https://arxiv.org/abs/1201.5597 . …