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Given a sequence $S$ of natural numbers, write ${\bf Gap}(S)$ for the set of differences between consecutive terms. (So $|{\bf Gap}(S)|=1$ precisely for arithmetic progressions, hence the connection to Erdős–Newman–Mirsky.)

Question 1: Can one partition the natural numbers into finitely many sequences $S_i$ with all the ${\bf Gap}(S_i)$ collectively disjoint and with no ${\bf Gap}(S_i)$ containing 1?

Question 2: If so, can one do it with $|{\bf Gap}(S_i)|<\infty$ for all $i$?

Question 3: If Question 2 comes up negative, can one prove this in something like the usual way, via generating functions and their poles? (I believe that a counter-example plus the pigeon-hole principle would yield a periodic counter-example.)

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  • $\begingroup$ Sinai Robins suggests asking a related question (which I pass along without much consideration) generalizing a related conjecture of Erdős, one only recently settled by Hough: For all $N$, can one cover the natural numbers, not necessarily disjointly, with sequences $S_i$ having disjoint $G(S_i)$ and $\min \cup_i G(S_i)>N$? $\endgroup$
    – David Feldman
    Apr 28, 2015 at 23:14

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I've now found that extending cyclically the pattern 12131214121312413151412131215 gives a partition of the natural numbers into five sets with disjoint gap sets. No examples exist with fewer than five cells in the partition.

So any possible generalization of Erdős–Newman–Mirsky will require some additional hypothesis.

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