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Apr 12, 2023 at 20:45 comment added Karl Fabian @bof: This is exactly what $c(p,a)$ is supposed to achieve :)
Apr 11, 2023 at 5:25 comment added bof Can't you just enumerate the infinite arithmetic progressions, construct a very rapidly increasing sequence of numbers that meets all of them, and take the complement?
Dec 11, 2022 at 19:09 comment added Karl Fabian @Bill Bradley: $a>1$ implies $\pi_a$ odd.
Dec 11, 2022 at 13:29 comment added Bill Bradley Per the statement $|c(p_1,a_1)-c(p_2,a_2)|>1$: what if we have a Mersenne prime and the corresponding adjacent power of two? (Perhaps we are only considering odd $\pi_a$?)
Nov 11, 2022 at 12:10 history answered Karl Fabian CC BY-SA 4.0