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As an example, it is certainly trivial to verify that Fermat's Last Theorem is true for all numbers up to 10^10 using a reasonably powerful computer. We now have mathematical proof of that theorem for all numbers, but are there other examples showing the opposite? E.g. a certain theorem is known to be true for all numbers up to 10^20, but there is also a known counter example at a much larger value.

Obviously it's easy to make up such a theorem for the purpose of the question, so I would like to limit the scope to theorems that were actually researched by the mathematical community.

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marked as duplicate by M. Winter, Dietrich Burde, Community Oct 7 '17 at 11:12

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    $\begingroup$ One famous example is the Chebyshev's bias. See also this paper. $\endgroup$ – lhf Oct 7 '17 at 11:03
  • $\begingroup$ Related to math.stackexchange.com/questions/194879/…. $\endgroup$ – lhf Oct 7 '17 at 11:04
  • $\begingroup$ The Goodstein-sequences are even more mind-boggling! Even with start value $4$, the sequence grows so fast, that brute force calculation would never show that the sequence eventually reaches $0$. But we know that every such sequence terminates ending with $0$. Surprisingly, the peano axioms cannot prove that every Goodstein-sequence eventually reaches $0$, but in $ZFC$, there is a relatively easy proof. $\endgroup$ – Peter Oct 7 '17 at 17:24
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Consider the difference between $\pi(x)$ and $\operatorname{li}(x)$, whose sign changes infinitely many times, as was proved by Littlewood.

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  • $\begingroup$ Also called "the story of Skewes' number", see this question. $\endgroup$ – Dietrich Burde Oct 7 '17 at 11:12
  • $\begingroup$ @DietrichBurde Thank you. I was not aware. $\endgroup$ – José Carlos Santos Oct 7 '17 at 17:09
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    $\begingroup$ Although the bound has been drastically reduced, it is still out of reach for brute force calculation! (+1) $\endgroup$ – Peter Oct 7 '17 at 17:15

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