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I know that exhaustive search was done to test numbers up to 2^68.

This seems like a big number but when looking at Collatz function as a Turing machine manipulating some input bit sequence, only sequences up to 68 bits were tested. Maybe something interesting happens with numbers that have ~1000000 bits (that is in range ~2^1000000)?

I couldn't find any project searching for Collatz counter-examples this way, do you know about any?

Also, when checking random numbers, are there any results that prove that only some numbers can become such counter-examples? I found somewhere that it is sufficient to check numbers x that (x mod 6) = 2 but I don't know if it was rigorously proven.

PanJanek
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    @MorganRodgers : yes, that is quite simple in languages like python, java or c# - you cannot use typical "integer" variables, because they use CPU registers limited to 64bits, but there are libraries for manipulating arbitrary big integers. 1000000bits is only about 125Kb of RAM. Much bigger numbers could be tested by modern computers. – PanJanek Aug 15 '21 at 10:58
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    It's like looking for a needle in a haystack. – DaBler Aug 15 '21 at 11:39
  • @MorganRodgers Yes, I know how many 1000000bit numbers there are (to the extent human brain can comprehend it :). I'm not going to try them all, just some random sample. I've read that if the cycle exists it's at least bilion steps long, so.... maybe somwhere, in range of this huge numbers, very long cycle exists, hitting number being part of this cycle could be easier then. – PanJanek Aug 15 '21 at 13:04
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    To demonstrate how hopeless this project is : $(1)$ The density of possible counterexamples is known to be extremely small, for sufficient large $x$ , the number of counterexamples below $x$ is bounded by $x^{0.16}$. I do not know what "sufficiently large" means in this context, but I am convinced your numbers will be "sufficiently large". – Peter Aug 15 '21 at 13:18
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    $(2)$ If you hit a number leading to a diverging sequence, you cannot show this with the computation. You would have to prove that the number is actually a counterexample. The only hope is therefore to hit a counterexample (if it exists) leading to a nontrivial cycle which will be extremely long (since we know that a nontrivial cycle must be extremely long). This is more hopeless than to try to factor an RSA number with $200$ digits by checking random primes whether they divide it. – Peter Aug 15 '21 at 13:21
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    A bit more promising would be to check numbers slightly above the claimed verified range. If there is a counterexample, you should try a range with a reasonable chance to hit one. Why should there be no counterexample upto $2^{100}$ ? This would still be like searching a needle in a haystack, but at least in a much smaller haystack. – Peter Aug 15 '21 at 13:31
  • @DaBler: Concerning the needle and the haystack: that would be a very small needle or a very large haystack... , I think. I assume (without calculating it) something like "an atom in the sun" or so? (P.S.: https://link.springer.com/article/10.1007/s11009-006-7287-0) – Gottfried Helms Aug 16 '21 at 09:53
  • (....) Due to wolfram alpha: a sewing needle has volume of $80 mm^3 = 8 \cdot 10^{-8} m^3$, a heystack of about $3 m^3$ allows thus $37´500´000 $ possible locations for a needle to hide. $\$ On the other hand, one number between all numbers of $10000$ digits means analoguously: one needle in about $2.66... E9992$ heystacks (remember, to compare, the visible universe has estimated about $1 E80$ atoms and think how many universes would be required...) – Gottfried Helms Aug 16 '21 at 10:09
  • About "needle in the haystack": you assume that there is only one counter-example number in the entire range of possible random numbers with 1000000bits. We don't know that. There could be billions+ of billions-length cycles – PanJanek Aug 16 '21 at 11:19
  • @PanJanek - of course! With an odd counterexample $a_1$ another counterexample is $a_2 = 3a_1+1$ and $a_3= a_1+ 1/2(a_1+1)$ ... and many more... – Gottfried Helms Aug 16 '21 at 11:34
  • @Peter I can't find the result you're referencing. Where is that bound from? – Luke Miles Feb 14 '23 at 02:02

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