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If there is a divergent integer in the collatz conjecture then there must be a smallest divergent number by the WOP. We can observe some properties of this number such as it must be odd because if it was even then the next number in the series $n/2$ is smaller which is a contradiction. Are there any other known facts about the smallest number?

shai horowitz
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    If $n\equiv 1\pmod 4$ then $3n+1\equiv 0\pmod 4$, so the smallest $n$ must have $n\equiv 3\pmod 4$. If $n\equiv 3\pmod {16}$ then $(3n+1)/2\equiv 5\pmod 8$ and thus $3(3n+1)/2+1\equiv 0\pmod 8$ so $[3(3n+1)/2+1]/8=(9n+5)/16$ is smaller than $n$ for such $n$ – Thomas Andrews Jul 01 '16 at 22:38
  • So we have n = 3 mod 4, and n =! 3 mod 16. – shai horowitz Jul 02 '16 at 05:20
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    Yes, you can continue to higher moduli of powers of $2$. For example, $n\not\equiv 7\pmod {32}$, because $n_1=(3n+1)/2\equiv 1\pmod {16}$ and $n_2=(3n_1+1)/4\equiv 1\pmod{4}$ and $n_3=(3n_2+1)/4<n$. This technique lets you eliminate some numbers, but it is very hard to see how it works in the long run to eliminate all of them. – Thomas Andrews Jul 02 '16 at 07:16
  • @GottfriedHelms Thats not particularly helpful it just sets a min lower bound to N – shai horowitz Jul 02 '16 at 11:30
  • Shai - oh, sorry. I posted this because it is a generalization of the "3 (mod 4)" observation of Thomas Andrews. But I can delete the comment if it is not helpful/not wanted. – Gottfried Helms Jul 02 '16 at 12:50
  • @GottfriedHelms Do you know if 15 mod 16 takes longer relatively then other modular residue? – shai horowitz Jul 02 '16 at 18:37
  • We know that Mersenne numbers go large and also powers of 3. Is it possible that some power of 3 is also a Mersenne number? – Robert Frost Jul 12 '16 at 07:26
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    @shaihorowitz We know that it is greater than some enormous number like $10^{50}$. We know that it is less than every $f^n$ of itself, where $f$ is the collatz function. And we know it's less than every $f^-n$ of itself. We know that if you append $01$ to its binary string you have a number that leads to it. We know that it's not of the form $2$ $(mod 3)$ otherwise a smaller number would lead to it. – Robert Frost Jul 14 '16 at 15:22
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    See my current answer at https://math.stackexchange.com/a/4171527/1714 for some more details. – Gottfried Helms Jun 13 '21 at 08:02

2 Answers2

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Such a number, call it $k$, cannot be congruent to $2 \;(\bmod\; 3)$, because otherwise it is possible to create a new counterexample.

For example, consider $\frac{2k-1}{3}$.

$$\frac{2k-1}{3}<k$$

For $k > -1$.

As $\frac{2k-1}{3}$ is odd, we iterate to $2k$. Being even, $2k$ goes to $k$. Therefore, not only is $k$ a counterexample, so is $\frac{2k-1}{3}$. As it is lesser, this contradicts, $k$ as the least counter-example.

Therefore, $\frac{2k-1}{3}$ cannot exist. This happens when $k \not\equiv 2 \;(\bmod\; n)$.

This logic passes the "$3n-1$" test, as if we take $7$, a known counter-example, we get $5=\frac{2\times7+1}{3}$, and $5$ is also a known counterexample.

Roskiller
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Such a number must be of the form $4n-1$ since if it was of the form $4n+1$ you would lead to $3n+1$ which is smaller.

Ryan Cole
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