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Yesterday afternoon I met professor Halfbrain at an art gallery. The professor looked tired and exhausted. He told me that he had spent many working days and many sleepless nights with lengthy calculations. Here is what the professor did:

  1. The professor picked an arbitrary integer $n\ge2$ as starting number.
  2. If $n$ was even, he replaced it by $n/2$.
    If $n$ was odd, he replaced it by $3n-1$.
  3. If the professor reached the number $1$, then the process terminated.
    Otherwise, the professor iterated this replacement procedure (over and over again).

The professor had tested thousands of possible starting numbers, and for each of his test cases the replacement process eventually led him to the number $1$ and terminated.

Hence the professor formulated the following number-theoretic conjecture:

Professor Halfbrain's conjecture:
For every starting integer $n\ge2$, the Halfbrain replacement procedure must eventually reach the number $1$.

Is the professor's conjecture indeed true, or has the professor once again made one of his well-known mathematical blunders?

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It appears to me this is

not true

Because

By example, start with $14$. First iteration, $14/2 = 7$. Second iteration $7*3-1 = 20$. Third iteration $20/2 = 10$. Fourth iteration $10/2= 5$. Fifth iteration $5*3-1 = 14$.

And

we are back where we started and will continue to loop.

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  • $\begingroup$ I think it works for every number that is divisible by 7 $\endgroup$ – David Feb 19 at 13:54
  • $\begingroup$ @David: 77 reaches 1 in 14 steps: 77, 230, 115, 344, 172, 86, 43, 128, 64, 32, 16, 8, 4, 2, 1. $\endgroup$ – Jaap Scherphuis Feb 19 at 14:41
  • $\begingroup$ @JaapScherphuis whoops, you're right $\endgroup$ – David Feb 19 at 14:43
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It is:

False.

For example:

$17, 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, 272, 136, 68, 34, 17$

This is the Collatz conjecture extended to the negative integers, where it is false.

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  • $\begingroup$ Yes, but the OP explicitly says n ≥ 2, so negative numbers would not count. $\endgroup$ – Darrel Hoffman Feb 20 at 14:26
  • $\begingroup$ When $n\lt0$, $3n+1\to-3n+1$ which is the negative of $3n-1$. $n=17\ge2$ in this case.@DarrelHoffman $\endgroup$ – JMP Feb 20 at 14:28
  • $\begingroup$ Okay, yeah, sorry, I misread it, I realized it shortly after I posted. $\endgroup$ – Darrel Hoffman Feb 20 at 14:31
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As JMP pointed out, this is a modified

Collatz conjecture, which should be 3n+1 when n is odd.

It appears that the professor has simply forgotten to cross the minus symbol for this to work correctly.

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It's false, because if any number smaller than the odd number $n$ leads to 1, then $n\to3n-1\to(3n-1)/2\to(9n-5)/2\dots$.
For example, if $(9n-5)/2=4n$, then $n=5$ leads to a loop.

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