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I noticed something about the Collatz Conjecture, (I was literally obsessed with trying to prove it). I of course have NO intention of trying to prove it, clearly it is beyond my reach and I hope not to offend anyone by what may be a nonsensical observation, but I was a bit curious.

This is what I've noticed or "my conjecture". Pick any $\textbf{odd number}$ n. Then $4n+1$ takes exactly $+2$ more steps than $n$.

I also came across this $4n+1$ rule a lot of times in other observations, I'm not sure if this is any important or completely nonsensical. If you expand the set out you get:

$25, 101, 405, 1612$
$23, 93, 373, 1413$
$17, 69, 277, 1109$
$11, 45, 181, 725$
$9, 29, 117, 469, 1877$
$7, 29, 117, 469, 1877$
$3, 13, 53, 213$
$1, 5, 21, 85, 341$

I checked this for all values up to 2000001, so nothing concrete at all, but just a bit curious at any explanations for this pattern and (potentially) an explanation at a function to find the number of steps it takes for a odd number to reach 1?

The problem is that the number of steps it takes for the left-most number seems random...., thus making it impossible to determine the other number's behaviour without knowing the "basis".

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  • $\begingroup$ What definition of "one step" are you using? At least under the more "normal" definitions I know of, $5$ takes either $4$ or $5$ steps whereas $1$ takes no steps. $\endgroup$ Dec 7 '15 at 3:22
  • $\begingroup$ 2 takes 1 step, 5 takes 5 steps, 16 takes 7 steps... etc. $\endgroup$
    – q.Then
    Dec 7 '15 at 3:24
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    $\begingroup$ $3(4n+1)+1=12n+4=2\times2\times(3n+1)$ would help. $\endgroup$
    – Element118
    Dec 7 '15 at 3:24
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    $\begingroup$ You should definitely try to prove the CC, just don't be too sad when you fail. Trying to prove famous conjectures can be very educational. You will be walking in the footsteps of others, but the view can still be spectacular. $\endgroup$ Dec 7 '15 at 10:30
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If $n$ is odd, we have $4n+1 \to 12n+4 \to 6n+2 \to 3n+1$, while $n \to 3n+1$ so the streams join with $4n+1$ taking two extra steps.

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  • $\begingroup$ Oh, I failed to notice that $\endgroup$
    – q.Then
    Dec 7 '15 at 3:25
  • $\begingroup$ Sorry, but I'm having a bit of trouble for why it's the same for 4n+2 if n is even? $\endgroup$
    – q.Then
    Dec 7 '15 at 3:37
  • $\begingroup$ You specified that $n$ is odd. I used that in the $n \to 3n+1$ step, but nowhere else. $\endgroup$ Dec 7 '15 at 3:38
  • $\begingroup$ I meant the rule also works for 4n+2 if n is even. As in if n is even, 4n+2 takes 2 more steps. I think its a simple derivation like yours, but I can't seem to get it $\endgroup$
    – q.Then
    Dec 7 '15 at 3:38
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    $\begingroup$ This is wrong. $4 \to 2 \to 1, 18 \to 9 \to 28 \to 14 \to 7 $ is longer than four steps. The point is that $n$ even does not constrain how many divide by $2$'s you do, while $4n+2$ says you will only do one. $\endgroup$ Dec 7 '15 at 3:46

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