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I'm not mathematician, so forgive me if I make wrong assumptions. I was wondering what the importance of the $3n$ is in the Collatz Conjuncture.

If you just do $n + 1$, it seems you'll end up at $1$ also.

This would seem logical to me, because you'll do $n + 1$ when it's an uneven number. The effect is that now you've made it an even number. Any you'll be able to reduce any even number to $2$ (possible wrong assumption here?). And $\frac{2}{2}$ is $1$.

So instead of:

$\quad$ Even number: $\frac{n}{2}$
$\quad$ Uneven number: $3n + 1$

Couldn't it be:

$\quad$ Even number: $\frac{n}{2}$
$\quad$ Uneven number: $n + 1$

Or has someone else already answered this question? Again, I'm no mathematician, so not really familiar with this.

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up vote 15 down vote accepted

The case of factor $3$ is more interesting because with factor $1$ it's easy to prove that the sequence hits $1$. We can prove this by induction: For $n_0=1,2$ this is clearly true. Suppose that it is true for all starting values $1,2,\dots,n_0-1$. Then if $n_0$ is even, the next number will be $n_0 / 2 < n_0$ and thus the sequence will hit $1$. If $n_0$ is odd, the next number will be $n_0 + 1$ and the number after it will be $\frac{n_0 + 1}{2} < n_0$ and thus the sequence will hit $1$ again by induction.

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+1 Beat me to it by 2 minutes :-) – Jyrki Lahtonen Nov 28 '11 at 10:21

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