# Will the Collatz conjecture work for $m \cdot (n)+1$ for an odd number, where m is any odd number?

The Collatz conjecture asks you to:

When '$n$' is the given number,

1) Divide $n$ by $2$ if the number is even.

2) Do $3n+1$ when the number is odd, and you will reach the series $4->2->1$.

Can we generalize this as:

1) Divide $n$ by $2$ if the number is even.

2) Do $m \cdot (n)+1$ when the number is odd...($m$ is any odd number)

Will this always work? If not, then why? The only possible implications I can deduce from this are that the value of the peak number in the resulting series of any number will vary drastically.

• Since when is this a theorem? Did somebody find a proof while I wasn't looking? – Henning Makholm Jun 26 '16 at 14:55
• The $5n+1$ problem is already false ($13$ is periodic). – lulu Jun 26 '16 at 15:01
• @lulu: beat me by 5 seconds. – barak manos Jun 26 '16 at 15:01
• @HenningMakholm I'll change that in a bit. Sorry. – Saunved Mutalik Jun 26 '16 at 16:25
• Please explain, what $m(n)$ in the above means. As far as I read q&a on the collatz-problem, there was always the base line to look at $m \cdot n+1$ with some constant $m$ like $3$ or $5$. Now you let it be an arbitrary function $m(n)$, which moreover is dependend on the current value of $n$ - so how could this make sense in a "Collatz-discussion"? – Gottfried Helms Jun 26 '16 at 23:49

As said in the comments, the $5n+1$ problem is false since $13$ is periodic. This leads to the question of what is a suitable generalization of the Collatz conjecture? Zhang Zhongfu and Yang Shiming suggest the following:

First define the mapping $T_n : \mathbb{N} \to \mathbb{N}$

$$T_n(x) = \begin{cases} \dfrac{x}{p_{i_1} \dots p_{i_k}} & p_{i_j} \text{ divides } x, i_j \leq n\\ p_{n+1} x + 1 & \text{no prime p_i \leq p_n divides x} \end{cases}$$

The conjecture is that for any $p_{n+1}$ and $x_0$, repeated iteration of $T_n(x)$ on $x_0$ will eventually yield $1$ or enter one of finitely many trivial cycles.

Note that a solution to this generalization will not prove the classical $n=1$ case because we allow non-trivial cycles.

• Is there any progress on the Collatz conjecture ? – user230452 Jun 26 '16 at 15:29

The numbers $m$ of the forms $m=2^k-1$ and $m=2^k+1$ play a special role, because they allow 1-step (or trivial) cycles on $a=1$ or $a=-1$. The example $m=5$ has additional cycles for odd $a$ in the positive integers, while $m=3$ has additional cycles for odd $a$ in the negative integers.
There is one more case known which has nontrivial cycles in the odd positive numbers, namely $m=181$ (which has something to do with the fact, that $181$ is very near the square-root $\sqrt{128 \cdot 256} =\sqrt{2^7 \cdot 2^8} = 2^{15/2}$) .

But this is all which can be derived with amateur's math. It is nearly nothing known about the existence of further cycles and especially nearly nothing about the existence (and possible properties) of divergent trajectories.