# Modifying the Collatz conjecture so that when $n$ is odd: $3n + x$

Modifying the Collatz conjecture so that when $$n$$ is odd: $$3n + x$$ and where $$x$$ is odd and $$x > 1$$.

For any odd $$x$$, starting the sequence at $$n = x$$ will always lead to a loop:

$$x \to 3x + x = 4x \to 2x \to x$$

But will there always be sequences that lead to infinite loops where $$n \neq x$$ (for every $$x > 1$$)?

Example, $$x=17$$, $$n=27$$:

$$27,98,49,164,82,41,140,70,\underline{35},122,61,200,100,50,25,92,46,23,86,43,146,73,236,118,59,194,97,308,154,77,248,124,62,31,110,55,182,91,290,145,452,226,113,356,178,89,284,142,71,230,115,362,181,560,280,140,70,\underline{35}$$

Example, $$x=11$$, $$n=37$$:

$$37,122,61,194,97,302,151,464,232, \underline{116},58,29,98,49,158,79,248,124,62,31,104,52,26,13,50,25,86,43,140,70,35,\underline{116}$$

Example, $$x=7$$, $$n=27$$:

$$27,88,44,\underline{22},11,40,20,5,\underline{22}$$

• Where does the extra $x$ come from in $x \to 3x + x$? Commented Aug 28, 2020 at 10:20
• its a sequence where $n=x$
– EMN
Commented Aug 28, 2020 at 10:22
• The second example is the same loop as the $n=x$ loop (note that $3$ is in the loop). The first does not contain $17$ though.
– lulu
Commented Aug 28, 2020 at 10:37
• I guess I'd start by testing the first $100$ values of $x$, or more if you can. As in my prior comment, I think you want to add the condition that the new loop does not contain $x$....after all, the analog of the Collatz conjecture for your process would be "show that every starting point eventually reaches the loop generated by $x$", if I have understood the spirit of your question correctly. Your first example shows that this is not true for $x=35$.
– lulu
Commented Aug 28, 2020 at 10:41
• Typo: in my prior comment, I meant to refer to $x=17$, with starting point $35$.
– lulu
Commented Aug 28, 2020 at 10:46

We can replace $$3n+1$$ with $$3n+x$$ (taking $$x>0$$ for simplicity) in the definition of the Collatz function, and we get Collatz-like behavior as long as $$x$$ is odd. When $$x=3y$$, there's no point using it; it will just act like $$3n+y$$, but with everything multiplied by $$3$$. Thus, the good choices for $$x$$ are odd numbers that are not multiples of $$3$$, i.e., numbers relatively prime to $$6$$, i.e., numbers adjacent to multiples of $$6$$, such as $$1,5,7,11,13,17,19,\ldots$$.

It turns out that replacing $$1$$ with $$x$$ is precisely equivalent to applying the original Collatz function to fractions with denominator $$x$$, and only writing down the numerators. Observe:

Using $$3n+1$$:

$$\frac15\to\frac85\to\frac45\to\frac25\to\frac15\to\cdots$$

Using $$3n+5$$:

$$1\to 8\to 4\to 2\to 1\to\cdots$$

When you plug in $$n=x$$, you have simply plugged in the number $$1$$, and you'll get familiar behavior, but with all numbers multiplied by $$x$$. Thus:

Using $$3n+1$$:

$$1\to 4\to 2\to 1\to\cdots$$

Using $$3n+5$$:

$$5\to 20\to 10\to 5\to\cdots$$

This works as a special case of the above, because $$1=\frac55, 4=\frac{20}5, 2=\frac{10}5$$. For this reason, it suffices to plug in values of $$n$$ that have no common factors with $$x$$. If you have $$x=35$$, and you want to know what happens with starting values such as $$n=7, 21, 49, 63$$, simply set $$x=5$$ instead, and consider starting values $$n=1,3,7,9$$.

Anyway, it appears that, for each admissible $$x$$, there exists a set of loops into which all trajectories fall. There is no value of $$x$$ for which anyone has discovered a divergent trajectory, and there is no value of $$x$$ for which anyone has proved that a finite set of loops will catch all inputs.

For some values of $$x$$, there are multiple loops available for trajectories to fall into; for others (such as $$x=1$$), we seem to have only a single loop capturing all trajectories coprime to $$x$$. Examples of the latter, besides $$x=1$$, are $$x=7, 19, 31, 41, 43, 49, 53$$, and $$65$$