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

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

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    $\begingroup$ Where does the extra $x$ come from in $x \to 3x + x$? $\endgroup$
    – Toby Mak
    Commented Aug 28, 2020 at 10:20
  • $\begingroup$ its a sequence where $n=x$ $\endgroup$
    – EMN
    Commented Aug 28, 2020 at 10:22
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Commented Aug 28, 2020 at 10:37
  • $\begingroup$ 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 $x=35$. $\endgroup$
    – lulu
    Commented Aug 28, 2020 at 10:41
  • $\begingroup$ Typo: in my prior comment, I meant to refer to $x=17$, with starting point $35$. $\endgroup$
    – lulu
    Commented Aug 28, 2020 at 10:46

1 Answer 1

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The answer to your question is unknown.

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 $65$

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