1
$\begingroup$

Inevitably for any amateur mathematician, I've been playing with the Collatz Conjecture. I have found it's easier to examine and to generate theorems if we use this equivalent statement:

$$c_{n+1}\equiv\begin{cases} \begin{array}{c} \frac{c_{n}}{2};\;c_{n}\in \; Even\\ \frac{3c_{n}+1}{2};\;c_{n}\in \; Odd \end{array}\end{cases}$$

and

$$\forall \{c_{0}>1\}\;\exists\; z\;\text { s.t.}\; c_{z}<c_{0}$$

In other words, I only prove that the series drops below its initial value. This is true iff the Collatz Conjecture is true.

Let's further define: $L(c_{0})$ as the number of steps this takes. For example $L(11)=5$ because $c_{n<5}\geqq11$ and $c_{5}<11$ $$11\rightarrow17\rightarrow26\rightarrow13\rightarrow20\rightarrow10$$

So here's my problem: When I binned up values of $L(c_{0})$ for lots of starting values, I found that there seem to be no values of $c_{0}$ where $L(c_{0})\epsilon\{3,6,9,11,14,17,19,22,25,28,...\}$. The members of this series have the form $1+\Big\lfloor\frac{n}{1-\frac{ln2}{ln3}}\Big\rfloor$. I googled the series and found the link below which says “These numbers appear in connection with the 3x+1 problem.”

https://oeis.org/A054414

But I haven't been able to prove this theorem or find any discussion of connection online.

So, how would we prove this theorem, that there exist no $c_{0}$ s.t. $L(c_{0})=1+\Big\lfloor\frac{n}{1-\frac{ln2}{ln3}}\Big\rfloor$ for integer n?

It's easy enough to prove for small values. For example, a Mathematica script will quickly verify that for values of the form $8m+{0,1,2,3,4,5,6,7}$, $L(c_{0})$ is indeed either <3 or >3, never =3. But done this way, the proof scales up in powers of 2. Is it possible to prove it in general?

$\endgroup$
  • 1
    $\begingroup$ OEIS has more info: the linked entry says "Except for 1, A054414 is the complement of A020914." That is, every integer (other than 1) appears in either A054414 or A020914 exactly once. Moreover, the OEIS entry for A020914 states: "This sequence represents allowable values of the "dropping time" in the Collatz (3x+1) problem when iterated according to the function f(n) := n/2 if n is even, (3n+1)/2 otherwise, as tabulated in A126241. There is one exception which is A126241(1), which has been set to zero by convention." So that may clarify the Collatz connection. $\endgroup$ – Semiclassical Mar 31 at 20:56
  • $\begingroup$ @Semiclassical Well, at least it confirms that what I "discovered" is already a known result. :-) And apparently the L() I defined above is called the "dropping time" with quotes. Thanks for the info. $\endgroup$ – Jerry Guern Mar 31 at 21:24
  • $\begingroup$ Your proposed theorem is not equivalent to the Collatz conjecture, because it fails for $c_0=1$. $\endgroup$ – Servaes Mar 31 at 21:45
0
$\begingroup$

Mike Winkler on arXiv "Mike Winkler - The algorithmic structure of the finite stopping time behavior of the 3x + 1 function" (arxiv.org/abs/1709.03385) has a lot of material, I think he has also a proof for the floor()-formula that you refer to.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.