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Apologies on the length of this question. I found it surprisingly difficult to take even this baby step with the Collatz Conjecture.

If you see any step that is unclear, please let me know in a comment and I will update.

Let:

  • gcd$(a,b)$ be the greatest common divisor of $a$ and $b$

  • $C(x) = \dfrac{3x+1}{2^w}$ where $w$ is the highest power of $2$ that divides $3x+1$

  • $x_1>1, x_2>1, \dots, x_n>1$ be the sequence of $n$ distinct odd integers for each application of $C(x_i)$ so that:

  • for $i > 1$, $x_i = C(x_{i-1})$
  • $x_i > 1$
  • $C_1(x) = C(x)$
  • $C_n(x) = C(C_{n-1}(x))$
  • for each $x_i$, there exists $w_{x_i,1}\ge 1, w_{x_i,2} \ge 1, \dots w_{x_i,n} \ge 1$ such that:

$$C_n(x_i) = \dfrac{3^n x_i + 3^{n-1} + \sum\limits_{i=1}^{n-1}\left(3^{n-i-i}2^{\left(\sum\limits_{k=1}^i w_{x_i,k}\right)}\right)}{2^{\left(\sum\limits_{k=1}^n w_{x_i,k}\right)}}$$

Note 1: Details for this equation can be found here.

  • $m \ge n$ be an integer with $m = \sum\limits_{k=1}^{n}w_{x_1,k}$

  • Let integers $c_1 > 0, c_2 > 0, \dots, c_n > 0$ form an n-cycle so that each $c_{i+n} = c_i$

  • $\text{avg}(c_1, c_2, \dots, c_n) = \dfrac{\sum\limits_{k=1}^n c_k}{n}$

Observation:

  • Given an n-cycle: $c_1, c_2, \dots, c_n$, there exists $1 \le k \le n$ such that for all $k \le j < k+n$:

$$\sum\limits_{i=k}^{j} c_i \le \text{avg}(c_1, c_2, \dots, c_n)$$

Argument

  • Base Case: $n=2$: either $c_1 \le \text{avg}(c_1, c_2)$ or $c_2 \le \text{avg}(c_1,c_2)$
  • Assume that $k$ exists for any $n$-cycle up to $n \ge 2$
  • Inductive Case:
  • Let $d_1, d_2, \dots, d_n, d_{n+1}$ be an $(n+1)$-cycle with $d_{n+1+i} = d_i$
  • There exists $1 \le m \le n$ with $d_m < \text{avg}(d_1, d_2, \dots, d_n, d_{n+1})$. Otherwise, all values are equal to $\text{avg}(d_1, d_2, \dots, d_{n+1})$ and any $1 \le i \le n$ will serve as $k$.
  • Let $c_1, c_2, \dots, c_n$ be an $n$-cycle such that: $$c_i = \begin{cases} d_i, & i < m\\ d_{i+1}-\text{avg}(d_1,\dots,d_{n+1}) + d_i, & i = m\\ d_{i+1}, & i > m\\ \end{cases}$$
  • Since $c_1, c_2, \dots, c_n$ forms an $n$-cycle, there exists $1 \le k \le n$ such that for all $k \le j < k+n$:

$$\sum\limits_{i=k}^{j} c_i \le \text{avg}(c_1, c_2, \dots, c_n) = \text{avg}(d_1, d_2, \dots, d_n, d_{n+1})$$

  • Case 1: $k = m$ $$d_k = d_m < \text{avg}(d_1, d_2, \dots, d_{n+1})$$
  • Case 2: $1 \le j \le n$ and $k+j < m$
  • By assumption: $$\sum\limits_{i=k}^{k+j} c_i = \sum\limits_{i=k}^j d_i \le \text{avg}(c_1, c_2, \dots, c_n) = \text{avg}(d_1, d_2, \dots, d_n, d_{n+1})$$
  • Case 3: $1 \le j \le n$ and $k+j = m$ $$\sum\limits_{i=k}^{k+j} d_i = \left(\sum\limits_{i=k}^{k+j-1} d_i\right) + d_m \le \text{avg}(d_1, d_2, \dots, d_n, d_{n+1})$$
  • Case 4: $1 \le j \le n$ and $k+j > m$ $$\sum\limits_{i=k}^{k+j} d_i = \left(\sum\limits_{t=k}^{k+j-1} c_t\right) - \text{avg}(d_1,d_2,\dots,d_{n+1}) + d_{m} \le \text{avg}(d_1, d_2, \dots, d_{n+1})$$

Question:

Does it now follow that if $x_1, x_2, \dots, x_n$ form an n-cycle, then either $2^{m-1} < 3^n$ or there exists $x_i$ where $x_i < n$

If yes, is there a simpler or more straight forward way to make the same argument?

Argument:

(1) Assume that $x_1, x_2, \dots, x_n$ forms an n-cycle.

(2) For each $x_i$, it follows that:

$$x_i = C_n(x_i) = \dfrac{3^n x_i + 3^{n-1} + \sum\limits_{i=1}^{n-1}\left(3^{n-i-i}2^{\left(\sum\limits_{k=1}^i w_{x_i,k}\right)}\right)}{2^{\left(\sum\limits_{k=1}^n w_{x_i,k}\right)}}$$

Which implies that:

$$x_i\left(2^{m}-3^n\right) = 3^{n-1} + \sum\limits_{i=1}^{n-1}\left(3^{n-i-i}2^{\left(\sum\limits_{k=1}^i w_{x_i,k}\right)}\right)$$

(3) $2^m > 3^n$

This follows since $2^m - 3^n = \dfrac{3^{n-1} + \sum\limits_{i=1}^{n-1}\left(3^{n-i-i}2^{\left(\sum\limits_{k=1}^i w_{x_i,k}\right)}\right)}{x_i}$

Since, clearly: $\dfrac{3^{n-1} + \sum\limits_{i=1}^{n-1}\left(3^{n-i-i}2^{\left(\sum\limits_{k=1}^i w_{x_i,k}\right)}\right)}{x_i} > 0$

(4) Assume that $2^{m-1} > 3^n$

(5) $2^m - 3^n > 2^m - 2^{m-1} = 2^{m-1}$

(6) The average of each $w_{x_i,k}$ is $\dfrac{m}{n}$ with $2^{\frac{m}{n}} > 3$ since:

  • $m \ln 2 > n \ln 3$
  • $\frac{m}{n}\ln 2 > \ln 3$
  • $2^{\frac{m}{n}} > 3$

(7) Since $x_1, x_2, \dots, x_n$ forms an $n$-cycle, from the Observation above, there exists an $x_i$ such that for each $1 \le u \le n$, $\left(\sum\limits_{k=1}^{u} w_{x_i,k}\right) \le \dfrac{um}{n}$

Note: The argument in the observation is derived from the solution to the well known gas stations on a circular trek problem.

(8) $2^{m-1}n > 3^{n-1} + \sum\limits_{i=1}^{n-1}\left(3^{n-i-i}2^{\left(\sum\limits_{k=1}^i w_{x_i,k}\right)}\right)$ since:

  • $2^{m-1} > 3^{n-1}$ from step(3) above
  • $2^{m-1} \ge 2^{\left(\sum\limits_{k=1}^{n-1} w_{x_i,k}\right)}$
  • $2^{m-1} > 2^{(n-1)\frac{m}{n}} > 3\times2^{\left(\sum\limits_{k=1}^{n-2} w_{x_i,k}\right)}$ since: $\dfrac{m}{n} > 1$ from $2^{\frac{m}{n}} > 3$ and $\frac{m}{n} + (n-1)\frac{m}{n} = m < m-1 + \frac{m}{n}$
  • $2^{(n-1)\frac{m}{n}} \ge 3^2\times2^{\left(\sum\limits_{k=1}^{n-3} w_{x_i,k}\right)}$
  • $\dots$
  • $2^{(n-1)\frac{m}{n}} \ge 3^{n-2}\times2^{w_{x_i,1}}$

(9) $x_i < \dfrac{2^{m-1}n}{(2^m - 3^n)} < \dfrac{2^{m-1}n}{2^{m-1}} = n$

Edit 1:

I found a mistake in my reasoning which led me to change the title slightly and change the question in order to fix the mistake in reasoning.

Edit 2:

I made changes based on comments from John Omielan.

Larry Freeman
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  • I haven't finished reading all of your question yet, but note your "$C_n(x) = C_1(C_2(\dots(C_n(x))\dots))$" is self-referential, and unclear, as stated. For $n=1$, we get $C_1(x) = C_1(x)$. For $n = 2$, we get $C_2(x) = C_1(C_2(x))$, i.e., $C_2(x)$ must be a value so $C_1$ of it returns that value. However, $C_1$ is not explicitly defined. Based on your previous question & the context, I believe you mean $C_n(x)$ is the composition of $C(x)$ $n$ times. One explicit, unambiguous way to state this is "$C_1(x)=C(x)$" and "$C_n(x)=C(C_{n-1}(x)) ; \forall ; n \gt 1$". – John Omielan Sep 27 '20 at 00:43
  • Your "Note 2" of "Each set of $w_{x_i,1}, w_{x_i,2}, \dots w_{x_i,n}$ will consist of the same set of values with a shifted order ...", plus the later note of "$x_1$ was used but this sum will be the same for all $x_i$" will only necessarily be true if you have an $n$-cycle starting at $x_1$. You don't state this previously as being a requirement, with your question later asking about if we have $n$-cycle. It seems you're implicitly assuming an $n$-cycle initially, so if this is your intent, you should state this then & reword the rest of the question accordingly. – John Omielan Sep 27 '20 at 01:09
  • You're welcome. Here's one more issue. In your (7) list of bullet points, there's "Since $v_{x_1,1}, \dots, v_{x_1,n}$ is a sequence of $n$ values, by assumption, there exists $j$ where $v_{x_j,1}, v_{x_j,2}, \dots v_{x_j,n}$ have the properties and $\dfrac{\sum\limits_{k=1}^n v_{x_j,k}}{n} = \text{avg}$". However, the assumption is only valid for a set of $n$ positive integers adding to $m$. In your case, this requires $\text{avg} = w_{x_j,l}-w_{x_j,n+1}$, where $\text{avg}$ is as defined earlier, which will usually be false. Also, you don't define it, but I assume $w_{x_j,n+1}=w_{x_j,1}$. – John Omielan Sep 27 '20 at 02:03
  • @JohnOmielan Great points. I will attempt to revise the argument to address your comments. – Larry Freeman Sep 27 '20 at 02:49

1 Answers1

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from $(3+\frac{1}{e_0})(3+\frac{1}{e_1})...(3+\frac{1}{e_n})=\frac{e_{n+1}}{e_0}\prod_{k=0}^n2^{\nu_2(3e_k+1)}$ you can see that for a cycle: $$2^m\leq (3+\frac{1}{x_{min}})^n$$

If you state that $2\cdot3^n<2^m$ than you have

$$2\cdot3^n< (3+\frac{1}{x_{min}})^n$$ $$2^\frac{1}{n}\cdot3<3+\frac{1}{x_{min}}$$ $$x_{min}<\frac{1}{3(2^\frac{1}{n}-1)}<n$$

Collag3n
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