# A Brief Analysis of the Collatz Conjecture

Algorithm for the Collatz Conjecture.

In short the Collatz problem is simple enough that anyone can understand it, and yet relates not just to number theory (as described in other answers) but to issues of decidability, chaos, and the foundations of mathematics and of computation. That’s about as good as it gets for a problem even a small child can understand.” Matt.

Remark: If n = 1, the algorithm terminates all computations.

Crucial Idea: R+ is a dense set.

Remark: R+ is the set of all positive real numbers.

We claim

1. $n_{t} = (\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} * r = 1$

over the Collatz sequence of positive odd integers, from $n_{0}$ to $n_{t} = 1$ where the index, t, is the number of trials it takes the Collatz sequence of odd integers to converge to one.

Remark: There are no infinite (nontrivial) cycles of any length ( $(\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} \rightarrow 0$ as $t \rightarrow \infty$) in the Collatz sequence since the index, t, is clearly finite.

In addition, our claim is fundamentally based on the distribution of maximum divisors, $2^{i_{max}}$, for the set of all positive even integers…

The positive real number, r, is determined by the algorithm for the Collatz conjecture. And therefore, its computation is generally complicated.

The variable, k, is determined by the maximum positive even number, $e_{max}$, in the Collatz sequence: $k = \left \lfloor{\frac{log(e_{max })}{log(2)}}\right \rfloor$.

Can we refute our analysis?

We assume maximum divisors, $2^{i}$, for any positive even integer.

For any appropriate positive odd integer, $n_{0} > 2^{68} -1$, we have $n_{2} = \frac{3n_{0} +1}{2^{i_{1}}} = \frac{3}{2^{i_{1}}} * r_{2}$ where $r_{2} = n_{0} +\frac{1}{3}$. $n_{3} = \frac{3(\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}} = \frac{3}{2^{i_{1}}} *\frac{3}{2^{i_{2}}} * r_{3}$

where $r_{3} = \frac{1}{9}(2^{i_{1}} + 9n_{0} +3)$. $n_{4} = \frac{3(\frac{3(\frac{3n_{0} +1}{2^{i_{1}}}) + 1}{2^{i_{2}}}) + 1}{2^{i_{3}}} = \frac{3}{2^{i_{1}}} *\frac{3}{2^{i_{2}}} *\frac{3}{2^{i_{3}}} *r_{4}$

where $r_{4} = \frac{2^{i_{1}}}{27}*(2^{i_{2}} + 3) + n_{0} +\frac{1}{3}$.

Remarks: The values, $2^{i_{j}}$, are distributed according to equation one: $n_{t} = (\frac{3}{2} )^{\frac{t}{2}} * (\frac{3}{4} )^{\frac{t}{4}} * (\frac{3}{8} ) ^{\frac{t}{8}} * ... *( \frac{3}{2^{k}} ) ^{\frac{t}{2^{k}}} * r = 1$.

Moreover, $r = r(n_{0})$ is a positive real number.

Example: If we let $n_{0} = 57$, then we compute $e_{max} = 196$, $k = 7$, and $t = 10$.

Therefore, $r = r(57) = 1/.0841394 = 11.8850384$.

Dave’s Conjecture: $r = r(n_{0 }, t) =$ O(t) or $r = c_{t} * t$ for some real number, $c_{t}$, such that either $c_{t} > 1$ or $0 < c_{t} < 1$.

Example: If we have $t = 1$ for some $n_{0 }$, then as $k \rightarrow \infty$, $r = \frac{1}{ \prod_{i=1}^{k }(\frac{3}{2^{i}})^{\frac{t}{2^{i}}}} \rightarrow \frac{4}{3 } = c_{1 }$.

Therefore, $r = r(n_{0 }, 1) \approx \frac{4}{3}$ for infinitely many positive odd integers, $n_{0} > 1$.

Questions: What are the values for, $c_{2}, c_{3}, c_{4}, ...$?