Collatz Funnel

Given any number \(n\),

$$n_{next} =\begin{cases} E(n)=n/2,  & \text{if $n$ is even} \\ O(n)=3n+1, & \text{if $n$ is odd}\end{cases}$$

The proof needed was that all numbers are reduced to \(1\) using these mappings.

Given that a number, \(n\) is odd, the operation,

\(O(n)=3n+1\) 

will always provide a even number,

P(O(n)=even)=1

P(O(n)=odd)=0

Given that a number \(n\) is even, the operation,

\(E(n)=\cfrac{n}{2}\)

can given either a even or odd number.

P(E(n)=odd)=1/2

P(E(n)=even)=1/2

The probability of any operation, producing an odd number is P(odd)=\(\cfrac{1}{4}\) 

and the probability of any operation, producing an even number is P(even)=\(\cfrac{3}{4}\)

This system of equations is bias towards producing even number, and so tends to reduce \(n\).

Given that a large number of lower numbers have already been found to reach \(1\), this shows that the system has a tendency to reach \(1\). This is because when a large unproven number is reduced to a smaller number proven to reach \(1\) then this large number has just been proven to reach \(1\) also.

Consider this,

This graph is symmetric about \(tan(45^o)\) line through the origin.  The label 'odd' and 'even' show how any selected number enters into the 'funnel' and is reduced.  The red lines deflect all odd numbers to the black lines that divide the number.  At this point the resulting number can be odd or even.  If even, the number is further reduced by the other black line.  If odd, the number is deflected to the red lines.

Number caught between the black lines eventually reach \(1\).  The red lines increase the numbers deflected off them and are always even.  These lines do not produce the needed \(1\).

Does this prove the Collatz Conjecture.  No, just that, the more lower number proven, the more likely the conjecture is true.