Even Expectation

 Every proven $n$ in Collatz Conjecture leads a path to $1$.  So. there can be a number $n_{max}$, below which all numbers have been proven true for Collatz Conjecture. (This can be created computationally).  The next immediate number after $n_{max}$, $n_{p}$ must be a odd integer, because an even integer will immediately be divided into $2$ and be send into the pack of proven numbers, as the conjecture dictate. 

This odd number $n_{p}$ will be move to bigger value to $3n_{p}+1$.  This number being even is divided by $2$, 

$n_e=\cfrac{3n_{p}+1}{2}$

this number can be odd or even.

Given that the probability of an even number by the operations in the Conjecture is $P(even)=\cfrac{3}{4}$ and the probability of an odd number after any operation is $P(odd)=\cfrac{1}{4}$.

The expected value for $n_e$ is then

$E(n_e)=\cfrac{3(3n_{p}+1)}{8}$     when $n_e$ is even

$E(n_e)=\cfrac{3n_{p}+1}{8}$     when $n_e$ is odd

These are expected value of $(n_e)$ given large numbers where the probability provide good approximation.

The next value of $n_e$ when it is even as we apply $\cfrac{n}{2}$ is,

$\cfrac{1}{2}\cfrac{3(3n_{p}+1}{8}=\cfrac{3(3n_{p}+1)}{16}=\cfrac{9n_p}{16}+\cfrac{3}{16}\lt n_p$

for large value of $n_{max}$, this number is below $n_p$ and so a proven number.

The next value of $n_e$ when it is odd, we apply $3n+1$

$3\cfrac{3n_{p}+1}{8}+1=\cfrac{9n_p}{8}+\cfrac{11}{8}$

as this is always s even number, we divide by $2$,

$\cfrac{1}{2}\left(\cfrac{9n_p}{8}+\cfrac{11}{8}\right)=\cfrac{9n_p}{16}+\cfrac{11}{16}\lt n_p$.

which is also in the pack below $n_p$

As such with a large continuous number of Collatz number, of maximum $n_{max}$, as starting point, $n_{max}$ implies the next integer $n_p$ is also a Collatz number, then by induction all integer $n\ge 2$ are Collatz numbers.

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