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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