Is this a valid proof of the Collatz theorem?

Question

Is this a valid proof of the Collatz Conjecture?

For all even numbers x >= 20, if all numbers 1 through x are in Collatz set, show x+1 and x+2 are in set. Then by induction, all positive integers are in set.

For all even numbers x >= 20, if all numbers 1 through x are in Collatz set, then all even numbers between x and 2x are in the set. Therefore x+2 is in the set.

For all even numbers x >= 20, ((x+1)*3) + 1) / 2 is an even number between x and 2x. Therefore x+1 is in the set.

We already know that all numbers between 1 and 20 are in the set.

(I picked 20 sort of at random.)

Answer 1

For all even numbers $x \ge 20$, $\dfrac{(x+1)\times3 + 1}{2}$ is an even number between $x$ and $2x$

is not correct if $x$ is a multiple of $2$ but not a multiple of $4$

since if $x=4n+2$ then $\dfrac{(x+1)\times3 + 1}{2}= 6n+5$, which is an odd number

Answer 2

No as other answers point out. Other results exists that don't ultimately prove it. For example, all numbers of form: $$ax+b$$ where $a$ has a higher exponent on 2, than division by 2's in the sequence starting at $b$; letting $c$ be the number of divisions by 2 in the sequence starting at $b$, and $d$ the number of multiply by 3 and add 1 steps, will fall to ${a\cdot 3^d\over 2^c}x+1$ guaranteed.

That shows that for example that $2055=2048x+7$ falls to $244=243x+1=4f$, in fact that plus a few divides by 2, gets us to $f=61=32y+29$, which then gets us to $53=27y+26=32z+21$ before the 32 previous runs out. Then going again with the next representation takes us to $3z+2$ which is just 5 in this case.

We can get much higher numbers down through a similar process, This process came to me a while back, after watch Numberphile's video on the Josephus problem ( how it used power of two plus something representations inspired me a bit). This still won't prove all numbers get to 1 if I recall correctly. Think someone took the search to $10^{21}$ using parts of it. Anyways have fun.