The proof of the Collatz Problem

The proof : by collatz function

             a_n/2,   if a_n=0  (mod 2)

a_n=

           3a_n+1,  if a_n=1  (mod 2)

The first step in my proof is compress the multi-function in one formula

a_n+1=(2^2*s-1+s)*a_n+s   , a_n=s (mod 2)

The second step is generated the function

a_n+1=(2^2*s1-1+s₁)*a_n+s₁   , a_n=s₁ (mod 2)

a_n+1=(2^2*s1-1+s₁)*a_n+s₁)*(2^2*s2-1+s₂)+s_{2   ,}a_n=s₂(mod 2) …

We can find a general formula for this expression:

 

The third step is practiced the Collatz problem

Then By My Formula

  and   we omit it because a_n dosen’t equal zero

And  2^2*s(j-i)-1 doesn’t equal zero.

  and 

The forth step we suppose that at  ,a_n+1=1 if it holds a_n+1 at any number less than  

The fifth step is analysed

a_n=1 and =1 it will be omit because s_j-i=0 either or 1 only

a_n=2 and =1/2  ,s_j-i=0 then a_n=2 and it is right

But is 1=?

==0

 

<0.5j

And by Terras (1976, 1979) function

 

                                                         t_n/2         if  t_n=0 (mod 2)

                                      t_n+1=

                                                                        (3t_n+1)/2   if   t_n=1 (mod 2)

We find that for all 1 step by 3t_n+1 there is two steps by divide on 2 then (1/3)j=

And we find that 1/3j<1/2jà 1/3<1/2

And

<0.5j

then  and Collatz conjecture be right.