1. Introduction:

Collatz conjecture are define that with G(n)={ n÷2 n≡0(mod2)

3n+1 n≡1(mod2)

conjecture said that there will alway exist k that Gk(n)=1.

2. backround of the proof

the proof will base on the position of odd number in the chart. the problem of this situation is about the input and the out put of the equation.

3. proof

by observing G(n)≡0(mod2) => G(n)= 2k×m

m=2 =>G(n)k+1=1

m≡1(mod2) => G(n)k=m

base on that fact

we will mainly focus on the behavior of G(n) when m≡1(mod2)

let O be the group of odd number

O={O1, O2, O3, …, OL,…}

where OL+1-OL =2

O1 =1

L: the position of odd start from 1

=>OL= 2L-1

by another observation

G(OL)-G(OL-1)=3 (since O1-O2 =2)

=> the output of the equation as O1-> infinity is alternating odd and even.

since G(1)≡0 (mod2)

=> G(O2L-1)≡0 (mod2) and G(O2L)≡1(mod2)

same agument, we will just forcus on output odd.

since G(O2L)≡1 (mod2) => G(O2L)= Osomething

we know that G(1)=G(O1)=2, G(OL)-G(OL-1)=3 and OL=2L-1

=> other way of writting for the equation of n ≡1(mod2)

G(O2L) =2+3(2L-1)= 6L-1= O3L

O3L=G(O2L)

now we look at G(odd)≡0(mod2)

G(n)≡0(mod2)

=>G(n)=2k×m

m≡1(mod2) (did mention)

m=2 =>G(n)=2k+1

2k+1=2+3L

=>L=2k+1-2÷3

=> 2k≡1(mod3)

=>k≡0(mod2)

with the position of odd that

L=(2k-2)/3 k≡1(mod2)

G(x)->1

since there are infinite k≡1(mod2)

=> there are infinite L that make G(x)->1

=> the conjecture is true

https://arxiv.org/abs/1608.03600 ]]>

http://pastebin.com/raw/q5DyHGiz

Yuval Levental

http://linkedin.com/in/yuvallevental

http://arxiv.org/abs/1608.03600 ]]>

http://arxiv.org/abs/1608.03600 ]]>