Collatz conjecture states that
choose a number greater than 1
Let that number be the x-If x is an even number then divide it by 2
-If x is an odd number then multiply it by 3 add the product to 1The answer will be the new number(x) and do the same method.
It states that all number will always end up to 1.
It can be formulated.
x mod 2=0 x/2
f(x) ={
x mod 2=1 3x+1But is fⁿ(x)=1?
n is the number of steps you did until you reach 1, if it will alaways reach 1Example
Let x=5
f¹(5)=3(5)+1
=16
f²(5)=(16)/2
=8
f³(5)=(8)/2
=4
f⁴(5)=(4)/2
=1As you can see above in 4 we have reached 1
The other way to formulate it is
x mod 2=0 x/2
f(x) ={
x mod 2=1 (3x+1)/2Since 3x+1 will always end up even number if x is odd number.
Using the formula itself we can formulate that
Let x sub 0 an even number
such that
x sub 0/2ⁿ=x (x is odd number)y[k] =y sub k
fⁿ(x) =(3ⁿ+3ⁿ¯¹x+3ⁿ¯²2^{y[1]}+3ⁿ¯³2^{y[1]+y[2]}+...+3¹2^{y[1]+y[2]+...+y[n-2]}+2^{y[1]+y[2]+...+y[n-1]})/2^{y[1]+y[2]+...+y[n]}
The first approach we might conclude is
Proving 3ⁿ+3ⁿ¯¹x+3ⁿ¯²2^{y[1]}+3ⁿ¯³2^{y[1]+y[2]}+...+3¹2^{y[1]+y[2]+...+y[n-2]}+2^{y[1]+y[2]+...+y[n-1]}
is equal to
2^{y[1]+y[2]+...+y[n]}And also including finding the relation of x and n
Like
x|n2|1 if x is 2 then n is 1.we only did is one step, the 2/2
We will use the original formula the.(3x+1 without /2)3|7
4|2
5|5
6|8
7|16
8|3
9|19
10|6
11|14The other approach is finding the numbers in the set of col
Let col() is the set of numbers in which using the formula it will end up to 1Like finding if x=col() such that x is possitive integer greater than 1
It is obvious that
2ⁿ=col()
Using this approach can we solve now the unsolved math problem for many years?
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