Start from n, any odd number

**Step 1.**

Multiply by 3 and add 1 = 3n+1

**Step 2**

Multiply by 3 = 9n+ 3

Factor as (2^a)(an odd number)

Add 2^a = 9n+3+2^a

**Step 3...**

Multiply by 3 = 27n + 9 + 3(2^a)

Factor as (2^b)(an odd number)...

Add 2^b = 27n + 9 + 3(2^a) + 2^b

Repeat until we reach 2^z(n) or 2^z

**As the total increases, the cumulative total is**

**[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^z...**

The
power of 3 falls by one in each term, until we reach 3^0 in the final
term

The power of 2 increases in each term but can rise by more than 1 at a time – so in the example above d>c>b>a.

The power of 2 increases in each term but can rise by more than 1 at a time – so in the example above d>c>b>a.

**If the Collatz conjecture is true, we will always reach an equation of the form**

**[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^d = 2^z**

**For there to be a 3n+1 loop, which would be a counterexample to the conjecture, we would need this equation to be true:**

**[3^y x n] + [3^(y-1)] + [2^a x 3^(y-2)] + [2^b x 3^(y-3)] …. + [2^c x 3]+ 2^d = 2^z(n)**

do you study math aswell??

ReplyDeleteWonderful Blog! Do you Matlab to construct these grids or any other program?

ReplyDeleteCheers prime