Collatz Conjecture - Simple Solution

[Guest]

Hi,

Lets start to series with an odd number (actually it does not matter it even or odd)

(with a probabilistic approach)

(an odd) produce an even number (%100 percent)
(an even) produce 1 even and 1 odd (%50 %50)
(1 even and 1 odd) produce (1 even and 1 odd) and (2 even) (total 3 even and 1 odd) (from now i'm not giving percentage because it can be confusing)
(3 even and 1 odd) produce (3 even and 3 odd) and (2 even) (total 5 even and 3 odd) (1 odd number produce 1 even number so i multiply it by 2 for balancing the probablity comes from the even number)
(5 even and 3 odd) produce (5 even and 5 odd) and (6 even) (total 11 even and 5 odd)
(11 even and 5 odd) produce (11 even and 11 odd) and (10 even) (total 21 even and 11 odd)

while iteration goes on like this the probablity of (even number / odd number) is ( (2n+1)/n or (2n-1)/n interchangeably) (n is the odd number count at current iteration level)

at last iteration we reached 21/11 which is quite close to 2/1

2/1 is the ultimate ratio (while n goes to infinity)

so Collatze function produce 2 even number for every 1 odd number (as probabilistic )

because of even numbers are multiplied by 1/2 and odd numbers multiplied by 3 for Collatze function:

average multiplier of the series is (1/2)*(1/2)*3 = 3/4 for each iteration

because 3/4 is smaller then 1 this series will eventually reach to smallest possible number (which is 1)

(any number which is multiplied by (3/4)^n while n goes to infinity, goes to 0, here it goes to 1 for Collatze function)


this approach also confirms that why (5n+1) version of Collatze function goes to infinity unlike (3n+1) (if we use 5n+1 instead of 3n+1)

(1/2)*(1/2)*5 = 5/4 which is bigger than 1, so (5/4)^n goes to infinity (while n goes to infinity)

regards,