## 12.8.16

### Powers of Collatz Part 2

In my previous blog entry I talked about how The Collatz Conjecture might have something to do with generating powers of (n+1)/2 mod n, especially when n is the product of distinct primes.

I conjecture that the algorithm behind The Collatz Conjecture always reaches some power of (n+1)/2 mod n for some n, and this power is not necessarily a power of 2 in Z.

So what happens when the algorithm behind the Collatz conjecture is applied to (n+1)/2 starting from ((n+1)/2)2 mod n?

Powers of Collatz Algorithm:

Input: n = p*q where p, q are distinct odd primes greater than 3

1. Calculate (n+1)/2
2. Calculate k = ((n+1)/2)2 mod n
3. Do k = k/2 if k is even
If k is odd then stop and return k = 3*k + 1 mod n
While k is not equal to 1

Output: not necessarily a power of (n+1)/2 mod n, in many cases the output is either p, q, or a multiple of p or q.

So I designed an experiment to see which products of primes are affected by this phenomenon up to some small limit, which is easy to calculate by a mobile calculator. For this experiment I take all odd primes greater than 3 and smaller than 41.

5, 7, 11, 13, 17, 19, 23, 29, 31, 37

The following products of primes are affected:

5*7,5*11,7*11, 5*19, 7*13, 7*19, 7*23, 7*29, 7*31, 11*37

35, 55, 77, 85, 91, 133, 161, 203, 217, 259, 407

(I did this experiment late last night so I might have missed a few.)

When n = 35 then k = 18, k^2 = 9, which is odd so multiply by 3 and add 1 mod 35, 3*9 + 1 = 28 = 3*7

When = 55 then k = 28 k^2 = 14, which is even so divide by 2,  14/2 =  7, which is odd so 3*7 + 1 = 22 = 2*11

And so on.