# 3x+1: Searching for 3−cycles

Looking for solutions to x = s/K, where

K = 2a1+a2+...+an - 3n
s = 3n-1 + 3n-2*2a1 + 3n-3*2a1+a2 + ... + 2a1+a2+...+a(n-1)

when n = 3, we can construct the following table, organised by m = ∑ai for 'a' values up to 4.

### m = 3, K = −19

 a1 a2 a3 s K s/K 1 1 1 19 −19 −1

### m = 4, K = −11

 a1 a2 a3 s K s/K 1 1 2 19 −11 −19/11 1 2 1 23 −11 −23/11 2 1 1 29 −11 −29/11

### m = 5, K = 5

 a1 a2 a3 s K s/K 1 1 3 19 5 19/5 1 2 2 23 5 23/5 1 3 1 31 5 31/5 2 1 2 29 5 29/5 2 2 1 37 5 37/5 3 1 1 49 5 49/5

### m = 6, K = 37

 a1 a2 a3 s K s/K 1 1 4 19 37 19/37 1 2 3 23 37 23/37 1 3 2 31 37 31/37 1 4 1 47 37 47/37 2 1 3 29 37 29/37 2 2 2 37 37 1 2 3 1 53 37 53/37 3 1 2 49 37 49/37 3 2 1 65 37 65/37 4 1 1 89 37 89/37

### m = 7, K = 101

 a1 a2 a3 s K s/K 1 2 4 23 101 23/101 1 3 3 31 101 31/101 1 4 2 47 101 47/101 2 1 4 29 101 29/101 2 2 3 37 101 37/101 2 3 2 53 101 53/101 2 4 1 85 101 85/101 3 1 3 49 101 49/101 3 2 2 65 101 65/101 3 3 1 97 101 97/101 4 1 2 89 101 89/101 4 2 1 121 101 121/101

### m = 8, K = 229

 a1 a2 a3 s K s/K 1 3 4 31 229 31/229 1 4 3 47 229 47/229 2 2 4 37 229 37/229 2 3 3 53 229 53/229 2 4 2 85 229 85/229 3 1 4 49 229 49/229 3 2 3 65 229 65/229 3 3 2 97 229 97/229 3 4 1 161 229 161/229 4 2 2 121 229 121/229 4 3 1 185 229 185/229

(c) John Whitehouse 2011-2013