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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