3‐Cycles

Looking for solutions to x = s/K, where 

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

When n=3, we can construct the following table, organised by m=Σa¬i. Green cells represent integer solutions, blue cells where the largest s/K is greater than 1. The green cells correspond to the A‐Lists [1,1,1] and [2,2,2] and correspond to going around the 1‐Cycle three times.

Σa¬i K Lists tested
3 −19
a1 a2 a3 s K s/K
1 1 1 19 −19 −1
4 −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
5 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
6 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
7 101
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
8 229
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

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(c) John Whitehouse 2011 - 2023