Searching for 3−cycles

Looking for solutions to x = s/K, where

K = 2^a1+a2+...+an - 3ⁿ
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 = ∑a_i 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

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