Let us consider a much easier problem: Are there sequences starting at a positive integer x and ending at x + 1 with a parity vector of the form 1^s 0^{l-s} (l \geq s \geq 1)? They can be viewed as quasi-periodic circuits or quasi-circuits (the expression “near-circuits” was already used in this thread with a different meaning). The only ones are given below:
- 1 \rightarrow 2,
- 3 \rightarrow 5 \rightarrow 8 \rightarrow 4 .
The proof goes as follows:
Writing that T^l(x) = x + 1 leads to the rational solution
x = \cfrac{2^s}{3^s-2^l}-1.
Since the above denominator is odd, assuming x to be a positive integer implies 3^s-2^l = 1 which only occurs in two cases:
- l=s=1,
- l=3, s=2.
There are various elementary proofs for this fact, one of which goes back to the 14th century! Here is a short proof. Amazingly, this argument was already used to show the nonexistence of high cycles in @mathkook 's paper.
Note that the equation 3^s-2^l = -1 has only one solution with s\geq 1, so there is just one such quasi-circuit of negative integers: -3 \rightarrow -4 \rightarrow -2.
The case of quasi-circuits ending at x-1 instead of x+1 seems as difficult as the case of circuits. So I leave it to the reader! Sometimes, a minor change in the statement of a problem is sufficient to drastically change its difficulty.