I was wondering if different rational Collatz cycles can have the same sum of members. (Since we were talking about summing cycle members at some point.)
These do:
73/175 \rightarrow 197/175 \rightarrow 383/175 \rightarrow 662/175 \rightarrow
331/175 \rightarrow 584/175 \rightarrow 292/175 \rightarrow 146/175 \rightarrow 73/175
Total = 2668/175 \approx 15.25
824/175 \rightarrow 412/175 \rightarrow 206/175 \rightarrow 103/175 \rightarrow
242/175 \rightarrow 121/175 \rightarrow 269/175 \rightarrow 491/175 \rightarrow 824/175
Total = 2668/175 \approx 15.25
Probably it’s well known, but it seems it’s because their parity sequences (11101000 and 00010111) are reverses of each other.
At every cycle length k, every parity vector v can be paired with its reverse v^R to get a matching sum-of-members. If v and v^R are rotations of each other (“palindromic”), the matching is trivial.
For k=10, there are 108 distinct cycles. Of these, there are 48 palindromic ones, plus 30 pairs with matching sum-of-members. Overall, the number of palindromic cycles is:
2^{(k+1)/2} if k is even
3 \cdot 2^{k/2-1} if k is odd
As k increases, the proportion of palindromic cycles to non-palindromic ones rapidly goes to 0. So, almost all cycles have a distinct “mirror” with the same sum-of-members.
Also… the members of the infamous k=11 integer cycle (-17 \ldots -17) obviously sum to an integer, but so do the members of its mirror cycle, even though that cycle is not itself composed of integers.
Can a cycle and its (distinct) mirror both be integer cycles, conceivably?