I recently proposed a sequence on the OEIS (A$\texttt{358059}$) and have many questions about it that maybe someone can help with. The definition is:
Lexicographically earliest infinite sequence such that if the terms in the sequence thus far are arranged in a circle, every number can be visited just once according to the following rule: for any value $a(n)$ in the circle, you can move $a(n)$ spaces around the circle in any direction.
This is, at least partly, a Hamiltonian path problem. Finding new terms is difficult because there's a lot of backtracking involved, so one question is how do you know when a term is already certain/won't change (is there some strategy)? Also, is it even possible for the sequence to be infinite? One editor (Gleb) has computed $22$ terms, which are likely to change still due to backtracking: $$1, 2, 3, 5, 4, 6, 8, 11, 13, 7, 19, 9, 37, 29, 53, 25, 24, 10, 12, 122, 28, 50, 34$$
For clarification, here is an example of how the sequence works:
At $n=6$, we have $1,2,3,5,4,6$. There are two potential paths if we arrange these numbers in a circle. Beginning with the number $4$, we can jump $4$ numbers to reach $1$, from $1$ we jump $1$ space to $2$, from $2$ to $5, 5$ to $3, 3$ to $6$. So our path was: $4,1,2,5,3,6$. Another potential path is: $1,2,5,4,3,6$. You can jump in either direction around the circle at any point (it's a modular system). The next term is simply the smallest number not already used for which such a path exists (with no repeating numbers) and which will not eventually lead to the sequence ending.
Can anyone make any progress on this or have any insights? Thanks in advance.
Clarification (at request of @kabenyuk): I'm honestly also surprised by how quickly the OEIS editors catch on without needing clarification. It's fairly simple though. What you do is write down each term in the sequence so far, one by one, around a clockface. Your goal is to visit each number only once (find a Hamiltonian path). You can jump from one term to another (either clockwise or counterclockwise) based on how many terms or "hours" away the term is (if the term is k, you can jump k terms around clock to next term). From any term, you can get to at most 2 other terms by going in one direction or the other. So you start from tol(1) is 1 since it's an LES (lexicographically earliest seq). You can put 2 as the next hour because there's a path: 1 to 2. Then you can put 3 as the next hour. Several possible paths here. You can jump from 1 to 2, and then counterclockwise from 2 to 3. Or from 2 clockwise to 1 and 1 counter to 2. Note that 3 must be the final destination here because it forms a loop (since tol(n)=n). The next term is 5 because 4 doesn't work (you get stuck because 2 has to lead to 4 and 1 or 3 is left in the dust).