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Sep 4 at 14:01 vote accept m3tro
Sep 4 at 13:14 comment added David E Speyer Glad to see it worked out! If you don't want to quotient by reflections, you can just use the rotations terms in the Burnside computation, and divide by $2n$ (order of the rotation group) instead of $4n$.
Sep 4 at 12:30 comment added m3tro @PeterTaylor thanks, but see my first comment to David above: "The reason I care about necklaces and not bracelets is because I don't want to quotient by reversal of the paths, paths are oriented". I see now that David has assumed quotient by reversal in his answer, but he could easily modify it to account for both possibilities.
Sep 4 at 10:48 comment added Peter Taylor @m3tro, A054499.
Sep 4 at 10:19 comment added m3tro Another neat result of the chord diagram approach is that one can immediately see which paths are "non-returning" (like path number 2 for 2D and paths number 3 and 4 in 3D in the image above) by checking whether the chord diagram is connected. The number of connected diagrams for different n is given in OEIS A018225 and, starting from n=2, is 1, 2, 6, 31...
Sep 4 at 10:15 comment added m3tro I was about to post this as an answer right now, when I came across your answer. You did get to a solution of the problem (can you confirm that your approach would give A007769?) so I am happy accept your answer if you updated it with the additional information.
Sep 4 at 10:07 comment added m3tro Hi David, thanks for the awesome answer. There has been in the meantime substantial discussion about this problem on the /r/math subreddit, see reddit.com/r/math/comments/1f7ui06/… I did figure out the connection to chord diagrams and realized that what I am looking for is OEIS A007769 "Number of chord diagrams with n chords; number of pairings on a necklace". Starting from n=2, we get 2, 5, 18, 105... The reason I care about necklaces and not bracelets is because I don't want to quotient by reversal of the paths, paths are oriented.
Sep 4 at 9:32 history edited David E Speyer CC BY-SA 4.0
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Sep 4 at 9:26 history edited David E Speyer CC BY-SA 4.0
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Sep 4 at 9:21 history answered David E Speyer CC BY-SA 4.0