%I #37 Feb 06 2024 12:59:25
%S 1,1,3,5,17,53,260,1466,10915,93196,917898,10015299,119914982,
%T 1557364352,21797494987,326930305166,5230756117008,88922108947567,
%U 1600594738591550,30411281088326498,608225534389576956,12772735698577492558
%N Consider a clock face with 2*n "hours" marked around the dial; a(n) = number of ways to match the even hours to the odd hours, modulo rotations and reflections.
%H Barry Cipra, <a href="/A357442/a357442.pdf">Illustration for a(5) = 17</a>
%H N. J. A. Sloane, <a href="/A357442/a357442.png">Sketch illustrating a(1) = a(2) = 1, a(3) = 3</a>
%H Philip Todd, <a href="https://arxiv.org/abs/2401.13002">Theorem Discovery Amongst Cyclic Polygons</a>, arXiv:2401.13002 [cs.CG], 2024.
%F See PARI code for the formula. - _Max Alekseyev_, Nov 10 2022
%o (PARI) { a357442(n) = ( sumdiv(n,d,(n\d)!*d^(n\d)*eulerphi(d)) + n*sum(k=0,n\2,n!\k!\2^k\(n-2*k)!) + if(n%2, n*((n-1)\2)!*2^((n-1)\2) + sumdiv(n,d, eulerphi(d)*sum(k=0,n\d\2,(n\d)! \ (2*k+1)! \ ((n\d-1)\2-k)! * (d/2)^((n\d-1)\2-k) ))) )\n\4; } \\ _Max Alekseyev_, Nov 10 2022
%Y Cf. A000031, A000699, A007769, A059375.
%K nonn
%O 1,3
%A _N. J. A. Sloane_, Nov 06 2022, based on an email from _Barry Cipra_, Oct 26 2022
%E Terms a(7) onward from _Max Alekseyev_, Nov 10 2022