%I #19 Sep 23 2019 09:57:32
%S 1,2,4,18,48,150,324,882,1920,4536,9900,22506,48240,106470,227556,
%T 490950,1044480,2228190,4708368,9961434,20950800,44037378,92229588,
%U 192937938,402549120,838860000,1744617420,3623864832,7515733680
%N Number of 2n-bead balanced binary strings of fundamental period 2n, rotationally equivalent to reversed complement.
%H Andrew Howroyd, <a href="/A045664/b045664.txt">Table of n, a(n) for n = 0..1000</a>
%F For n >= 1, a(n) = n*A027375(n) = 2*n*A000740(n) = n^2*A001037(n).
%F a(n) = 2*n*Sum_{d|n} mu(n/d)*2^(d-1) for n > 0. - _Andrew Howroyd_, Sep 14 2019
%t a[n_] := If[n == 0, 1, 2n Sum[MoebiusMu[n/d] 2^(d-1), {d, Divisors[n]}]];
%t a /@ Range[0, 30] (* _Jean-François Alcover_, Sep 23 2019, from PARI *)
%o (PARI) a(n)={if(n<1, n==0, 2*n*sumdiv(n, d, moebius(n/d)*2^(d-1)))} \\ _Andrew Howroyd_, Sep 14 2019
%Y Cf. A000740, A001037, A007727, A027375, A045662, A045663.
%K nonn
%O 0,2
%A _David W. Wilson_