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Number of 2n-bead balanced binary strings of fundamental period 2n, rotationally equivalent to reverse.
7

%I #13 Sep 23 2019 09:50:42

%S 1,2,4,6,32,50,204,266,1024,1224,4900,5522,21600,23998,95508,102750,

%T 409600,437546,1747152,1847522,7380000,7758870,31027876,32449826,

%U 129752064,135207500,540783100,561628620,2246337184,2326762742

%N Number of 2n-bead balanced binary strings of fundamental period 2n, rotationally equivalent to reverse.

%H Andrew Howroyd, <a href="/A045662/b045662.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = 2*n*A045680(n) for n > 0.

%F a(n) = 2*n*Sum_{d|n} mu(n/d) * binomial(2*floor(d/2), floor(d/2)) for n > 0. - _Andrew Howroyd_, Sep 14 2019

%t a[n_] := If[n == 0, 1, 2n Sum[MoebiusMu[n/d] Binomial[d - Mod[d, 2], Quotient[d, 2]], {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) * binomial(d-d%2, d\2))); \\ _Andrew Howroyd_, Sep 14 2019

%Y Cf. A007727, A045663, A045664, A045680.

%K nonn

%O 0,2

%A _David W. Wilson_