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Number of 2n-bead balanced binary strings, rotationally equivalent to complement.
10

%I #25 Jul 22 2024 14:28:29

%S 1,2,6,8,22,32,72,128,278,512,1056,2048,4168,8192,16512,32768,65814,

%T 131072,262656,524288,1049632,2097152,4196352,8388608,16781384,

%U 33554432,67117056,134217728,268451968,536870912,1073774592,2147483648

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

%H Andrew Howroyd, <a href="/A045654/b045654.txt">Table of n, a(n) for n = 0..1000</a>

%H R. Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences ...</a>

%H R. Stephan, <a href="/A079944/a079944.ps">Table of generating functions</a>

%F a(0)=1, a(2n) = a(n)+2^(2n), a(2n+1) = 2^(2n+1). - _Ralf Stephan_, Jun 07 2003

%F G.f.: 1/(1-x) + sum(k>=0, t(1+2t-2t^2)/(1-t^2)/(1-2t), t=x^2^k). - _Ralf Stephan_, Aug 30 2003

%F For n >= 1, a(n) = sum(0<=k<=A007814(n), 2^(n/2^k)). - _David W. Wilson_, Jan 01 2012

%F Inverse Moebius transform of A045663. - _Andrew Howroyd_, Sep 15 2019

%o (PARI) a(n)={if(n==0, 1, my(s=0); while(n%2==0, s+=2^n; n/=2); s + 2^n)} \\ _Andrew Howroyd_, Sep 22 2019

%o (Python)

%o def A045654(n): return sum(1<<(n>>k) for k in range((~n & n-1).bit_length()+1)) if n else 1 # _Chai Wah Wu_, Jul 22 2024

%Y Cf. A000984, A045653, A045655, A045656, A045663.

%K nonn

%O 0,2

%A _David W. Wilson_