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A045654 Number of 2n-bead balanced binary strings, rotationally equivalent to complement.

%I A045654 #58 Jul 30 2025 22:09:45
%S A045654 1,2,6,8,22,32,72,128,278,512,1056,2048,4168,8192,16512,32768,65814,
%T A045654 131072,262656,524288,1049632,2097152,4196352,8388608,16781384,
%U A045654 33554432,67117056,134217728,268451968,536870912,1073774592,2147483648,4295033110,8589934592
%N A045654 Number of 2n-bead balanced binary strings, rotationally equivalent to complement.
%H A045654 Andrew Howroyd, <a href="/A045654/b045654.txt">Table of n, a(n) for n = 0..1000</a>
%H A045654 Tilman Piesk, <a href="/A045654/a045654_4.txt">Row sums of triangle derived from A385665</a>
%H A045654 Ralf Stephan, <a href="/somedcgf.html">Some divide-and-conquer sequences ...</a>
%H A045654 Ralf Stephan, <a href="/A079944/a079944.ps">Table of generating functions</a>
%F A045654 a(0)=1, a(2n) = a(n)+2^(2n), a(2n+1) = 2^(2n+1). - _Ralf Stephan_, Jun 07 2003
%F A045654 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 A045654 For n >= 1, a(n) = Sum_{k=0..A007814(n)} 2^(n/2^k). - _David W. Wilson_, Jan 01 2012
%F A045654 Inverse Moebius transform of A045663. - _Andrew Howroyd_, Sep 15 2019
%F A045654 a(n) = 2*A127804(n-1) for n > 0. - _Tilman Piesk_, Jul 05 2025
%F A045654 a(n) = Sum_{k=1..n} 2 * n * A385665(n,k) / k. - _Tilman Piesk_, Jul 07 2025
%e A045654 From _Andrew Howroyd_, Jul 06 2025: (Start)
%e A045654 The a(1) = 2 length 2 balanced binary strings are: 01, 10.
%e A045654 The a(2) = 6 strings are: 0101, 1010, 0011, 0110, 1100, 1001.
%e A045654 The a(3) = 8 strings are: 010101, 101010, 000111, 001110, 011100, 111000, 110001, 100011. (End)
%p A045654 a:= proc(n) option remember;
%p A045654       2^n+`if`(n::even and n>0, a(n/2), 0)
%p A045654     end:
%p A045654 seq(a(n), n=0..33);  # _Alois P. Heinz_, Jul 01 2025
%o A045654 (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 A045654 (Python)
%o A045654 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 A045654 Cf. A000984, A045653, A045655, A045656, A045663, A127804, A385665.
%K A045654 nonn,easy
%O A045654 0,2
%A A045654 _David W. Wilson_