A045654 Number of 2n-bead balanced binary strings, rotationally equivalent to complement.
1, 2, 6, 8, 22, 32, 72, 128, 278, 512, 1056, 2048, 4168, 8192, 16512, 32768, 65814, 131072, 262656, 524288, 1049632, 2097152, 4196352, 8388608, 16781384, 33554432, 67117056, 134217728, 268451968, 536870912, 1073774592, 2147483648, 4295033110, 8589934592
Offset: 0
Examples
From _Andrew Howroyd_, Jul 06 2025: (Start) The a(1) = 2 length 2 balanced binary strings are: 01, 10. The a(2) = 6 strings are: 0101, 1010, 0011, 0110, 1100, 1001. The a(3) = 8 strings are: 010101, 101010, 000111, 001110, 011100, 111000, 110001, 100011. (End)
Links
- Andrew Howroyd, Table of n, a(n) for n = 0..1000
- Tilman Piesk, Row sums of triangle derived from A385665
- Ralf Stephan, Some divide-and-conquer sequences ...
- Ralf Stephan, Table of generating functions
Programs
-
Maple
a:= proc(n) option remember; 2^n+`if`(n::even and n>0, a(n/2), 0) end: seq(a(n), n=0..33); # Alois P. Heinz, Jul 01 2025 -
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 -
Python
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
Formula
a(0)=1, a(2n) = a(n)+2^(2n), a(2n+1) = 2^(2n+1). - Ralf Stephan, Jun 07 2003
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
For n >= 1, a(n) = Sum_{k=0..A007814(n)} 2^(n/2^k). - David W. Wilson, Jan 01 2012
Inverse Moebius transform of A045663. - Andrew Howroyd, Sep 15 2019
a(n) = 2*A127804(n-1) for n > 0. - Tilman Piesk, Jul 05 2025
a(n) = Sum_{k=1..n} 2 * n * A385665(n,k) / k. - Tilman Piesk, Jul 07 2025