A138954 Number of complement symmetries in the rotations of the binary expansion of a number.

0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Author

Max Sills, Apr 03 2008

Keywords

• easy
• nonn

Comments

It seems that the number of complement rotational symmetries is nonzero iff #0 = #1 in the binary expansion of a number.

The above statement is true in only one direction. It is clearly necessary for the number of 1 bits to equal the number of 0 bits. However, this is not sufficient. The first counterexample is n = 37 with binary expansion 100101 and complement 011010. Values of n for which a(n) is nonzero are therefore a proper subset of A031443. - Andrew Howroyd, Jan 12 2020

Examples

			a(2) = 1 because 2 has binary expansion 10 and the complement shows up once in rotations;
a(10) = 2 because 10 has binary expansion 1010 and its complement shows up twice in rotations.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1023

Crossrefs

Cf. A031443, A138904.

Programs

PARI

a(n)={my(s=0); if(n, my(b=logint(n,2)+1); if(2*hammingweight(n)==b, my(w=2^b-1-n); for(i=2, b, w=if(w%2, w+2^b, w)\2; if(w==n, s++)))); s} \\ Andrew Howroyd, Jan 12 2020

Extensions

Missing a(8) inserted and terms a(21) and beyond from Andrew Howroyd, Jan 12 2020