[seqfan] Re: A045655

[David Wilson]

I analyzed A045655.

It counts 2n-bead 2-color strings of beads that are rotationally 
equivalent to their reversed complement.

Consider the string 011100, where 0 is white and 1 is black.
Complement 011100 and you get 100011, reverse this and you get 110001.
So the reversed complement of 011100 is 110001.
Notice that you can rotate 011100 and you also get 110001 (in this case, 
rotate left two places).
This means that 011100 is rotationally equivalent to its reversed 
complement 110001.

If we look at all 64 possible 6-bead strings, we find 20 that are 
rotationally equivalent to their reversed complements:

000111  001011  001101  001110  010011
010101  010110  011001  011010  011100
100011  100101  100110  101001  101010
101100  110001  110010  110100  111000

Of these 20 strings, many are rotationally equivalent to each other, for 
example 001011 and 010110. If two strings are
rotationally equivalent, then they represent different rotations of the 
same necklace. If we group the above 20 strings
into sets (equivalence classes) of strings that are rotationally 
equivalent to one another, we find that there are really
just four necklaces:

{000111, 001110, 011100, 100011, 110001, 111000}
{001011, 010110, 011001, 100101, 101100, 110010}
{001101, 010011, 011010, 100110, 101001, 110100}
{010101, 101010}

So there are 20 6-bead strings, but only 4 6-bead necklaces, that are 
rotationally equivalent to their reversed
complements. This is why A045655(3) = 20 and A011782(3) = 4.

On 12/31/2011 10:16 AM, GEOFFREY CRITZER wrote:
> David
> I am interested in your sequence A045655.  I do not understand your 
> description.
> Any reply is greatly appreciated.
> Best Regards
> Geoffrey