[seqfan] Re: A045655
David Wilson
davidwwilson at comcast.net
Sun Jan 1 21:52:25 CET 2012
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
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