[seqfan] Re: A045655

[franktaw at netscape.net]

As I understand the definition of A045655, you take a string of length 
2*n, say (for n=3) 001101. Complement it bit-wise: 110010, then reverse 
it left-to-right: 010011. Now, is this rotationally equivalent to the 
original string? In this case, yes: rotate left two places and you get 
back 001101.

The comment by Geoffrey Critzer may be correct, but it is not at all 
obvious. In my opinion, if it is correct, at least a sketch of the 
proof ought to be included or referenced.

Franklin T. Adams-Watters

-----Original Message-----
From: Maximilian Hasler <maximilian.hasler at gmail.com>

it does not matter whether you add "reversed complement" or not,
the count will be the same whatever bijection (from { 0,...,2^n-1 }
into itself) you apply to the first and/or second component of the
pairs.

as far as I understand, the "rotations" are to be taken digit-wise in
n-bit binary words,
e.g. 011 > 110 > 101.

Maximilian



On Sun, Jan 1, 2012 at 1:18 AM, Ed Jeffery <lejeffery7 at gmail.com> wrote:
> David,
>
> It seems as if Geoffrey Critzer contradicted your definition in your 
title
> for A045655 by dropping your reference to "reversed complement." The
> question is, what do you mean by that terminology: are you referring 
to
> "ones complement" from binary arithmetic? If so, then for longer 
strings
> the symmetry you seem to be suggesting will be lost.
>
> If you are referring to dihedral symmetry, then, as you know, two 
objects
> in the plane are either (or they are not) congruent up to rotations 
or they
> are (or are not) reflections in a line. So, in terms of sequences of 
digits:
>
> Are you ordered pairs (a,b) supposed to be such that either (i) a 
equals b,
> or (ii) b is a reflection of a (i.e., b takes the digits of a in 
reverse
> order)? If so, then evidently Geoffrey Critzer's definition must be 
the
> correct one.
>
> It is a bit confusing, but I like your sequence.
>
> Regards,
>
> Ed Jeffery
>
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>
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