Palindromic Binary Sequence,Ratio,Plaid

[Leroy Quet]

I posted this to sci.math and rec.puzzles. As well as the sequence 
itself, it has some related material which is interesting if a little 
off-topic for this group.

----------- 

Consider this sequence of 1's and 0's which is "super-palindromic", and 
has an interesting property if converted to a binary-represented real.

Let each A be a finite sequence of 0's and 1's, 
where:

A(1) = 1;

for k >= 2,

A(k) = {A(k-1), A(k-1)}  if a(k-1) = 0;

A(k) = {A(k-1), 0, A(k-1)}  if a(k-1) = 1;


where the sequences in the {} are concatenated,

and a(k) is the k_th term of the sequence:

limit{k->oo}  A(k).

The sequence begins:

1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,
1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,0,1,0,1,1,0,1,...


Now, consider the real x, where

x = sum{k=1 to oo} a(k)/2^k,

the binary-represented x having the k_th digit right of the decimal-point 
(the binary-point?) equal to a(k).

ie.
x = .101101010110101011010101101... 

Then the ratio of number of 0's in this total sequence ("total sequence" 
= the limit of A(k) as k approaches infinity) to the number of 1's in 
this total sequence is...
x.

---
There is at least one other real, y, with the ratio-property: its binary 
representaion has a ratio of y = (number of 0's in binary representation) 
/(number of 1's in binary representation).

Even though it is not palindromic, the y-sequence can be generated in a 
similar manner to the x-sequence.

B(1) = 1;

for k >= 2,

B(k) = {B(k-1), B(k-1)}  if b(k-1) = 0;

B(k) = {B(k-1), B(k-1), 0}  if b(k-1) = 1;

b(k) = k_th element of {limit k-> oo} B(k).

(I think this is right...)(??)

What is known about all reals with the "ratio property"?

---
Back to the palindromic sequence:
We can extend this to multiple dimensions.

Super-Palindromic Plaid:

Take a grid of infinite size, taking up, say, the +/+ quadrant.

We can color the grid's squares as follows: 

Color (1,1), the square closest to origin, blue.
So A(1) = single blue square.

For k >= 2,

If (k,k), the k_th square along the diagonal, is yellow, then A(k) =

-----------------
!       !       !
!A(k-1) !A(k-1) !
!       !       !
-----------------   (View with fixed-width font.)
!       !       !
!A(k-1) !A(k-1) !
!       !       !
-----------------

And so, A(k) is a square of double the side-length of A(k-1).

If (k,k) is blue, then A(k) =

-------------------
!       !r!       !
!A(k-1) !r!A(k-1) !
!       !r!       !
-------------------
!rrrrrrr!y!rrrrrrr!
!------------------
!       !r!       !
!A(k-1) !r!A(k-1) !
!       !r!       !
-------------------

where y is a single yellow square, 
and the r squares form 4 single-square-width lines colored red.

A(k) is a square of side-length = (1 + twice that of A(k-1)).


So, the sequence of (k,k)s, the diagonal squares, corresponds to 
the sequence of a(k)'s above, with:

a(k) = 0 -> (k,k) = yellow

a(k) = 1 -> (k,k) = blue

I bet this makes an interesting design.
Here it is to begin with: (ascii-approximate)

# * # # * # * # * # # * #
* + * * + * * * + * * + *
# * # # * # * # * # # * #
# * # # * # * # * # # * #
* + * * + * * * + * * + *
# * # # * # * # * # # * #
* * * * * * + * * * * * *
# * # # * # * # * # # * #
* + * * + * * * + * * + *
# * # # * # * # * # # * #
# * # # * # * # * # # * #
* + * * + * * * + * * + *
# * # # * # * # * # # * #


I have seen such recursive plaid-designs made somewhere else before. 

Thanks,
Leroy Quet