Palindromic Binary Sequence,Ratio,Plaid

[Leroy Quet]

Is this just a coicidence, or is "my" sequence simply the sequence 
{1-b(k)}, where my b(k) is the k_th term of the following?:
(Note: sequence below starts one term before mine.)


ID Number: A022925
Sequence:  0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,
           1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,
           1,0,1,0,0,1,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,1,0,1,0,0,1,0,
           1,0,1,0,0,1,0,1,0
Name:      Number of 5^m between 2^n and 2^(n+1).
Keywords:  nonn
Offset:    0
Author(s): Clark Kimberling (ck6 at evansville.edu)

Thank you,
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