Palindromic Binary Sequence,Ratio,Plaid
Leroy Quet
qqquet at mindspring.com
Wed Mar 26 03:19:42 CET 2003
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
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