Note-taker next week: Hal
Snack-bringer next week: Emilie
Meeting begins with : Any questions?
Sam: (to Jim) Do you know anything about shifted number fences?
Here's his example: using the pattern
A E=(C+BD)/A
B C D
E
Dana Scott's recurrence
a_{n}*a_{n-4}=a_{n-1}*a_{n-3}+a_{n-2}
the sequence is
1,1,1,1,2,3,5,13,22,41,111...
When you use the above pattern you can get this-
1 1 2 3 5 ...
2 3 5 13 22 ...
5 13 22 41 111 ...
22 41 111 191 361 ...
. . . . .
. . . . .
. . . . .
Compare the above to Reid Barton's work with the sequence
b_{n+1}*b_{n-1}=b_{n}^2+b_{n}
with 1's as initial conditions, which also satisfies
the linear recurrence
b_{n}=5*b_{n-1}-5*b_{n-2}+b_{n-3}.
Using the same pattern as above we get-
1 1 1 1 1 1...
2 2 2 2 2 2...
6 6 6 6 6 6...
. . . . . .
. . . . . .
. . . . . .
Is there any significance to the fact that the above pattern is shifted,
and this one is not shifted??
Jim: Well not quite sure about that, but here is some stuff that might help
answer the question.
We'll look at two sequences and use the pattern:
A
B C AD=BC+1
D
The first sequence is given by b_{n+1}*b_{n-1}=b_{n}^2+1.
You get the following pattern when using the pattern:
1 1 1 1 1 1 1
1 1 1 1 1 1
2 2 2 2 2
5 5 5 5
13 13 13
34 34
And it continues on. Well if you replace all the 1s by variables and do the
pattern again then you get Laurent polynomials. These polynomials correspond
to the perfect matchings of hexagon snakes.
**Jim will prepare this in more detail, how they exactly correspond to the
matchings of snakes**
The second recurrence is a_{n}*a_{n-4}=a_{n-1}*a_{n-3}+1
And you get the following pattern:
1 1 2
1 1 3
1 1 2 4
1 1 3 9
1 1 2 4 14
1 1 3 9 19
1 1 2 4 14 43
1 3 9 19
4 14 43
Replace the zig-zag row of 1s with variables and do the pattern you get Laurent
polynomials again corresponding to perfect matchings of hexagon snakes.
**Jim will prepare this also**
**Jim will make available a copy of a homework problem from math 192 and
solution having to do with the recurrence a_{n}*a_{n-3}=a_{n-1}*a_{n-2}+1 **
Back to Sam's question and Dana Scott recurrence:
Replace the ones with variables and you will get laurent polynomials again.
These polynomials (probably) correspond to something similar to what Reid
Barton came up with, only the graphs may be tilted or something.
Hal: To compare to what we did on Tuesday, did you(Jim) have a picture in mind
when figuring out what k values to put into the F(n,k) formulas??
Jim: I had similar things in mind.
For example, the zigzaggy recurrence above could be written as:
1 1 1 1 1 1
2 2 2 2 2
3 3 3 3
4 4 4
9 9
Using the pattern
A AE=BD+1
B
C
D
E
this is one way to skew the array.
Here's a question:
would this recurrence give integers?
a_{n}*a_{n-3}=a_{n-1}*a_{n-2}+a_{n-1}+a_{n-2}
1,1,1,3,7,31,85,393,1093,5071...
:) this could be something to study
Here's some more to consider:
use this pattern
A
B C D E=(C^2+BD)/A
E
1 1 1 1 1
1 1 1 1 1
2 2 2 2 2
8 8 8 8 8
64 64 64 64 64
These number count the perfect matchings of the aztec diamond. This
recurrence is really a collapsing of a 3d recurrence. Imagine three
sheets of paper, the top is the same as the bottom so I won't draw
in the bottom sheet.
________________
/ /
/ / note: A is on the top sheet
/ A /______ B, C, C', and D are on the middle sheet
/________C'____/ / E (not shown) is on the bottom sheet
/ B D /
/ C / rule: AE = BD + CC'
/__________________/
Make each sheet to have free variables with each sheet the same as all
the other sheets. Now iterate the pattern and you'll get Laurent
polynomials again!
Here's a rule:
a_{n-1}*a_{n+1}=a_{n}^2+a_{n}^2 <---one dimensional, you get numbers
(or monomials with two variables
and a huge coefficients, if
you start with initial conditions
a_0 = x, a_1 = y)
E=(C^2+BD)/A <----two dimensional, you get polynomials with large
coefficients (but no as large as with the one-dimensional
recurrence
the plane thing above <-------three dimensional, you get polynomials with
coefficients of 1
if you lift something and coefficients bigger than 1 then you haven't
lifted it far enough.
how far you lift is related to how fast the one dimensional recurrence grows.
***break time *****
Now we're going to name the Markoff Brothers:
a^2+b^2+c^2=3abc <-----'Arpo (Harpo)
a^2+b^2+2*c^2=4abc <------Chico
a^2+2*b^2+3*c^2=6abc <------Groucho
(the letters "A", "C", and "G" relate to the grown-up ways of
understanding certain tilings of the plane by triangles, in
terms of Lie algebras)
**Jim will send a link to neil herriots work**
Hal wants to talk about the tangent circle thing.
if you have two circles tangent OO <---those are circles
and you keep flipping them over each other you can get all the odd numbers
as their curvatures. the relationship between curvatures is:
c=a+2b where
c=new curvature
a=circle you are flipping with respect to
b=circle that you are actually flipping
now that we know this, we can hypothesize that we expect integers when
flipping over a large chain of circles OOOOOOOOOOOOOOOOO
but we have not proven it.
Jim: let's leave this cuz it's not really combinatorics.
**Jim will look at David Boyd's work and tell us about it (appolonian
gasket related stuff)**
Carl (and Sam): We were looking at induction on Newton's method. You have to
multiply 2 double sums and it's hard.
Jim: go to a simpler case like sqrt(2) or golden ratio and see if you can
prove it there. it might give more info.
Sam: could you explain number windows Jim?
Jim: I dont really understand them myself. Look at the Intro to the
Encyclopedia of Integer Sequences or google for "Fred Lunnon"+FRS+cookbook.
Someone asked how to see if a sequence can be satisfied by a linear equation.
Jim: look at the determinant of a hankel or toplitz matrix
Hankel Matrix for fibonacci:
| 1 2 3 |
| 2 3 5 | = 0
| 3 5 8 |
Toplitz Matrix for fibonacci:
| 3 5 8 |
| 2 3 5 | = 0
| 1 2 3 |
Here's some maple code that Jim gave us to run the octahedron recurrence:
F := proc(n,i,j) option remember; if n < 2 then x(n,i,j) else
simplify((F(n-1,i-1,j)*F(n-1,i+1,j)+F(n-1,i,j-1)*F(n-1,i,j+1))
/F(n-2,i,j));fi;end;
or we can do a simpler recurrence:
G := proc(n,k) option remember; if n<2 then x(n,k) else
simplify((G(n-1,k-1)*G(n-1,k+1)+1)/G(n-2,k));fi;end;
Now we have an interlude, Steven thinks that he has a proof of how many
solutions there are to 'Arpo Markoff mod(p). Martin tried to explain it
(cuz steven had to leave), but he did it too quick and I didn't get it.
Jim want to look at this recurrence.
A
B C D AE=C^3+BD
E
s(n+1) = (s(n)^3+s(n)^2)/s(n-1)
1,1,2,12,936,68408496....
we're not sure if this is a good sign or not. only 6 terms, but they are
integers.
here's the code.
F := proc(n,i,j) option remember; if n<2 then x(n,i,j) else
simplify((F(n-1,i-1,j)*F(n-1,i+i,j)+F(n-1,i,j-1)*F(n-1,i,j+1)*F(n-1,i,j))
/(F(n-2,i,j)));fi; end;
Carl is going to check the laurentness of this recurrence.
That's it for today.