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.