Notes 2004-03-30, by Hal Present were: Jim, Stephen, Hal, Emilie, Brendan, Carl, John. Thursday's notetaker: Emilie. Snacks on Thursday: Carl John ran 10 miles this morning. John wore a shirt that says "for a good prime, call 555-2357!" T-shirts with slogans are dumb. Everyone is being paid. Writeups: Brendan is updating the Newton's Recurrence (NR) paper [http://www.sit.wisc.edu/~bcyounger/] No need to CVS to do versioning. Brendan has written up to the "fractional linear transformation" part. What if the roots are the same in NR? Can we just appeal to continuity? (Aside: if f(x)=a(x-r)^2, then N^n(x)=r-(r-x)/2^n. Neat!) Jim says: You can show that successive Fibonacci numbers are relatively prime by induction. x \mapsto p(x)/p'(x) p(x) = x^2-x-1 x \mapsto (x^2+1)/(2x-1) 1/1 \mapsto 2/1 \mapsto 5/3 \mapsto 34/21 \mapsto ... (1,1,2,3,5,21,34,... are fibb. numbers) m/n \mapsto (m^2+n^2)/(n(2m-n)) (*) and the m/n formulation comes from the x formulation, but it will also give back the x formulation. Jim says something about affine vs. projective. x vs. m/n P_n and Q_n are relatively prime. Can we prove that? Is it true that m/n reduced gives (m^2+n^2)/(n(2m-n)) in reduced form? x \mapsto 1+1/x gives 1/1 \mapsto 2/1 \mapsto 3/2 \mapsto 5/3 \mapsto ... every fib. number! Is equiv to: m/n \mapsto (m+n)/m THEOREM: gcd(m,n)=1 ==> gcd(m^2+n^2,n(2m-n))=1 Jim gave an partial proof by contradiction. Stephen claims that the easier way to prove it would be to use arithmetic mod p. Jim's proof begins: Assume gcd(m,n)=1. Assume exists a prime \$p\$ that divides m^2+n^2 and n(2m-n). We want to show that such a prime must also divide m and n, violating our assumption. Since p|m^2+n^2 and p|2mn-n^2, then p|m^2+2mn=m(m+2n) (the difference). Then, p|m OR p|m+2n. Proceed on a case-by-base basis. Break. I note that (*) gives: for F_0=0 and F_1=1, F_{2n+1} = F_{n+1}^2 + F_{n}^2 and F_{2n} = F_n * ( 2 * F_{n+1} - F_{n} ) Switch gears to Somos-type sequences: Lifting is hard. We have a 2-d lift for the sequence family Emilie and Paul discovered (EP). Need a combinatorial model for the EP sequences. Analogy to s_n * s_{n-4} = s_{n-1} * s_{n-3} + s_{n-2}^2 which Hal says result from a hexagon-shaped matching --- but Jim meant to create an analogy with the slightly but crucially different recurrence s_n * s_{n-4} = s_{n-1} * s_{n-3} + s_{n-2} (not that there's a non-quadratic term). What does the graph look like? It's in Reid Barton's paper; someone should check. Refer to Paul's recursion: s_n * s_{n-5} = s_{n-1} + s_{n-2} * s_{n-3} + s_{n-4} [Note: During the meeting, I (Jim) called this the "Bernard LeCloitre recursion", but his first name is Benoit, his last name is Cloitre, and his recursion is s_n * s_{n-3} = s_{n-1} * s_{n-2} + s_{n-1} + s_{n-2} so I was triply wrong!] BC can be lifted to a 2-d lattice: . . . . . . . . . . . . . . . . . . D E F . . . . . A B C . . . . . . . . . . . F = (B+C*D+E)/A or a_{i,j} * a_{i-2,j-1} = a_{i-2,j} * a_{i,j-1} + a_{i-1,j} + a_{i-1,j-1} This recurrence lifts both BC and EP. I expect that there are ways to extend this recurrence to one of the form a_{i,j} * a_{i-2,j-1} = a_{i-2,j} * a_{i,j-1} + U a_{i-1,j} + V a_{i-1,j-1} where U and V carry subscripts (linear functions of i and j), in such a way that we get not just faithful polynomials but homogeneous faithful polynomials. These subscripted U and V variables should correspond to edges of a graph. At least, that's the hope! U and V are edge variables. Insert values here to find out something about the edges in the resulting graph. \ / u\ /v if degrees of u,v,and w all add \ / up to 1 in each term, then the Y picture probably looks like | that (they share a vertex)! |w | Edge variables are better than face variables. We recess to the other room.