SSL Notes for 2/19 Today's notetaker: Martin Today's snackbringer: Carl Tuesday's notetaker: Paul Tuesday's snackbringer: Martin Today, we had a visitor, Chris Henley of Cornell, a mathematical physicist who's done work on some statistical mechanics models that also are of interest to combinatorialists. In the beginning of the meeting, there were two groups working on things: John, Sam, and Carl talking about the combinatorial interpretation of Newton's method, and Paul and Emilie talking about some Somos-like recurrences. Jim mentioned how he began working with undergraduates, and that he found that if he worked with several undergraduates at once, they would work with each other and their mutual learning would benefit each other much more than working alone. Then, each student gave a short talk on what they were doing. Paul and Emilie looked at this recurrence: a(n)a(n-5) = a(n-1)a(n-4) + a(n-2) + a(n-3) = 1,1,1,1,1,3,5,9,17,65,117,227,449,173e7,3137,... If the first 5 terms are taken to be variables, the denominator appears to be Laurent. If the first 5 terms are 1, it has been verified to be integral for the first thousand terms or so. There were several properties which were cyclic for the number 5. For example, the ratio a(n) / a(n-1) for the integer version appears to cycle among 5 numbers. [Actually, later on it turned out that this is not true: the ratios get attracted to a limit-cycle of period 5, but don't actually hit the values they're converging towards. -- Jim] The number of terms multiplied in the denominator of the polynomial version never appears to be a multiple of five. Also, the coefficients of the polynomial in the denominator are low; even though the 14th iteration has over 1700 terms, the coefficients are no larger than 20. This may indicate a more difficult lift to two dimensions. Sam and Carl were looking at Brendan's formula, which seemed to indicate the following: 2^n 2^n \-- \-- / 2^n - k - t - 1 \ 1 lim > > | | x^k = ----------- n -> infty /__ /__ \ 2^n - k - 2t / 1 - x - x^2 k=0 t=0 = 1 + x + 2x^2 + 3x^3 + ... + F_ix^i + ... Or the more general formula P_n = sum(sum((-1)^t*(2^n choose k)* (2^n-k-t-1 choose 2^n-k-2t)*x^k,k=0..2^n),t=0..2^n) They had trouble with it unless they defined the binomial coefficient in the following way for negative numbers: -n choose 0 = 1 -n choose k = 0 (k not equals 0) Jim wondered if this 2^n thing was arbitrary and could be replaced by n, thus creating a new function P'_n where P'_(2^n) = P_n. Sam was also interested in the tiling of the plane with 30-30-120 degree triangles. He sent out an email about his understanding of Herriot's work here. Snacks consisted of Rice Krispy (R) treats made with real butter. After going to the lab, Emilie and Paul defined a more general version of their function: a(n,k)a(n-k) = a(n-1)a(n-(k-1)) + a(n-(k-1)/2) + a(n-(k+1)/2) (k odd) = a(n-1)a(n-(k-1)) + a(n-k/2) (k even) They have verified integrality for k up to 20 or so. Sam and Carl are somewhat convinced that P'_n is well-behaved, though there is still some disagreement.