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.