SSL Meeting
02/24/04

3:45 pm meeting adjourns

Sam is the snack provider today
Paul is the note taker today
Hal is the snack provider on 02/26/04
Emilie is the note taker on 02/26/04

Jim likes the motivation that we left off with last time, so he opens the 
computer room to kick off the meeting.

Brendan, Carl and Sam have been working on trying to prove the Newton's 
Method coefficient identities analytically.
Brendan says he thinks it might be easier to prove such an identity 
combinatorially, since the computation involved is intense.

Emilie, Hal and Paul are trying to show that the sequence given by the 
recurrence a(n)a(n-7)=a(n-1)a(n-6)+a(n-3)+a(n-4) give integers when 
beginning with a(0)=..=a(6)=1.  Tried by induction and relative primality.  
Then found out that relative primality couldn't work in the general case, 
since it didn't even work in specific examples.

(Break for snack)

Carl: We are looking now more at Newton's Method for particular values of 
a, b and c.  Sam had an intuition as to why we get fibonacci numbers when 
looking at the expansion of the sum over t and k of 
	binomial(2^n-k-t-1,t+1)*x^(2^n-k).  
It has something to do with the fact that we can get fibonacci numbers by 
taking the sum over i of binomial(k-i,i)x^k.

Jim: If we can prove this formula, it will show why we get the nice prime 
factorizations that we do, because t and k uniquely define the term.

Sam: We have been looking at specific values for a, b and c of 1,0,and -m 
respectively.  In the specific case where m=2, we get F(n)=sum over k of 
m^k*x^(2k) which is related to P(n).

Jim suggests proof by machine.  Wilf and Zielberger have created a program 
to write up proofs of binomial coefficient identities that we should take 
a look at so that we can keep moving.  There are probably references to it 
in Wilf's book A=B and also Zielberger's webpage.

Carl says he will look into it.

Emilie has been looking at proving integrality for the sequence given by 
a(n)a(n-7)=a(n-1)a(n-6)+a(n-3)+a(n-4) with a(0)=..=a(6)=1.  We can't prove 
it using relative primality, as Jim suggested, since the numbers are not 
relatively prime.

Jim suggests that, since we can see the 1500th term in the sequence in 
Maple, that the sequence must be growing simple exponentially, and so a 
possible proof of integrality would be to find a linear recurrence that 
satisfies the given sequence.

Paul says he will look into the David Gale article to see if the general 
sequences a(n)a(n-k)=a(n-1)a(n-k+1)+a(n-(k-1)/2)+a(n-(k+1)/2) for k odd 
and a(n)a(n-k)=a(n-1)a(n-k+1)+a(n-k/2) for k even are known already.  
Also, he mentions that a(n)a(n-6)=a(n-2)a(n-4)+a(n-3) gives an integer 
sequence that does not necessarily increase from a(n) to a(n+1). (i.e. 
monotonic sequence).

(Back to the computer lab)

Jim finds that 
	a(n)a(n-7)=a(n-2)a(n-5)+a(n-1)+a(n-6) 
and 
	a(n)a(n-7)=a(n-3)a(n-4)+a(n-2)+a(n-5) 
also give integers to 100 terms if the starting terms are all 1's.  He 
suggests that it would be interesting to find a generalization of that 
for higher odd values of k.

(end meeting)