Notetaker for today = Hal Notetaker for thursday = Brendan Notetaker for tuesday = Carl Snacks for today = Martin Snacks for thurdsday = Carl Brendan will email his URL to Jim. New markers are now in B107. While individual interrogations were going on: Sam and Hal looked at binomial(2*N,m)*Fib(2*N-m-1) = add(binomial(N,i)*binomial(N,(m-i))* (2*Fib(N-i-2)+Fib(N-i-1))*Fib(N-(m-i)-1),i=0..m); Is it hypergeometric? Maybe. Since Fib() is a hypergeometric sum, that sum is a sum of hypergeometric sums. Ick. Carl and Brendan have been looking at M\"obius tranformtions for the Newton Recurrence. Carl summarized what he did. a couple formulas for the notes... M is the complex mobius transformation taking the roots r1 -> 0 r2 -> inf we get: F(z) = M^(-1)([M(z)]^2) = (z^2 - r1*r2) / (2z - r1 - r2) the nth iterate of this is F^n(z) = M^(-1)([M(z)]^(2^n)) which (should) give us an explicit formula for the nth application of Newton's method (!) [..more next time..] Emilie and Paul got an exact formula for nth term of the the k=3 sequence. S(n) = (-1/11) * (-1)^n + (6/11) * (1)^n + (1/22*11^(1/2)-1/110*15^(1/2)*11^(1/2)-1/22*15^(1/2)+3/22) * (-1/2*15^(1/2)-1/2*11^(1/2))^n + (-1/22*11^(1/2)-1/110*15^(1/2)*11^(1/2)+1/22*15^(1/2)+3/22) * (1/2*15^(1/2)+1/2*11^(1/2))^n + (3/22-1/22*11^(1/2)-1/22*15^(1/2)+1/110*15^(1/2)*11^(1/2)) * (-1/2*15^(1/2)+1/2*11^(1/2))^n + (1/22*11^(1/2)+1/22*15^(1/2)+3/22+1/110*15^(1/2)*11^(1/2)) * (1/2*15^(1/2)-1/2*11^(1/2))^n If R is the original recurrence: R:=proc(n) option remember; if n<3 then 1 else return((R(n-1)*R(n-2)+R(n-1)+R(n-2))/R(n-3)); fi; end: They solve the corresponding polynomial, to get the function S(n), then test to see that S(n)=R(n). sols := { solve(x^6-14*x^4+14*x^2-1,x) }: a := sols[1]: b := sols[2]: c := sols[3]: d := sols[4]: e := sols[5]: f := sols[6]: assign(solve({r+s+t+u+v+w=R(0), r*a+s*b+t*c+u*d+v*e+w*f=R(1), r*a^2+s*b^2+t*c^2+u*d^2+v*e^2+w*f^2=R(2), r*a^3+s*b^3+t*c^3+u*d^3+v*e^3+w*f^3=R(3), r*a^4+s*b^4+t*c^4+u*d^4+v*e^4+w*f^4=R(4), r*a^5+s*b^5+t*c^5+u*d^5+v*e^5+w*f^5=R(5)},{r,s,t,u,v,w})): S:= proc(n) option remember; simplify(r*a^n+s*b^n+t*c^n+u*d^n+v*e^n+w*f^n); end: printf("S(n) = (%a) * (%a)^n + (%a) * (%a)^n + (%a) * (%a)^n \ + (%a) * (%a)^n + (%a) * (%a)^n + (%a) * (%a)^n\n\n",r,a,s,b, t,c,u,d,v,e,w,f); for n from 0 to 30 do printf("R(%a) - S(%a) = %a - %a = %q\n",n,n,R(n),S(n),R(n)-S(n)); end do: evalb(S(n)*S(n-3)-S(n-1)*S(n-2)-S(n-1)-S(n-2)=0); NEAT!