Franklin T. Adams-Watters wrote:
I asked, what happens if the 1 in that equation is replaced by some other constant, c: a(n+1) = (a(n)+c)/a(n-1). The case c=0 gives us period 6: starting with x, y, the sequence continues y/x, 1/x, 1/y, x/y, and then loops. ... When you graph a(n) vs. a(n+1), with suitable parameters the result is an ovoid. ... The obvious next step is to find a non-trivial function f_c which satisfies the functional equation f_c(x,y) = f_c(y,(y+c)/x). ... This is where I got stuck - I wasn't able to find such an f_c. Maybe somebody with better analytic skills than I have will be able to.
Let f_c(x,y) = (c + (c+1)(x+y) + x^2 + y^2 + x^2 y + x y^2)/(x y). I wish I could claim that I used great analytic skills, but all I did was look at some special cases, assume that the curves satisfied polynomial equations of low degree, and use Mathematica to solve for the coefficients. After a few cases, the pattern was obvious, and led to the invariant f_c. I did notice that this recurrence is related to a 7-th order Somos recurrence: Define s(0), ..., s(4) arbitrarily and then define s(n) for n>=5 so that a(n) = s(n) s(n+5) / (s(n+2) s(n+3)). Then s satisfies the Somos recurrence s(n) s(n-7) = s(n-1) s(n-6) + c s(n-3) s(n-4). I'm sure there must be an easy derivation of f_c from known facts about Somos sequences, but I couldn't find one. Both Franklin's recurrence and Somos's have been studied by many people, so this is probably all well-known to someone. David W. Cantrell wrote:
Here's an example. Suppose a(1) = 1 and a(2) = 2 and c = 1/2. The sequence itself is not periodic. The continuous function, of which the sequence is a restriction, has a period of roughly 5.248; a graph of one cycle is shown at <http://img453.imageshack.us/img453/744/recursioncu0.gif>. Its maximum and minimum are roughly 2.549 and 0.741, resp.
The max and min are roots of 4x^4 - 80x^3 + 228x^2 - 128x + 1 = 0 Numerically, they're about 2.549193378506151504365717868828 and 0.741012926112203188665166488474 Dean Hickerson dean@math.ucdavis.edu