Upper Bounds and Lower Bounds on the members of the orbit of the generalized Ladas Equation x[n+1]=(p+x[n]+x[n-1]))/x[n-2] By Shalosh B. Ekhad Let a,b,c,p be a positive real numbers, and consider the terms of the non-li\ near recurrence p + x[n - 1] + x[n] x[n + 1] = ------------------- x[n - 2] and in Maple notation x[n+1] = (p+x[n-1]+x[n])/x[n-2] Subject to the initial conditions x[1] = a, x[2] = b, x[3] = c Then x[n] is always between m and M are the maximum that X can take in the b\ ounded surface 2 2 2 2 2 2 2 2 X Z a b c - X Y Z a c + X Z a b c + X Y a b c - X Y Z a c - X Y Z a b 2 3 2 2 2 3 - X Y Z a c - X Y Z b - X Y Z b c - X Y Z b p + X Z a b c + Y a b c 2 2 2 + Y Z a b c + Y a b c p - X Y Z a b - X Y Z a c - X Y Z b - X Y Z b c 2 - X Y Z b p + X Y a b c + X Z a b c + Y a b c + Y Z a b c + Y a b c p = 0 and in Maple notation X^2*Z^2*a*b*c-X*Y*Z*a^2*c^2+X^2*Z*a*b*c+X*Y^2*a*b*c-X*Y*Z*a^2*c-X*Y*Z*a*b^2-X*Y *Z*a*c^2-X*Y*Z*b^3-X*Y*Z*b^2*c-X*Y*Z*b^2*p+X*Z^2*a*b*c+Y^3*a*b*c+Y^2*Z*a*b*c+Y^ 2*a*b*c*p-X*Y*Z*a*b-X*Y*Z*a*c-X*Y*Z*b^2-X*Y*Z*b*c-X*Y*Z*b*p+X*Y*a*b*c+X*Z*a*b*c +Y^2*a*b*c+Y*Z*a*b*c+Y*a*b*c*p = 0 Proof: It is readily seen, by pluging in the recurrence that for any three consecut\ ive terms of the orbit,x[n-2]=X,x[n-1]=Y,x[n]=Z we have the following invariant 2 2 2 2 2 (x[n] x[n - 2] + p x[n - 1] + x[n] x[n - 2] + x[n] x[n - 2] 2 2 3 + x[n] x[n - 1] + x[n - 2] x[n - 1] + x[n - 1] + p x[n - 1] 2 + x[n] x[n - 2] + x[n] x[n - 1] + x[n - 2] x[n - 1] + x[n - 1] )/(x[n] x[n - 1] x[n - 2]) and in Maple notation (x[n]^2*x[n-2]^2+p*x[n-1]^2+x[n]^2*x[n-2]+x[n]*x[n-2]^2+x[n]*x[n-1]^2+x[n-2]*x[ n-1]^2+x[n-1]^3+p*x[n-1]+x[n]*x[n-2]+x[n]*x[n-1]+x[n-2]*x[n-1]+x[n-1]^2)/x[n]/x [n-1]/x[n-2] Plugging in we n=3, we have that x[n-2]=X, x[n-1]=Y, and x[n]=Z must obey 2 2 2 2 2 3 2 2 2 (X Z + X Z + X Y + X Z + Y + Y Z + Y p + X Y + X Z + Y + Y Z + Y p)/(Z Y X) - ( 2 2 2 2 2 3 2 2 2 a c + a c + a b + a c + b + b c + b p + a b + a c + b + b c + b p) /(c b a) = 0 simplifying, we see that 2 2 2 2 2 2 2 2 X Z a b c - X Y Z a c + X Z a b c + X Y a b c - X Y Z a c - X Y Z a b 2 3 2 2 2 3 - X Y Z a c - X Y Z b - X Y Z b c - X Y Z b p + X Z a b c + Y a b c 2 2 2 + Y Z a b c + Y a b c p - X Y Z a b - X Y Z a c - X Y Z b - X Y Z b c 2 - X Y Z b p + X Y a b c + X Z a b c + Y a b c + Y Z a b c + Y a b c p = 0 Both lower bounds and upper bounds must be the smallest and largest values t\ hat X can take on the above bounded surface. Let's illustrate the theorem with p=5, a=3,b=7,c=11 The smallest and largest member among the first 20000 members are [1.186482953, 11.01831510] ------------------------------------- ------------------------------ This took, 1.520, seconds.