Super Golden Ratio

[Albert Chan]

ψ^3 = (ψ^2 + 1)
ψ^4 = (ψ^2 + 1)*ψ = (ψ^2 + 1) + ψ = (1*ψ^2 + ψ + 1)
ψ^5 = (ψ^2 + 1)*ψ^2 = (ψ^2 + ψ + 1) + ψ^2 = (2*ψ^2 + ψ + 1)
ψ^6 = (ψ^2 + 1)^2 = (ψ^2 + ψ + 1) + 2*ψ^2 + 1 = (3*ψ^2 + ψ + 2)
...
ψ^n = S(n-1)*ψ^2 + S(n-3)*ψ + S(n-2)
ψ^(2n) = S(2n-1)*ψ^2 + S(2n-3)*ψ + S(2n-2)
ψ^(3n) = S(3n-1)*ψ^2 + S(3n-3)*ψ + S(3n-2)

ψ^(3n) = k1*ψ^(2n) + k2*ψ^n + k3

k1 = round(ψ^n), for n ≥ 6      → ψ ≈ k1^(1/n)
k3 = (αβγ)^n = 1

For big n, S(3n-1)/S(3n-3) is better estimate for ψ^2, than S(2n-1)/S(2n-3).
Thus, we expected k2 ≠ 0. But, there are exceptions, when n is small.

n = 1: ψ^3 = ψ^2 + 1      // n=1 --> k2=0

n=11: S(3n-1)/S(3n-3) = 85626/39865 = (67*1278)/(67*595) = 1278/595 = S(2n-1)/S(2n-3)

ψ^33 = 67*ψ^22 + 1      // n=11 --> k2=0

ψ^11 = 67 + 1/ψ^22 = 67 + 1/(67 + 1/ψ^22)^2

(08-09-2022 10:45 AM)Gerson W. Barbosa Wrote:  Or, for more digits,

(67 + 1/(67^2 + 2/(67 + 3/(2×67^2 + 11/(3×67)))))^(1/11)
= 1.46557123187676802665673122475

This also explained why ψ^33 is very close to integer.
With cubing of roots, "linear" term is still small.

(-ψ^11)^3 + 67*(-ψ^11)^2 + 1 = 0

lua> b, c = 67, 0
lua> b*(b*b-3*c)+3, c*(c*c-3*b)+3
300766      3

(-ψ^33)^3 + 300766*(-ψ^33)^2 + 3*(-ψ^33) + 1 = 0

ψ^33 = 300766 - 3/ψ^33 + 1/ψ^66 ≈ 300766 - 3/300766 + 1/300766^2 ≈ 300765.9999900255