Super Golden Ratio

[Albert Chan]

(08-12-2022 05:46 PM)Thomas Klemm Wrote:  We assume that \(\alpha\), \(\beta\) and \(\gamma\) are the roots of \(x^3=x^2+1\).
Thus we have:

\(
\begin{align}
\alpha + \beta + \gamma &= 1 \\
\alpha \cdot \beta \cdot \gamma &= 1 \\
\end{align}
\)

For the direct formula \(S(n) = u \cdot \alpha^n + v \cdot \beta^n + w \cdot \gamma^n\) the constants \(u\), \(v\) and \(w\) are solved so that: \(S(1) = S(2) = S(3) = 1\)

This leads to:

\(
\begin{bmatrix}
\alpha & \beta & \gamma \\
\alpha^2 & \beta^2 & \gamma^2 \\
\alpha^3 & \beta^3 & \gamma^3 \\
\end{bmatrix}
\begin{bmatrix}
u \\
v \\
w \\
\end{bmatrix}
=
\begin{bmatrix}
1 \\
1 \\
1 \\
\end{bmatrix}
\)

This can be solved for \(u\) to get ...

Simpler approach is do partial fraction decomposition of generation function.
S(n) formula is coefficient of x^n, with RHS geometric series in normalized form.

\(\displaystyle
\frac{x}{(1 - α x)(1- β x) (1 - γ x)} =
\frac{u}{1 - α x} +
\frac{v}{1- β x} +
\frac{w}{1 - γ x}
\)

Mulitply both side by (1 - α x), then let x = 1/α, we solved u

\(\displaystyle u = \frac{1/α}{(1 - β/α) (1 - γ/α)}
= \frac{α}{(α-β)(α-γ)}\)

\(\displaystyle \frac{1}{u} = α - (β+γ) + \frac{βγ}{α} = α - (1-α) + (α-1) = (3α - 2)\)

\(\displaystyle \frac{1}{u} + \frac{1}{v} + \frac{1}{w}
= (3α\!-\!2) + (3β\!-\!2) + (3γ\!-\!2) = 3 - 6 = -3 \)

If we let x = 0, we get 0 = u + v + w \(\displaystyle \;⇒ \frac{1}{uv} + \frac{1}{uw} + \frac{1}{vw} = 0\)

\(\displaystyle \frac{1}{uvw}
= (3α\!-\!2)(3β\!-\!2)(3γ\!-\!2) = 27(αβγ) - 18(αβ\!+\!αγ\!+\!βγ) + 12(α\!+\!β\!+\!γ) - 8
= 27 + 12 - 8 = 31
\)

\(\displaystyle ⇒ (x-\frac{1}{u})(x-\frac{1}{v})(x-\frac{1}{w}) = x^3 + 3x^2 - 31\)