2 The family of metallic means

This is the set of positive irrational numbers obtained by taking $G(0)=$ $a=$ $G(1)=$ $b=1$ in equation (1.3) and considering different values for the parameters $p$ and $q$.

D e f i n i t i o n. The metallic means family (MMF) is the set of positive eigenvalues of the matrix (1.6) for different values of natural number $p$ and integer $q$ with $p^2+4q>0.$

All the members of this family are positive quadratic irrational numbers that are the positive solutions of quadratic equation (1.4).

Let us begin with $x^2 - px - 1 = 0$. Then it is very easy to find the members of the MMF that satisfy this equation, expanding them in continued fractions. In fact, if $p = q= 1$ then $x^2 = x + 1$, which can be written $x=1+\frac{1}{x}$. Replacing iteratively the value of $x$ of the second term, we have $x={\displaystyle 1+\frac{1}{\displaystyle 1 + \frac{1}{1+\frac{1}{\ddots }}}},$ that is, $x = [1,1,1,\ldots] = [\overline{1}]$, a purely periodic continued fraction that defines the Golden Mean,

$$\phi=[\overline{1}] =\frac{1+\sqrt{5}}{2}.$$ (2.1)

Analogously, if $p = 2$ and $q = 1$ we obtain the Silver Mean:

$$\sigma_{Ag}=\lim_{n\to\infty}\frac{G(n+1)}{G(n)}=[\overline{2}] = {1+\sqrt{2}},$$ (2.2)

another purely periodic continued fraction. If $p = 3$ and $q = 1$, we get the Bronze Mean:

$$\sigma_{Br}=\lim_{n\to\infty}\frac{G(n+1)}{G(n)}=[\overline{3}] = \frac{3+\sqrt{13}}{2}.$$ (2.3)

For $p = 4,\; q = 1$, the metallic mean $\sigma_p^q$ is $\sigma_4^1 = 2 + \sqrt 5 = [ \overline 4 ] = \phi^3,$ a striking result related to the continued fraction expansion of odd powers of the Golden Mean. It is interesting to mention that odd powers as well as even powers of the Golden Mean have interesting and different continued fraction expansions in terms of Lucas numbers $L(n)$ ( $L(n+1)=L(n)+L(n-1)$, as for $F(n)$ but $L(0)=2,\;L(1)=1$) [1]. The remaining metallic means are

$$\sigma_p^1=[\overline{p}]=\frac{p+\sqrt{p^2+4}}{2},\;p=5,6,7,\ldots,$$

purely periodic continued fraction expansions. The slowest converging one of all them is the Golden Mean, since all its denominators are the smallest possible (ones). An elegant way of stating this result is

The Golden Mean $\phi$ is the most irrational of all irrational numbers.

If, instead, we consider equation

$$x^2 - x - q = 0,$$ (2.4)

we have for $q = 1$ again the Golden Mean. If $p = 1$ and $q = 2,$ we obtain the Copper Mean, $\sigma_{Cu}=2=[2,\overline 0],$ a periodic continued fraction. If $p = 1$ and $q = 3,$ we get the Nickel Mean, $\sigma_{Ni} =\lim_{n \to \infty } \frac{G(n + 1)}{G(n)} = \frac{1 + \sqrt{13} }{2}=[2, \overline 3 ].$ Analogously

$$\begin{array}{l} \sigma_1^4=[2,\overline{1,1,3}];\quad\sigma... ...{1,5}];\quad \sigma_1^{12}=4= [4,\overline 0 ] \\ \end{array}$$

and all these members of the MMF are of the form $[m,\overline {n_1, n_2 ,\ldots}].$

Precisely,

$${\rm if} \;n^2-n\le q<n^2+n,\; {\rm then}\; \ \sigma _1^q=[n,\overline {n_1,n_2,\ldots}],$$

$${\rm if} \;q=n^2-n\; {\rm then}\; \sigma _1^q=[n,\overline {0}],$$

$${\rm if} \;q=n^2-n+1,\; {\rm then}\; \ \sigma _1^q=[n,\overline {2n-1}],$$

$${\rm if} \;q=n^2+n,\; {\rm then}\; \ \sigma _1^q=[n,\overline {1,2n-1}].$$

Indeed, we have $4n^2-4n+1\le 1+4q<4n^2+4n+1$, whence $n\le\frac{1+\sqrt{1+4q}}{2}=\sigma _1^q<n+1$, which proves the first assertion. The second is obvious. Let $q=n^2-n+1$. Then set $x=n+\frac{1}{y}.$ From (2.4) we obtain $\frac{2n-1}{y}+\frac{1}{y^2}-1=0$ and $y=2n-1+\frac{1}{y},$ $y=[\overline {2n-1}],$ $x=\sigma _1^q=[n,\overline {2n-1}].$ Finally, let $q=n^2+n-1$. Then set again $x=n+\frac{1}{y}$ and obtain $\frac{2n-1}{y}+\frac{1}{y^2}-(2n-1)=0$ and $(2n-1)y=(2n-1)+\frac{1}{y},$ $y= 1+\frac{1}{(2n-1)y}= 1+\frac{1}{\displaystyle 2n-1+\frac{1}{\displaystyle y}} = [\overline {1,2n-1}],$ $x=\sigma _1^q=[n,\overline {1,2n-1}].$