Metallic mean

The metallic mean (also metallic ratio, metallic constant, or noble means ) of a natural number $n$ is a positive real number, denoted here $$S_n,$$ that satisfies the following equivalent characterizations: [n;n,n,n,n,\dots] = n + \cfrac{1}{n+\cfrac{1} {n+\cfrac{1} {n+\cfrac{1} {n+\ddots\,}}}} $$
 * the unique positive real number $$x$$ such that $x=n+\frac 1x$
 * the positive root of the quadratic equation $$x^2-nx-1=0$$
 * the number $\frac{n+\sqrt{n^2+4}}2$
 * the number whose expression as a continued fraction is

Metallic means are generalizations of the golden ratio ($$n=1$$) and silver ratio ($$n=2$$), and share some of their interesting properties. The term "bronze ratio" ($$n=3$$), and terms using other metals names (such as copper or nickel), are occasionally used to name subsequent metallic means.

In terms of algebraic number theory, the metallic means are exactly the real quadratic integers that are greater than $$1$$ and have $$-1$$ as their norm.

The defining equation $$x^2-nx-1=0$$ of the $n$th metallic mean is the characteristic equation of a linear recurrence relation of the form $$x_k=nx_{k-1}+x_{k-2}.$$ It follows that, given such a recurrence the solution can be expressed as
 * $$x_k=aS_n^k+b\left(\frac{-1}{S_n}\right)^k,$$

where $$S_n$$ is the $n$th metallic mean, and $a$ and $b$ are constants depending only on $$x_0$$ and $$x_1.$$ Since the inverse of a metallic mean is less than $1$, this formula implies that the quotient of two consecutive elements of such a sequence tends to the metallic mean, when $k$ tends to the infinity.

For example, if $$n=1,$$ $$S_n$$ is the golden ratio. If $$x_0=0$$ and $$x_1=1,$$ the sequence is the Fibonacci sequence, and the above formula is Binet's formula. If $$n=1, x_0=2, x_1=1$$ one has the Lucas numbers. If $$n=2,$$ the metallic mean is called the silver ratio, and the elements of the sequence starting with $$x_0=0$$ and $$x_1=1$$ are called the Pell numbers. The third metallic mean is sometimes called the "bronze ratio".

Geometry


The defining equation $x=n+\frac 1x$ of the $n$th metallic mean induces the following geometrical interpretation.

Consider a rectangle such that the ratio of its length $n$ to its width $n$ is the $n$th metallic ratio. If one remove from this rectangle $L$ squares of side length $W$, one gets a rectangle similar to the original rectangle; that is, a rectangle with the same ratio of the length to the width (see figures).

Some metallic means appear as segments in the figure formed by a regular polygon and its diagonals. This is in particular the case for the golden ratio and the pentagon, and for the silver ratio and the octagon; see figures.

Powers
Denoting by $$S_m$$ the metallic mean of m one has
 * $$ S_{m}^n = K_n S_m + K_{n-1} ,$$

where the numbers $$K_n$$ are defined recursively by the initial conditions $K_{0} = 0$ and $K_{1} = 1$, and the recurrence relation
 * $$ K_n = mK_{n-1} + K_{n-2}. $$

Proof: The equality is immediately true for $$n=1.$$ The recurrence relation implies $$K_2=m,$$ which makes the equality true for $$k=2.$$ Supposing the equality true up to $$n-1,$$ one has
 * $$\begin{align}

S_m^n & = mS_m^{n-1}+S_m^{n-2} &&\text {(defining equation)}\\ & = m(K_{n-1}S_n + K_{n-2})+ (K_{n-2}S_m+K_{n-3}) &&\text{(recurrence hypothesis)}\\ & = (mK_{n-1}+K_{n-2})S_n +(mK_{n-2}+K_{n-3}) &&\text{(regrouping)}\\ & = K_nS_m+K_{n-1} &&\text{(recurrence on }K_n). \end{align}$$ End of the proof.

One has also
 * $$ K_n = \frac{S_m^{n+1} - (m-S_m)^{n+1}}{\sqrt{m^2 + 4}} . $$

The odd powers of a metallic mean are themselves metallic means. More precisely, if $n$ is an odd natural number, then $$S_m^n=S_{M_n},$$ where $$M_n$$ is defined by the recurrence relation $$M_n=mM_{n-1}+M_{n-2}$$ and the initial conditions $$M_0=2$$ and $$M_1=m.$$

Proof: Let $$a=S_m$$ and $$b=-1/S_m.$$ The definition of metallic means implies that $$a+b=m$$ and $$ab=-1.$$ Let $$M_n=a^n+b^n.$$ Since $$a^nb^n =(ab)^n=-1$$ if $n$ is odd, the power $$a^n$$ is a root of $$x^2- M_n-1=0.$$ So, it remains to prove that $$M_n$$ is an integer that satisfies the given recurrence relation. This results from the identity
 * $$\begin{align}a^n+b^n &= (a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+a^{n-2})\\

&= m(a^{n-1}+b^{n-1})+(a^{n-2}+a^{n-2}). \end{align}$$ This completes the proof, given that the initial values are easy to verify.

In particular, one has
 * $$\begin{align}

S_m^3 &= S_{m^3 + 3m} \\ S_m^5 &= S_{m^5 + 5m^3 + 5m} \\ S_m^7 &= S_{m^7 + 7m^5 + 14m^3 + 7m} \\ S_m^9 &= S_{m^9 + 9m^7 + 27m^5 + 30m^3 + 9m} \\ S_m^{11} &= S_{m^{11} + 11m^9 + 44m^7 + 77m^5 + 55m^3 + 11m} \end{align}$$ and, in general,
 * $$ S_m^{2n+1} = S_M,$$

where
 * $$M=\sum_{k=0}^n {{2n+1} \over {2k+1}} {{n+k} \choose {2k}} m^{2k+1}.$$

For even powers, things are more complicate. If $n$ is a positive even integer then
 * $$ {S_m^n - \left\lfloor S_m^n \right\rfloor} = 1 - S_m^{-n}. $$

Additionally,
 * $$ {1 \over {S_m^4 - \left\lfloor S_m^4 \right\rfloor}} + \left\lfloor S_m^4 - 1 \right\rfloor = S_{\left(m^4 + 4m^2 + 1\right)} $$
 * $$ {1 \over {S_m^6 - \left\lfloor S_m^6 \right\rfloor }} + \left\lfloor S_m^6 - 1 \right\rfloor = S_{\left(m^6 + 6m^4 + 9m^2 +1\right)}. $$

Generalization
One may define the metallic mean $$S_{-n}$$ of a negative integer $−n$ as the positive solution of the equation $$x^2-(-n)-1.$$ The metallic mean of $−n$ is the multiplicative inverse of the metallic mean of $n$:
 * $$ S_{-n}=\frac{1}{S_n}.$$

Another generalization consists of changing the defining equation from $$x^2-nx-1 =0 $$ to $$x^2-nx-c =0 $$. If
 * $$ R=\frac{n\pm\sqrt{n^2+4c}}{2}, $$

is any root of the equation, one has
 * $$ R - n= \frac{c}{R}. $$

The silver mean of m is also given by the integral
 * $$ S_m = \int_0^m {\left( {x \over {2\sqrt{x^2+4}}} + {{m+2} \over {2m}} \right)} \, dx. $$

Another form of the metallic mean is
 * $$ \frac{n+\sqrt{n^2+4}}{2} = e^{\operatorname{arsinh(n/2)}}. $$