The metallic mean (also metallic ratio, metallic constant, or noble means^[3]) of a natural number n is a positive real number, denoted here ${\displaystyle S_{n},}$ that satisfies the following equivalent characterizations:

• the unique positive real number ${\displaystyle x}$ such that ${\textstyle x=n+{\frac {1}{x}}}$
• the positive root of the quadratic equation ${\displaystyle x^{2}-nx-1=0}$
• the number ${\textstyle {\frac {n+{\sqrt {n^{2}+4}}}{2}}}$
• the number whose expression as a continued fraction is

  ${\displaystyle [n;n,n,n,n,\dots ]=n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+\ddots \,}}}}}}}}}$

Golden ratio within the pentagram and silver ratio within the octagon.

Metallic means are generalizations of the golden ratio (${\displaystyle n=1}$) and silver ratio (${\displaystyle n=2}$), and share some of their interesting properties. The term "bronze ratio" (${\displaystyle n=3}$), and terms using other metals names (such as copper or nickel), are occasionally used to name subsequent metallic means.^{[4] [5]}

As the golden ratio is connected to the pentagon (first diagonal/side), the silver ratio is connected to the octagon (second diagonal/side). As the golden ratio is connected to the Fibonacci numbers, the silver ratio is connected to the Pell numbers, and the bronze ratio is connected to OEIS: A006190. Each Fibonacci number is the sum of the previous number times one plus the number before that, each Pell number is the sum of the previous number times two and the one before that, and each "bronze Fibonacci number" is the sum of the previous number times three plus the number before that. Taking successive Fibonacci numbers as ratios, these ratios approach the golden mean, the Pell number ratios approach the silver mean, and the "bronze Fibonacci number" ratios approach the bronze mean.

This section does not cite any sources. (August 2020)

Gold, silver, and bronze ratios within their respective rectangles.

Denoting by ${\displaystyle S_{m}}$ the metallic mean of m one has

${\displaystyle S_{m}^{n}=K_{n}S_{m}+K_{n-1},}$

where the numbers ${\displaystyle K_{n}}$ are defined recursively by the initial conditions K₀ = 0 and K₁ = 1, and the recurrence relation

${\displaystyle K_{n}=mK_{n-1}+K_{n-2}.}$

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

${\displaystyle {\begin{aligned}S_{m}^{n}&=mS_{m}^{n-1}+S_{m}^{n-2}&&{\text{(defining equation)}}\\&=m(K_{n-1}S_{n}+K_{n-2})+(K_{n-2}S_{m}+K_{n-3})&&{\text{(recurrence hypothesis)}}\\&=(mK_{n-1}+K_{n-2})S_{n}+(mK_{n-2}+K_{n-3})&&{\text{(regrouping)}}\\&=K_{n}S_{m}+K_{n-1}&&{\text{(recurrence on }}K_{n}).\end{aligned}}}$

End of the proof.

One has also ^{[citation needed]}

${\displaystyle K_{n}={\frac {S_{m}^{n+1}-(m-S_{m})^{n+1}}{\sqrt {m^{2}+4}}}.}$

The odd powers of a metallic mean are themselves metallic means. More precisely, if n is an odd natural number, then ${\displaystyle S_{m}^{n}=S_{M_{n}},}$ where ${\displaystyle M_{n}}$ is defined by the recurrence relation ${\displaystyle M_{n}=mM_{n-1}+M_{n-2}}$ and the initial conditions ${\displaystyle M_{0}=2}$ and ${\displaystyle M_{1}=m.}$

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

${\displaystyle {\begin{aligned}a^{n}+b^{n}&=(a+b)(a^{n-1}+b^{n-1})-ab(a^{n-2}+a^{n-2})\\&=m(a^{n-1}+b^{n-1})+(a^{n-2}+a^{n-2}).\end{aligned}}}$

This completes the proof, given that the initial values are easy to verify.

In particular, one has

${\displaystyle {\begin{aligned}S_{m}^{3}&=S_{m^{3}+3m}\\S_{m}^{5}&=S_{m^{5}+5m^{3}+5m}\\S_{m}^{7}&=S_{m^{7}+7m^{5}+14m^{3}+7m}\\S_{m}^{9}&=S_{m^{9}+9m^{7}+27m^{5}+30m^{3}+9m}\\S_{m}^{11}&=S_{m^{11}+11m^{9}+44m^{7}+77m^{5}+55m^{3}+11m}\end{aligned}}}$

and, in general,^{[citation needed]}

${\displaystyle S_{m}^{2n+1}=S_{M},}$

where

${\displaystyle M=\sum _{k=0}^{n}{{2n+1} \over {2k+1}}{{n+k} \choose {2k}}m^{2k+1}.}$

For even powers, things are more complicate. If n is a positive even integer then^{[citation needed]}

${\displaystyle {S_{m}^{n}-\left\lfloor S_{m}^{n}\right\rfloor }=1-S_{m}^{-n}.}$

Additionally,^{[citation needed]}

${\displaystyle {1 \over {S_{m}^{4}-\left\lfloor S_{m}^{4}\right\rfloor }}+\left\lfloor S_{m}^{4}-1\right\rfloor =S_{\left(m^{4}+4m^{2}+1\right)}}$

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

This section does not cite any sources. (April 2024)

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

${\displaystyle S_{-n}={\frac {1}{S_{n}}}.}$

Another generalization consists of changing the defining equation from ${\displaystyle x^{2}-nx-1=0}$ to ${\displaystyle x^{2}-nx-c=0}$. If

${\displaystyle R={\frac {n\pm {\sqrt {n^{2}+4c}}}{2}},}$

is any root of the equation, one has

${\displaystyle R-n={\frac {c}{R}}.}$

The silver mean of m is also given by the integral^{[citation needed]}

${\displaystyle S_{m}=\int _{0}^{m}{\left({x \over {2{\sqrt {x^{2}+4}}}}+{{m+2} \over {2m}}\right)}\,dx.}$

Another form of the metallic mean is^{[citation needed]}

${\displaystyle {\frac {n+{\sqrt {n^{2}+4}}}{2}}=e^{\operatorname {arsinh(n/2)} }.}$

The metallic mean for any given integer ${\displaystyle N}$ can be constructed geometrically in the following way. Define a right triangle with sides ${\displaystyle A}$ and ${\displaystyle B}$ having lengths of ${\displaystyle 1}$ and ${\displaystyle N/2}$, respectively. The ${\displaystyle N}$th metallic mean ${\displaystyle M}$ is simply the sum of the length of ${\displaystyle B}$ and the hypotenuse, ${\displaystyle H}$.^[7]

For ${\displaystyle N=1}$,

${\displaystyle H={\sqrt {(N/2)^{2}+1^{2}}}={\sqrt {5/4}}=1.1180339...}$

and so

${\displaystyle M=B+H=1/2+1.1180339...=1.6180339...=}$ φ.

Setting ${\displaystyle N=2}$ yields the silver ratio.

${\displaystyle H={\sqrt {(2/2)^{2}+1^{2}}}={\sqrt {2}}=1.4142135...}$

Thus

${\displaystyle M=B+H=2/2+1.4142135...=2.4142135...}$

Likewise, the bronze ratio would be calculated with ${\displaystyle N=3}$ so

${\displaystyle H={\sqrt {(3/2)^{2}+1^{2}}}={\sqrt {13/4}}=1.8027756...}$

yields

${\displaystyle M=B+H=3/2+1.8027756...=3.3027756...}$

Non-integer arguments sometimes produce triangles with a mean that is itself an integer. Examples include N = 1.5, where

${\displaystyle H={\sqrt {(1.5/2)^{2}+1^{2}}}={\sqrt {1.5625}}=1.25}$

and

${\displaystyle M=B+H=1.5/2+1.25=2}$

which is simply a scaled-down version of the 3–4–5 Pythagorean triangle.

• Constant
• Mean
• Ratio
• Plastic ratio