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Metallic mean

Generalization of golden and silver ratios

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Source updated July 22, 2026

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

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

Metallic means are (successive) derivations of the golden ( $n=1$ ) and silver ratios ( $n=2$ ), and share some of their interesting properties. The term "bronze ratio" ( $n=3$ ) (cf. Golden Age and Olympic Medals) and even metals such as copper ( $n=4$ ) and nickel ( $n=5$ ) are occasionally found in the literature.

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 nth 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 nth 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.

01Geometry

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

Consider a rectangle such that the ratio of its length L to its width W is the nth metallic ratio. If one remove from this rectangle n 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.

02Powers

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 K₀ = 0 and K₁ = 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

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

End of the proof.

One has also

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

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 $a^{n}b^{n}=(ab)^{n}=-1$ 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

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

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

In particular, one has

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

and, in general,

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

where

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

For even powers, things are more complicated. If n is a positive even integer then

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

Additionally,

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

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

For the square of a metallic ratio we have: $S_{m}^{2}=[m{\sqrt {m^{2}+4}}+(m+2)]/2=(p+{\sqrt {p^{2}+4}})/2$

where $p=m{\sqrt {m^{2}+4}}$ lies strictly between $m^{2}+1$ and $m^{2}+2$ . Therefore

$S_{m^{2}+1}<S_{m}^{2}<S_{m^{2}+2}$

03Generalization

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

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

04Relation to half-angle cotangent

A tangent half-angle formula gives $\cot \theta ={\frac {\cot ^{2}{\frac {\theta }{2}}-1}{2\cot {\frac {\theta }{2}}}}$ which can be rewritten as $\cot ^{2}{\frac {\theta }{2}}-(2\cot \theta )\cot {\frac {\theta }{2}}-1=0\,.$ That is, for the positive value of ${\textstyle \cot {\frac {\theta }{2}}$ , the metallic mean $S_{2\cot \theta }=\cot {\frac {\theta }{2}}\,,$ which is especially meaningful when ${\textstyle 2\cot \theta$ is a positive integer, as it is with some Pythagorean triangles.

Metallic Ratios in Primitive Pythagorean Triangles

05Relation to Pythagorean triples

For a primitive Pythagorean triple, a² + b² = c², with positive integers a < b < c that are relatively prime, if the difference between the hypotenuse c and longer leg b is 1, 2 or 8 then the Pythagorean triangle exhibits a metallic mean. Specifically, the cotangent of one quarter of the smaller acute angle of the Pythagorean triangle is a metallic mean.

More precisely, for a primitive Pythagorean triple (a, b, c) with a < b < c, the smaller acute angle α satisfies $\tan {\frac {\alpha }{2}}={\frac {c-b}{a}}\,.$ When c − b ∈ {1, 2, 8}, we will always get that $n=2\cot {\frac {\alpha }{2}}={\frac {2a}{c-b}}$ is an integer and that $\cot {\frac {\alpha }{4}}=S_{n}\,,$ the n-th metallic mean.

The reverse direction also works. For n ≥ 5, the primitive Pythagorean triple that gives the n-th metallic mean is given by (n, n²/4 − 1, n²/4 + 1) if n is a multiple of 4, is given by (n/2, (n² − 4)/8, (n² + 4)/8) if n is even but not a multiple of 4, and is given by (4n, n² − 4, n² + 4) if n is odd. For example, the primitive Pythagorean triple (20, 21, 29) gives the 5th metallic mean; (3, 4, 5) gives the 6th metallic mean; (28, 45, 53) gives the 7th metallic mean; (8, 15, 17) gives the 8th metallic mean; and so on.

06Numerical values

First metallic means
n	Ratio	Value	Name
$n$	${\frac {n+{\sqrt {4+n^{2}}}}{2}}={\frac {n}{2}}+{\sqrt {1+\left({\frac {n}{2}}\right)^{2}}}$
0	${\frac {0+{\sqrt {4}}}{2}}=0+{\sqrt {1}}$	1
1	${\frac {1+{\sqrt {5}}}{2}}$	1.618033988...	Golden
2	${\frac {2+{\sqrt {8}}}{2}}=1+{\sqrt {2}}$	2.414213562...	Silver
3	${\frac {3+{\sqrt {13}}}{2}}$	3.302775637...	Bronze
4	${\frac {4+{\sqrt {20}}}{2}}=2+{\sqrt {5}}$	4.236067977...
5	${\frac {5+{\sqrt {29}}}{2}}$	5.192582403...
6	${\frac {6+{\sqrt {40}}}{2}}=3+{\sqrt {10}}$	6.162277660...
7	${\frac {7+{\sqrt {53}}}{2}}$	7.140054944...
8	${\frac {8+{\sqrt {68}}}{2}}=4+{\sqrt {17}}$	8.123105625...
9	${\frac {9+{\sqrt {85}}}{2}}$	9.109772228...
10	${\frac {10+{\sqrt {104}}}{2}}=5+{\sqrt {26}}$	10.099019513...

07Relation to Aperiodic Order

The $k$ -th metallic mean serves as the inflation ratio for one-dimensional substitution tilings, such as $a\to a^{k}b$ and $b\to a$ . These sequences exhibit long-range aperiodic order. By applying an interval removal process to these tilings, one can construct self-similar Cantor sets where the Hausdorff dimension is determined by the metallic mean scaling factor.

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Sources and credits

This article is adapted from the Wikipedia article “Metallic mean”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

Images, from Wikimedia Commons:

Fibonacci spiral 34.svg by Dicklyon, Public domain
Gold, silver, and bronze rectangles.svg by Rubber Duck (☮ • ✍), Public domain
AgamRiaSanga.png by Bababoidbl, CC0

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Metallic mean

01Geometry

02Powers

03Generalization

04Relation to half-angle cotangent

05Relation to Pythagorean triples

06Numerical values

07Relation to Aperiodic Order

Sources and credits

Related topics

Constant (mathematics)

Average

Ratio