The metallic mean (also metallic ratio, metallic constant, or noble mean^[1]) 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}}={\frac {2}{{\sqrt {n^{2}+4}}-n}}}
  
• 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 \,}}}}}}}}}
  

Metallic means are (successive) derivations of the golden (👁 {\displaystyle n=1}
) and silver ratios (👁 {\displaystyle n=2}
), and share some of their interesting properties. The term "bronze ratio" (👁 {\displaystyle n=3}
) (Cf. Golden Age and Olympic Medals) and even metals such as copper (👁 {\displaystyle n=4}
) and nickel (👁 {\displaystyle n=5}
) are occasionally found in the literature.^[2][3][a]

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

The defining equation 👁 {\displaystyle x^{2}-nx-1=0}
of the nth metallic mean is the characteristic equation of a linear recurrence relation of the form 👁 {\displaystyle x_{k}=nx_{k-1}+x_{k-2}.}
It follows that, given such a recurrence the solution can be expressed as

👁 {\displaystyle x_{k}=aS_{n}^{k}+b\left({\frac {-1}{S_{n}}}\right)^{k},}

where 👁 {\displaystyle S_{n}}
is the nth metallic mean, and a and b are constants depending only on 👁 {\displaystyle x_{0}}
and 👁 {\displaystyle 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 👁 {\displaystyle n=1,}
👁 {\displaystyle S_{n}}
is the golden ratio. If 👁 {\displaystyle x_{0}=0}
and 👁 {\displaystyle x_{1}=1,}
the sequence is the Fibonacci sequence, and the above formula is Binet's formula. If 👁 {\displaystyle n=1,x_{0}=2,x_{1}=1}
one has the Lucas numbers. If 👁 {\displaystyle n=2,}
the metallic mean is called the silver ratio, and the elements of the sequence starting with 👁 {\displaystyle x_{0}=0}
and 👁 {\displaystyle x_{1}=1}
are called the Pell numbers.

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.

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 complicated. 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)}.}

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

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

👁 {\displaystyle S_{m^{2}+1}<S_{m}^{2}<S_{m^{2}+2}}

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)x-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^[4]

👁 {\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^[4]

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

A tangent half-angle formula gives 👁 {\displaystyle \cot \theta ={\frac {\cot ^{2}{\frac {\theta }{2}}-1}{2\cot {\frac {\theta }{2}}}}}
which can be rewritten as 👁 {\displaystyle \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 👁 {\displaystyle 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.

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.^[5]

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

The 👁 {\displaystyle k}
-th metallic mean serves as the inflation ratio for one-dimensional substitution tilings, such as 👁 {\displaystyle a\to a^{k}b}
and 👁 {\displaystyle 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.^[19]

• Constant
• Mean
• Ratio
• Plastic ratio

• Stakhov, Alekseĭ Petrovich (2009). The Mathematics of Harmony: From Euclid to Contemporary Mathematics and Computer Science, p. 228, 231. World Scientific. ISBN 9789812775832.

• Cristina-Elena Hrețcanu and Mircea Crasmareanu (2013). "Metallic Structures on Riemannian Manifolds", Revista de la Unión Matemática Argentina.
• Rakočević, Miloje M. "Further Generalization of Golden Mean in Relation to Euler's 'Divine' Equation", Arxiv.org.