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

The metallic means (also ratios or constants) of the successive natural numbers are the continued fractions:

${\displaystyle n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+{\cfrac {1}{n+\ddots \,))))))))=[n;n,n,n,n,\dots ]={\frac {n+{\sqrt {n^{2}+4))}{2)).}$

The golden ratio (1.618...) is the metallic mean between 1 and 2, while the silver ratio (2.414...) is the metallic mean between 2 and 3. The term "bronze ratio" (3.303...), or terms using other names of metals (such as copper or nickel), are occasionally used to name subsequent metallic means.^[1][2] The values of the first ten metallic means are shown at right.^[3][4] Notice that each metallic mean is a root of the simple quadratic equation: ${\displaystyle x^{2}-nx=1}$, where ${\displaystyle n}$ is any positive natural number.

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.

Properties

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

These properties are valid only for integers m. For nonintegers the properties are similar but slightly different.

The above property for the powers of the silver ratio is a consequence of a property of the powers of silver means. For the silver mean S of m, the property can be generalized as

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

where

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

Using the initial conditions K₀ = 1 and K₁ = m, this recurrence relation becomes

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

The powers of silver means have other interesting properties:

If n is a positive even integer:

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

Additionally,

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

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

Also,

${\displaystyle S_{m}^{3}=S_{\left(m^{3}+3m\right)))$

${\displaystyle S_{m}^{5}=S_{\left(m^{5}+5m^{3}+5m\right)))$

${\displaystyle S_{m}^{7}=S_{\left(m^{7}+7m^{5}+14m^{3}+7m\right)))$

${\displaystyle S_{m}^{9}=S_{\left(m^{9}+9m^{7}+27m^{5}+30m^{3}+9m\right)))$

${\displaystyle S_{m}^{11}=S_{\left(m^{11}+11m^{9}+44m^{7}+77m^{5}+55m^{3}+11m\right)}.}$

In general:

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

The silver mean S of m also has the property that

${\displaystyle {\frac {1}{S_{m))}=S_{m}-m}$

meaning that the inverse of a silver mean has the same decimal part as the corresponding silver mean.

${\displaystyle S_{m}=a+b}$

where a is the integer part of S and b is the decimal part of S, then the following property is true:

${\displaystyle S_{m}^{2}=a^{2}+mb+1.}$

Because (for all m greater than 0), the integer part of S_m = m, a = m. For m > 1, we then have

${\displaystyle S_{m}^{2}=ma+mb+1}$

${\displaystyle S_{m}^{2}=m(a+b)+1}$

${\displaystyle S_{m}^{2}=m\left(S_{m}\right)+1.}$

Therefore, the silver mean of m is a solution of the equation

${\displaystyle x^{2}-mx-1=0.}$

It may also be useful to note that the silver mean S of −m is the inverse of the silver mean S of m

${\displaystyle {\frac {1}{S_{m))}=S_{(-m)}=S_{m}-m.}$

Another interesting result can be obtained by slightly changing the formula of the silver mean. If we consider a number

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

then the following properties are true:

${\displaystyle R-\lfloor R\rfloor ={\frac {c}{R))}$ if c is real,

${\displaystyle \left({1 \over R}\right)c=R-\lfloor \operatorname {Re} (R)\rfloor }$ if c is a multiple of i.

The silver mean of m is also given by the integral

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

Another interesting form of the metallic mean is given by

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

Trigonometric expressions

N	Trigonometric expression[5]	Associated regular polygon
1	${\displaystyle 2\cos {\frac {\pi }{5))}$	pentagon
2	${\displaystyle \tan {\frac {3\pi }{8))}$	octagon
3	${\displaystyle 8\cos {\frac {\pi }{13))\cos {\frac {3\pi }{13))\cos {\frac {4\pi }{13))}$	tridecagon
4	${\displaystyle 8\cos ^{3}{\frac {\pi }{5))}$	pentagon
5	${\displaystyle 128\cos {\frac {\pi }{29))\cos {\frac {4\pi }{29))\cos {\frac {5\pi }{29))\cos {\frac {6\pi }{29))\cos {\frac {7\pi }{29))\cos {\frac {9\pi }{29))\cos {\frac {13\pi }{29))}$	29-gon
6	${\displaystyle {\frac {\sin {\frac {21\pi }{40))}{\sin {\frac {7\pi }{8))-\sin {\frac {5\pi }{8))\sin {\frac {\pi }{40))\csc {\frac {11\pi }{40))-\sin {\frac {19\pi }{20))\sin {\frac {9\pi }{40))\csc {\frac {7\pi }{10))))}$	tetracontagon
7
8	${\displaystyle {\frac {\sin {\frac {6\pi }{17))\sin {\frac {7\pi }{17))\sin {\frac {3\pi }{17))\sin {\frac {12\pi }{17))}{\sin {\frac {\pi }{17))\sin {\frac {9\pi }{17))\sin {\frac {4\pi }{17))\sin {\frac {2\pi }{17))))}$	heptadecagon
9
10

Geometric construction

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, ${\displaystyle H}$.^[6]

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.