Metallic mean

If one removes the largest possible square from the end of a gold rectangle one is left with a gold rectangle. If one removes two from a silver, one is left with a silver. If one removes three from a bronze, one is left with a bronze. Examine the dotted lines representing the boundaries of the perfect squares within each rectangle, and note that the number of said dotted lines always equals N.

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,

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

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

A golden triangle. The ratio a:b is equivalent to the golden ratio φ. In a silver triangle this would be equivalent to δ_S.

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

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

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 ${\displaystyle N=1}$,

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.

N = 2

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

Thus

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

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

N = 3

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

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

N = 1.5

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

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 number

 k. OEIS: A084844, Denominators of the continued fraction n + 1/(n + 1/...) [n times].

• 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.