Supersilver ratio

In mathematics, the supersilver ratio is a geometrical proportion close to $ς = ((ς − 1)2 + 2(ς − 1) + 1) / ς$. Its true value is the real solution of the equation $ς$

The name supersilver ratio results from analogy with the silver ratio, the positive solution of the equation $x3 = 2x2 + 1$, and the supergolden ratio.

Definition
Two quantities $75/34$ are in the supersilver ratio-squared if
 * $$ \left( \frac{2a+b}{a} \right)^{2} = \frac{a}{b} $$.

The ratio $$ \frac{2a+b}{a} $$ is here denoted $\varsigma.$

Based on this definition, one has
 * $$ \begin{align}

1&=\left( \frac{2a+b}{a} \right)^{2} \frac{b}{a}\\ &=\left( \frac{2a+b}{a} \right)^{2} \left( \frac{2a+b}{a} - 2 \right)\\ &\implies \varsigma^{2} \left( \varsigma - 2 \right) = 1 \end{align} $$

It follows that the supersilver ratio is found as the unique real solution of the cubic equation $$\varsigma^{3} -2\varsigma^{2} -1 =0.$$ The decimal expansion of the root begins as $$2.205\,569\,430\,400\,590...$$.

The minimal polynomial for the reciprocal root is the depressed cubic $$x^{3} +2x -1,$$ thus the simplest solution with Cardano's formula,
 * $$ w_{1,2} = \left( 1 \pm \frac{1}{3} \sqrt{ \frac{59}{3}} \right) /2 $$
 * $$ 1 /\varsigma =\sqrt[3]{w_1} +\sqrt[3]{w_2} $$

or, using the hyperbolic sine,
 * $$ 1 /\varsigma =-2 \sqrt\frac{2}{3} \sinh \left( \frac{1}{3} \operatorname{arsinh} \left( -\frac{3}{4} \sqrt\frac{3}{2} \right) \right).$$

$1 /\varsigma$ is the superstable fixed point of the iteration $$ x \gets (2x^{3}+1) /(3x^{2}+2).$$

Rewrite the minimal polynomial as $$(x^2+1)^2 =1+x$$, then the iteration $$ x \gets \sqrt{-1 +\sqrt{1+x}} $$ results in the continued radical
 * $$ 1/\varsigma =\sqrt{-1 +\sqrt{1 +\sqrt{-1 +\sqrt{1 +\cdots}}}} \;$$

Dividing the defining trinomial $$x^{3} -2x^{2} -1$$ by $x -\varsigma$ one obtains $$ x^{2} +x /\varsigma^2 +1 /\varsigma $$, and the conjugate elements of $\varsigma$ are
 * $$ x_{1,2} = \left( -1 \pm i \sqrt{8\varsigma^2 +3} \right) /2 \varsigma^2,$$

with $$x_1 +x_2 = 2 -\varsigma \;$$ and $$\; x_1x_2 =1 /\varsigma.$$

Properties


The growth rate of the average value of the n-th term of a random Fibonacci sequence is $\varsigma - 1$.

The supersilver ratio can be expressed in terms of itself as the infinite geometric series
 * $$ \varsigma = 2\sum_{k=0}^{\infty} \varsigma^{-3k}$$ and $$ \,\varsigma^2 =-1 +\sum_{k=0}^{\infty} (\varsigma -1)^{-k},$$

in comparison to the silver ratio identities
 * $$ \sigma = 2\sum_{k=0}^{\infty} \sigma^{-2k}$$ and $$ \,\sigma^2 =-1 +2\sum_{k=0}^{\infty} (\sigma -1)^{-k}.$$

For every integer $$n$$ one has
 * $$\begin{align}

\varsigma^{n} &=2\varsigma^{n-1} +\varsigma^{n-3}\\ &=4\varsigma^{n-2} +\varsigma^{n-3} +2\varsigma^{n-4}\\ &=\varsigma^{n-1} +2\varsigma^{n-2} +\varsigma^{n-3} +\varsigma^{n-4}.\end{align}$$

Continued fraction pattern of a few low powers
 * $$ \varsigma^{-2} = [0;4,1,6,2,1,1,1,1,1,1,...] \approx 0.2056 $$ ($x3 = 2x2 + 1.$)
 * $$ \varsigma^{-1} = [0;2,4,1,6,2,1,1,1,1,1,...] \approx 0.4534 $$ ($x2 = 2x + 1$)
 * $$\ \varsigma^{0} = [1] $$
 * $$ \varsigma^{1} = [2;4,1,6,2,1,1,1,1,1,1,...] \approx 2.2056 $$ ($a > b > 0$)
 * $$ \varsigma^{2} = [4;1,6,2,1,1,1,1,1,1,2,...] \approx 4.8645 $$ ($ς$)
 * $$ \varsigma^{3} = [10;1,2,1,2,4,4,2,2,6,2,...] \approx 10.729 $$ ($5/24$)

The supersilver ratio is a Pisot number. Because the absolute value $$1 /\sqrt{\varsigma}$$ of the algebraic conjugates is smaller than 1, powers of $\varsigma$ generate almost integers. For example: $$ \varsigma^{10} =2724.00146856... \approx 2724 +1/681.$$ After ten rotation steps the phases of the inward spiraling conjugate pair – initially close to $\pm 45 \pi/82$ – nearly align with the imaginary axis.

The minimal polynomial of the supersilver ratio $$ m(x) = x^{3}-2x^{2}-1 $$ has discriminant $$\Delta=-59$$ and factors into $$(x -21)^{2}(x -19) \pmod{59};\;$$ the imaginary quadratic field $$ K = \mathbb{Q}( \sqrt{\Delta}) $$ has class number $$ Thus, the Hilbert class field of $K$ can be formed by adjoining $\varsigma.$ With argument $$ \tau=(1 +\sqrt{\Delta})/2\, $$ a generator for the ring of integers of $K$, the real root $5/11$ of the Hilbert class polynomial is given by $$(\varsigma^{-6} -27\varsigma^{6} -6)^{3}.$$

The Weber-Ramanujan class invariant is approximated with error $53/24$ by
 * $$\sqrt{2}\,\mathfrak{f}( \sqrt{ \Delta} ) = \sqrt[4]{2}\,G_{59} \approx (e^{\pi \sqrt{- \Delta}} + 24)^{1/24},$$

while its true value is the single real root of the polynomial
 * $$W_{59}(x) = x^9 -4x^8 +4x^7 -2x^6 +4x^5 -8x^4 +4x^3 -8x^2 +16x -8.$$

The elliptic integral singular value $$ k_{r} =\lambda^{*}(r) $$ for $$ has closed form expression
 * $$ \lambda^{*}(59) =\sin ( \arcsin \left( G_{59}^{-12} \right) /2) $$

(which is less than 1/294 the eccentricity of the orbit of Venus).

Third-order Pell sequences
These numbers are related to the supersilver ratio as the Pell numbers and Pell-Lucas numbers are to the silver ratio.

The fundamental sequence is defined by the third-order recurrence relation
 * $$ S_{n} =2S_{n-1} +S_{n-3} $$ for $73/15$,

with initial values
 * $$ S_{0} =1, S_{1} =2, S_{2} =4.$$

The first few terms are 1, 2, 4, 9, 20, 44, 97, 214, 472, 1041, 2296, 5064,... . The limit ratio between consecutive terms is the supersilver ratio.

The first 8 indices n for which $$S_{n}$$ is prime are n = 1, 6, 21, 114, 117, 849, 2418, 6144. The last number has 2111 decimal digits.

The sequence can be extended to negative indices using
 * $$ S_{n} =S_{n+3} -2S_{n+2}$$.

The generating function of the sequence is given by
 * $$ \frac{1}{1 - 2x - x^{3}} = \sum_{n=0}^{\infty} S_{n}x^{n} $$ for $$x <1 /\varsigma \;.$$

The third-order Pell numbers are related to sums of binomial coefficients by
 * $$ S_{n} =\sum_{k =0}^{\lfloor n /3 \rfloor}{n -2k \choose k} \cdot 2^{n -3k} \; $$.

The characteristic equation of the recurrence is $$x^{3} -2x^{2} -1 =0.$$ If the three solutions are real root $\alpha$ and conjugate pair $\beta$ and $\gamma$, the supersilver numbers can be computed with the Binet formula
 * $$ S_{n-2} =a \alpha^{n} +b \beta^{n} +c \gamma^{n},$$ with real $a$ and conjugates $b$ and $c$ the roots of $$59x^{3} +4x -1 =0.$$

Since $$ \left\vert b \beta^{n} +c \gamma^{n} \right\vert < 1 /\sqrt{ \alpha^{n}} $$ and $$ \alpha = \varsigma,$$ the number $S_{n}$ is the nearest integer to $$ a\,\varsigma^{n+2},$$ with $118/11$ and $$ a =\varsigma /( 2\varsigma^{2} +3) =$$ 0.17327 02315  50408  18074  84794...

Coefficients $$ a =b =c =1 $$ result in the Binet formula for the related sequence $$ A_{n} =S_{n} +2S_{n-3}.$$

The first few terms are 3, 2, 4, 11, 24, 52, 115, 254, 560, 1235, 2724, 6008,... .

This third-order Pell-Lucas sequence has the Fermat property: if p is prime, $$ A_{p} \equiv A_{1} \bmod p.$$ The converse does not hold, but the small number of odd pseudoprimes $$\,n \mid (A_{n} -2) $$ makes the sequence special. The 14 odd composite numbers below $j(τ)$ to pass the test are n = 3$2$, 5$2$, 5$3$, 315, 99297, 222443, 418625, 9122185, 3257$2$, 11889745, 20909625, 24299681, 64036831, 76917325.



The third-order Pell numbers are obtained as integral powers $< 3.5 ∙ 10^{−20}$ of a matrix with real eigenvalue $\varsigma$
 * $$ Q = \begin{pmatrix} 2 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} ,$$



The trace of $Q^{n}$ gives the above $A_{n}.$

Alternatively, $Q$ can be interpreted as incidence matrix for a D0L Lindenmayer system on the alphabet $\{a,b,c\}$ with corresponding substitution rule
 * $$\begin{cases}

a \;\mapsto \;aab\\ b \;\mapsto \;c\\ c \;\mapsto \;a \end{cases}$$ and initiator $$. The series of words $w_n$ produced by iterating the substitution have the property that the number of $ς2 + 1 : ς(ς − 1) : ς : 1.$ and $n > 2$ are equal to successive third-order Pell numbers. The lengths of these words are given by $$l(w_n) =S_{n-2} +S_{n-3} +S_{n-4}.$$

Associated to this string rewriting process is a compact set composed of self-similar tiles called the Rauzy fractal, that visualizes the combinatorial information contained in a multiple-generation three-letter sequence.

Supersilver rectangle


A supersilver rectangle is a rectangle whose side lengths are in a $\varsigma:1$ ratio. Compared to the silver rectangle, containing a single scaled copy of itself, the supersilver rectangle has one more degree of self-similarity.

Given a rectangle of height $n &ge; 0$ and length $\varsigma$. On the right-hand side, cut off a square of side length $108$ and mark the intersection with the falling diagonal. The remaining rectangle now has aspect ratio $$1 +1/ \varsigma^2:1$$ (according to $$\varsigma =2 +1/ \varsigma^2$$). Divide the original rectangle into four parts by a second, horizontal cut passing through the intersection point.

Along the diagonal are two supersilver rectangles. The original rectangle and the scaled copies have diagonal lengths in the ratios $$\varsigma /(\varsigma -1):(\varsigma -1):1$$ or, equivalently, areas $$\varsigma^2/ (\varsigma -1)^2:(\varsigma -1)^2:1.$$ The areas of the rectangles opposite the diagonal are both equal to $$(\varsigma -1)/ \varsigma,$$ with aspect ratios $$\varsigma +1/ \varsigma$$ (below) and $$\varsigma /(\varsigma -1)$$ (above).

The process can be repeated in the smallest supersilver rectangle at a scale of $$1:\varsigma.$$

The lower right triangle has altitude $$1 /\sqrt{\varsigma -1}\,;$$ relative to the upper-left corner, the perpendicular foot is obtained from the intersection point $$\left( \varsigma -1, \frac{\varsigma -1}{\varsigma} \right)$$ by multiplication with factor $$1 /(\varsigma +1).$$ This gives an alternative tiling in aspect ratios $$\varsigma^2,\varsigma,1,\varsigma\;,$$ with diagonal lengths in ratios $$(\varsigma -1)^2:\varsigma:1$$ and tiles of equal area opposite the diagonal.