Approximation of harmonious numbers

Question

I have a question about an approximation of so-called "harmonious numbers". These are a generalisation of the golden ratio, the plastic number, and so on, (related to the Fibonacci and Padovan sequences respectively). Their decimal expansions to $5\ sf$ for illustration:

\begin{array}{|c|c|c|c|} \hline n& 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline χ_n & 1.6180 & 1.3247 & 1.2207 & 1.1673 & 1.1347 & 1.1128 & 1.0970 \\ \hline \end{array}

Studied by Dutch architect Dom Hans van der Laan in the 1920s, they are basically the positive real solutions to the equations

\begin{align} &x^n = x+1 \quad \quad \ (n\geq 2, \ n\in\mathbb{Z})\\ \end{align}

They appear to be quite nicely approximated by

\begin{align} &\frac{2 n - 1 + \log 2}{2 n - 1 - \log 2}\\ \end{align}

I was just wondering whether anyone could shed some light on why this might be the case?

ref: Plastic number: construction and applications L Marohnić, T Strmečki

Answer 1

If we consider the equation $$x^n = x+1$$ it means that we look for the inverse of $$n=\frac{\log (x+1)}{\log (x)}$$ Building the $[1,1]$ Padé approximant of the rhs around $x=1$, we have $$n=\frac{\log (2)+\frac{1}{2} (1+\log (2))(x-1) }{x-1 } $$ Solving for $x$ gives $$\color{red}{x_1=\frac{2 n - 1 + \log (2)}{2 n - 1 - \log (2)}}$$ which is the formula.

We could do better building the $[2,2]$ Padé approximant of the rhs around $x=1$ and obtain $$\color{red}{x_2=\frac{2 \sqrt{3} \sqrt{12n^2-12n+(4 \log ^2(2)-6 \log (2)+3 ) } +(8 \log (2)-3)}{12 n-(9+4 \log (2)) }}$$ Comparing the results $$\left( \begin{array}{cccc} n & x_1 & x_2 & \text{solution} \\ 2 & 1.6009462 & 1.6179521 & 1.6180340 \\ 3 & 1.3218811 & 1.3247134 & 1.3247180 \\ 4 & 1.2198076 & 1.2207433 & 1.2207441 \\ 5 & 1.1668856 & 1.1673038 & 1.1673040 \\ 6 & 1.1345022 & 1.1347241 & 1.1347243 \\ 7 & 1.1126441 & 1.1127757 & 1.1127757 \\ 8 & 1.0968972 & 1.0969815 & 1.0969816 \\ 9 & 1.0850130 & 1.0850702 & 1.0850703 \\ 10 & 1.0757254 & 1.0757661 & 1.0757661 \\ 11 & 1.0682673 & 1.0682972 & 1.0682972 \\ 12 & 1.0621466 & 1.0621692 & 1.0621692 \\ \end{array} \right)$$

Edit

You can have still better writing $$\color{red}{x_3=x_1-\frac{x_1 (x_1+1) (n \log (x_1)-\log (x_1+1))}{x_1 (n-1)+n}}$$

Now, may be the simplest $$x=1+t+(2n-1)\sum_{k=2}^\infty \frac{P_k(n)}{ 2^{k-1}\,k!}u^k $$ $$t=\frac{2\log(2)}{2n-1} \qquad \qquad u=\frac t{2n-1}$$ $$\left( \begin{array}{cc} k & P_k(n) \\ 2 & 4 n-1 \\ 3 & 16 n^2-4 n+1 \\ 4 & 64 n^3-8 n^2+20 n-1 \\ 5 & 256 n^4-32 n^3+264 n^2+52 n+1 \\ 6 & 1024 n^5-384 n^4+1568 n^3+1984 n^2+492 n-1 \\ 7 & 4096 n^6-1536 n^5-2112 n^4+17984 n^3+24672 n^2+4188 n+1\\ \end{array} \right)$$

Using the above terms, the goden ratio is reproduced within an absolute error of $6.21\times 10^{-7}$ and the plastic number within an absolute error of $3.12\times 10^{-9}$.

Update

There is another way to generate approximation considering $$f(x)=n\,\log(x)-\log(x+1)$$ Let $$x_{(k)}=\frac{\sum _{i=0}^k a_i\, n^i } {\sum _{i=0}^k b_i\, n^i }$$ and expand $f(x_{(k)})$ as a series for infinitely large values of $n$. This gives for example $$\color{blue}{x_{(2)}=\frac{24 n^2+12 (\lambda -2) n+\left(2 \lambda ^2-9 \lambda +6\right) }{24 n^2-12 (\lambda +2) n+\left(2 \lambda ^2+3 \lambda +6\right) }}\qquad (\lambda=\log(2))$$

Answer 2

Actually, there are only two morphic numbers. From this reference: Jan Aarts, Robbert Fokkink, Godfried Kruijtzer, Morphic numbers, NAW 5/2 nr. 2 maart 2001, we have the definition

A real number $p$ > 1 is called a morphic number if there exist natural numbers $k$ and $l$ such that $$p + 1 = p^k \text{ and } \ p − 1 = p^{−l}$$ The only two numbers that satisfy this condition are the golden ratio and the plastic number.

You may be thinking of the Metallic mean, which are given by

$$m = \frac{n+\sqrt{n^2+4}}{2}, \quad n\ge 0$$

EDIT: The OP has pointed out that he meant the harmonious numbers. I'm familiar with these as well. They derive from the equation $X+1=X^m$, for $m>1$. In addition, they arise from the sequence

$$f_k=f_{k-m+1}+f_{k-m},\quad k<m\\ f_{k\le1}=1\\ \lim_{k\to\infty}\frac{f_{k+1}}{f_k}=X $$

These constitute for about half of what I call the pseudomorphic numbers, the remainder of which are defined by $\chi-1=\chi^{-n}, n>1$, or

$$f_k=f_{k-1}+f_{k-1-n},\quad k>n+1\\ f_{k\le n+1}=1\\ \lim_{k\to\infty}\frac{f_{k+1}}{f_k}=\chi $$