Metallic sequences

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Suppose we have an infinite sequence

  \[ a_1,a_2,a_3,a_4,... \]    

and that for all $i>2$ we have

  \begin{equation} a_ i=na_{i-1}+a_{i-2}, \end{equation}   (1)

where $n>1$ is a positive integer.

In the main article we claimed that in this case the ratio of successive terms of the sequence converges to the metallic mean $\lambda _ n$. We’ll now give you a justification of this claim. Recall that

  \begin{equation} \lambda _ n=n+\frac{1}{\lambda _ n}.\end{equation}   (2)

Substituting the expression given by this equation for the $\lambda _ n$ in the denominator on the right hand side of the same expression gives

  \[ \lambda _ n=n+\frac{1}{n+\frac{1}{\lambda _ n}}.  \]    

Another substitution gives

  \[ \lambda _ n=n+\frac{1}{n+\frac{1}{n+\frac{1}{\lambda _ n}}}.  \]    

Continuing on in this vein, it is not too hard to prove that $\lambda _ n$ is equal to the infinite continued fraction

  \[ n+\frac{1}{n+\frac{1}{n+\frac{1}{n+...}}}.  \]    

(You can find out more about continued fractions here.)

Now back to our sequence. The ratio of successive terms is

  \[ \frac{a_ i}{a_{i-1}}. \]    

By equation (1) this is equal to

  \[ \frac{a_ i}{a_{i-1}}=\frac{na_{i-1}+a_{i-2}}{a_{i-1}}=n+\frac{a_{i-2}}{a_{i-1}}. \]    


  \[ \frac{a_ i}{a_{i-1}}=n+\frac{a_{i-2}}{a_{i-1}}=n+\frac{1}{\frac{a_{i-1}}{a_{i-2}}}. \]    

Applying equation (1) to the term $a_{i-1}$ we get

  \[ \frac{a_ i}{a_{i-1}}=n+\frac{1}{\frac{na_{i-2}+a_{i-3}}{a_{i-2}}}=n+\frac{1}{n+\frac{a_{i-3}}{a_{i-2}}}. \]    

You can see the picture: continuing to make substitutions according to equation (1), we end up with a finite continued fraction expression for $a_{i}/a_{i-1}$ of the form

  \[ a_{i}/a_{i-1}=n+\frac{1}{n+\frac{1}{n+\frac{1}{...+\frac{a_1}{a_2}}}}. \]    

Letting $i$ tend to infinity we can then prove that the ratio of successive terms in the sequence converges to infinite continued fraction

  \[ n+\frac{1}{n+\frac{1}{n+\frac{1}{n+...}}}.  \]    

Which, as we have sketch-proved above, is equal to $\lambda _ n.$

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