Metallic sequences

Submitted by Marianne on January 29, 2020

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

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

Now

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

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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