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Related to this question and inspired by this challenge on PPCG.

The challenge is as follows: for a given natural number $x \in \mathbb{N}$, find the first Fibonacci number $F_n$ that contains $x$. With containing I mean that the run of digits of the base-10 representation of $x$ is present in the base-10 representation of $F_n$.

For example: for $x = 32$, the first Fibonacci number containing $32$ is the $30^{th}$: $832040$.

The question is, is there a solution for any $x$?

JAD
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1 Answers1

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Yes, there is. We can even demand that the prescribed $x$ appears at the beginning of a Fibonacci number.

For large $n$ the Fibonacci number $F_n$ shares its all but possibly the least significant digit with $$F_n\approx\frac1{\sqrt5}\left(\frac{1+\sqrt5}2\right)^n$$ by Binet's formula.

Because $\theta=\log_{10}((1+\sqrt5)/2)$ is irrational the claim follows from Kronecker's density theorem. The argument is the same as here, where the question is about the string of the most significant digits in the decimal expansion of $2^n$.

For the sake of completeness here's the proof for the irrationality of $\theta$. Assume contrariwise that there exists a rational number $m/n$ such that $$ 10^{m/n}=\frac{1+\sqrt5}2. $$ Raising this to power $n$ would then imply that $$ 10^m=\left(\frac{1+\sqrt5}2\right)^n. $$ But here the left hand side is an integer where as the right hand side differs from an integer by the tiny amount $\left(\frac{1-\sqrt5}2\right)^n$.

A way of finding explicit (but not necessarily anywhere near the smallest) values of $n$ is then to find good rational approximations of $\theta$. Here $\theta=0.2089876\ldots$. A good rational approximation is then $$ \frac{14}{67}=0.20895522\ldots $$ This implies that $$ \left(\frac{1+\sqrt5}2\right)^{67}\approx 1.005013\cdot10^{14} $$ is very close to a power of ten. By Binet's formula this means that $F_{n+67}$ begins with a sequence of digits close to that of $F_n$, only slightly "bigger".

For example, if you want a Fibonacci number beginning $32\cdots$, you can do the calculations $$ \begin{aligned} x&=\log_{10}(32\cdot\sqrt5)\approx1.8546\ldots,\\ y&=67\theta-\lfloor 67\theta\rfloor\approx0.0021718967\ldots,\\ m&=(x-\lfloor x\rfloor)/y\approx 393.47\ldots \end{aligned} $$ Then (thanks to Mathematica) you can calculate $$ \begin{aligned} F_{393\cdot67}=3192055\ldots,\\ F_{394\cdot 67}=3200805\ldots,\\ F_{395\cdot67}=3224142\ldots. \end{aligned} $$ You see how incrementing $n$ by $67$ increases the value of the integer formed by seven most significant digits by about $0.5$ per cent.

When that half per cent resolution is not good enough to give you the string you want (happens inevitably for longer strings), you need to look for better rational approximations for $\theta$. The tool for that is given by the theory of continued fractions (and best rational approximations).

Jyrki Lahtonen
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  • Of course, I should also refer to the earliest incarnation. – Jyrki Lahtonen Jul 28 '17 at 08:17
  • So if I understand correctly, because $\log_{10}(\alpha)$ is irrational, we can pick $n$ to correspond with $x$? That's pretty neat. – JAD Jul 28 '17 at 08:21
  • Correct, @Jarko – Jyrki Lahtonen Jul 28 '17 at 08:21
  • Hmm, is it then straightforward to show how given an $x$ we would actually approach this to get a solution? – JAD Jul 28 '17 at 08:24
  • Some might ask for a proof that $\theta=\log_{10}((1+\sqrt5)/2)$ is irrational (not me). The good thing is that the method is entirely constructive, you can actually find the index, working with a continued fraction expansion of $\theta,$ for instance. –  Jul 28 '17 at 08:25
  • Thanks, @ProfessorVector. My first guess was exactly to try and use continued fractions to find an example. – Jyrki Lahtonen Jul 28 '17 at 08:35
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    @ProfessorVector I'm working on the numerics of an example, but I am also following a broadcast from an interesting bridge match, so I'm not firing on all cylinders at the moment.... If you can do this fast, go ahead and post it. As the question has a somewhat algorithmic (rather than pure existential) focus that should become the accepted answer. – Jyrki Lahtonen Jul 28 '17 at 08:41
  • @Jarko I added something like an algorithm for finding one $n$. You see that it won't produce the smallest such $n$, but I hope it clarifies a bit of how the rational approximation business can be used. – Jyrki Lahtonen Jul 28 '17 at 10:39
  • Interesting stuff. I'll try and see if I can wrap my head around it. – JAD Jul 28 '17 at 11:00
  • @JarkoDubbeldam There are steps missing. Like how to find those good rational approximations. That is covered in many textbooks, possibly also Wikipedia. – Jyrki Lahtonen Jul 28 '17 at 11:03