Is there any number other than 109 whose reciprocal contains the Fibonacci sequence? [closed]

Question

Let $p$ be any odd number, and compute $1/p$ to $p$ decimal places. Compare your answer with the string that is formed by appending all remainders of $(10^n\ \text{mod p}) \text{ mod p}$ where ${0 < n < p}$. You will find that all answers are the same for all ${p}$'s of the form ${10k + 9}$.

Here is a strange property of $109$. The decimal expansion of $1/109$ contains the Fibonacci sequence, the end of the period of the non-terminating periodic decimal expansion of $1/109$ is $...7247706422018348623853211$, from right to left is the Fibonacci sequence (if the sum $f_{n - 1} + f_{n - 2}$ is greater than 10, then carry $1$ and add it to the next term).

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MY QUESTIONS

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1. Is there any number $p$ other than $109$ whose reciprocal is a Fibonacci sequence?
2. Is there any published work about this?

Answer 1

Consider the related decimal

$x=\sum_{n=0}^{\infty}\frac{f_n}{10^n}=0.11235955056179775280898876404\cdots.$ Since $$f_n=\frac{\tau^n+(\frac{-1}{\tau})^n}{\sqrt{5}}$$ for $\tau=\frac{1+\sqrt{5}}2,$ We see that $x$ is the sum of an infinite geometric progression and can be evaluated in closed form (and can be seen ahead of time to be some rational number in $\mathbb{Q}(\sqrt{5})$ ).

However it is more elegant to note that $x=\frac{10}{89}$ because$$x+10x=1.235955056179775280898876404\cdots=100x-10$$ with the behavior explained by the Fibonacci recurrence.

Hence $10=(100-10+1)x=89x.$

Using some base other than $10$, such as $0.010102030508132134\cdots$ in base $100,$ would give a similar result.

To get the numerator to be $1$ just put a decimal point in front of the Fibonacci sequence $0.0112359\cdots$

What you have is slightly different so instead of $100-10+1=89$ the appropriate thing is $100+10-1=109.$ I understand this though I feel my explanation could be improved. I'll have the period end $\cdots532110$ which will turn out to give $\frac{10}{109}.$

If my unknown real is $$y=0.abc\cdots\mathbf{853211}0|abc\cdots$$ where the bar marks the end of the period, then $$y+10y=a.bc\cdots+0.abc\cdots=u.vabc\cdots\mathbf{385321}u|vabc\cdots$$ where the $u,v$ are digits to be determined because of some carrying. In fact the second $uv$ should be $10$ and the first $1.0$ . Details at the end if desired, but essentially everything got shifted one position to the right and the initial $1\ 0$ are the two base terms $0,1$ not obtained from the recurrence, in descending order.

So $y+10y=u.v+\frac{y}{10}=1.0+\frac{y}{10}$ which gives $110y=10+y$ and $y=\frac{10}{109}$

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Some details on $y+10y:$

Below is a portion of the addition around the end of the period.

The digits $\mathbf{385321}$ bolded above in $y+10y$ clearly result from the digits bolded above in $y$ and the following $0.$ But after them must be $uv=10$ to continue the pattern. Following that it starts again with $abc\cdots.$

$\begin{matrix} 8&5&3&2&1&1&0|&a&b&c&d \\ 5&3&2&1&1&0&a|&b&c \\ 3&8&5&3&2&1&u|&v&a&b& c& \end{matrix}$

so

$\begin{matrix} 8&5&3&2&1&1&0|&a&b&c&d \\ 5&3&2&1&1&0&a|&b&c&d \\ 3&8&5&3&2&1&\mathbf{1}|&0&a&b& & \end{matrix}$

We can get the digits by dividing out $\frac{10}{109}.$ But it is amusing to notice that the bolded $\mathbf{1}$ must be a carry meaning that $a=0$ and $b=9$ so $c=1$ and $d$ is either $8$ or $7$ and in fact $7$ since $e+d$ ends in a $1$ etc.