<s>[INST] Fibonacci numbers from $998999$ Is there a nice explanation of 
$$\frac{1}{998999}=0.000\,\underbrace{001}_{F_1}\,\underbrace{001}_{F_2}\,\underbrace{002}_{F_3}\,\underbrace{003}_{F_4}\,\underbrace{005}_{F_5}\,\underbrace{008}_{\ldots}\,013\,021\,034\,055\,089\,144\,233\,377\,...$$
or is this a mere coincidence?
The pattern breaks at $16$th Fibonacci number producing $988$ instead of $987$.
 [/INST] $$\begin{align}
\frac{1}{10^{2k}-10^k-1} &= 10^{-2k}\frac{1}{1 - 10^{-k}(1 + 10^{-k})}\\
&= 10^{-2k} \sum_{n=0}^\infty 10^{-nk}(1+10^{-k})^n\\
&= 10^{-2k}\sum_{n=0}^\infty 10^{-nk}\sum_{m=0}^n \binom{n}{m}10^{-mk}\\
&= \sum_{0\leqslant m\leqslant n} \binom{n}{m} 10^{-(m+n+2)k}\\
&= \sum_{r=0}^\infty \left(\sum_{m=0}^{\lfloor r/2\rfloor} \binom{r-m}{m}\right)10^{-(r+2)k}
\end{align}$$
So it remains to see that
$$F_{r+1} = \sum_{m=0}^{\lfloor r/2\rfloor} \binom{r-m}{m}.$$
In absence of an elegant idea, induction will have to do. The cases $r  = 0,1$ are easily checked. Then, for the induction step, we separate odd and even $r$,
$$\begin{align}
F_{2k+3} &= F_{2k+2} + F_{2k+1}\\
&= \sum_{m=0}^k \binom{2k+1-m}{m} + \sum_{m=0}^k \binom{2k-m}{m}\\
&= \binom{2k+1}{0} + \sum_{m=1}^k \left(\binom{2k+1-m}{m} + \binom{2k-(m-1)}{m-1}\right) + \binom{k}{k}\\
&= \binom{2k+2}{0} + \sum_{m=1}^k \binom{2k+2-m}{m} + \binom{k+1}{k+1},
\end{align}$$
which is the desired formula, and
$$\begin{align}
F_{2k+2} &= F_{2k+1} + F_{2k}\\
&= \sum_{m=0}^k \binom{2k-m}{m} + \sum_{m=0}^{k-1} \binom{2k-1-m}{m}\\
&= \binom{2k}{0} + \sum_{m=1}^k \left(\binom{2k-m}{m} + \binom{2k-1-(m-1)}{m-1}\right)\\
&= \binom{2k+1}{0} + \sum_{m=1}^k \binom{2k+1}{m},
\end{align}$$
which also is the desired formula. Hence
$$\frac{1}{10^{2k}-10^k-1} = \sum_{r=0}^\infty F_{r+1} 10^{-(r+2)k},$$
and you get the Fibonacci numbers until you reach the first with more than $k$ digits, which carries over into the previous.
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