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 $987$.
 A: It is not a coincidence. Let $s(x)$ be the generating function of Fibonacci numbers. Then we have $$s(x)=\frac{x}{1-x-x^2}=F_0+F_1x+F_2x^2+\dots,\ \mbox{for}\  |x|<\frac{1}{\varphi},$$ where $\varphi$ is the golden ratio. Now put $x:=10^{-3}$ and you easily get your equality.
A: It works in every base, but i suppose, that base 37 is interesting, because 35,36 and hence 0;0,1,1,2,3,5,8,13,21,35,... is a cube in that base.
In decimal, 0;0,1,1,2,3,6 is a square.  That is, $89*106^2=1000004.$  
A similar thing happens with 997999, which gives numbers near $\sqrt 2$.  The following numbers are 1000/997999, and 1001/997999, the ratio between the three-digit groups gets ever closer to square-root of 2.  
 1001      0.001 003 007 017 041 099 239 578  ...
 1000      0.001 002 005 012 029 070 169 408 ...

A: $$\begin{align}
\frac{1}{10^{2k}-10^k-1} &= 10^{-2k}\frac{1}{1 - 10^{-k}(1 + 10^{-k})}\\
&= 10^{-2k} \sum_{n=0}^\infty 10^{-nk}(1+10^{-k})^n\\
&= 10^{-2k}\sum_{n=0}^\infty 10^{-nk}\sum_{m=0}^n \binom{n}{m}10^{-mk}\\
&= \sum_{0\leqslant m\leqslant n} \binom{n}{m} 10^{-(m+n+2)k}\\
&= \sum_{r=0}^\infty \left(\sum_{m=0}^{\lfloor r/2\rfloor} \binom{r-m}{m}\right)10^{-(r+2)k}
\end{align}$$
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.
