TITLE: Fibonacci numbers from $998999$
QUESTION [36 upvotes]: 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 $987$.

REPLY [4 votes]: Imagine a mystery number written with digits grouped by three, like
$$M= 0.\ 000\ 001\ 001\ 002\ 003\ 005\ 008\ 013... = 0.|A|B|C|D|E|F... $$
Subtract from this number itself shifted three positions right,
$$M-0.001M= 0.\ 000 \ 000\ 001\ 001\ 002\ 003\ 005\ 008\ 013...\\=0.|A|B-A|C-B|D-C|E-D|F-E... $$
and six positions right.
$$M-0.001M-0.000001M= 0.\ 000 \ 001\ 000\ 000\ 000\ 000\ 000\ 000\ 000...\\=0.|A|B-A|C-B-A|D-C-B|E-D-C|F-E-D... $$
If there have been no carries, the results starts with $0.000001$ and all remaining decimals are $0$, then the groups follow the Fibonacci recurrence $A_{n+2}-A_{n+1}-A_n=0$, and
$$M=\frac{0.000001}{1-0.001-0.000001}=\frac1{998999}.$$
By the same reasoning you see why $99$ gives all ones, $9998$ gives the powers of $2$, and $997999$ gives approximations of $\sqrt2+1$ (see @wendy's post):
$$\frac1{99} = 0.\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01...:A_{n+1}-A_n=0$$
$$\frac1{9998} = 0.\ 0001\ 0002\ 0004\ 0008\ 0016\ 0032\ 0064\ 0128...:A_{n+1}-2A_n=0$$
$$\frac1{997999}=0.\ 001\ 002\ 005\ 012\ 029\ 070\ 169\ 408...:A_{n+2}-2A_{n+1}-A_n=0$$