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 $987$.

Solution 1:

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.

Solution 2:

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

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.

Solution 3:

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):

$$\frac1{99} = 0.\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01\ 01...:A_{n+1}-A_n=0$$ $$\frac1{9998} = 0.\ 0001\ 0002\ 0004\ 0008\ 0016\ 0032\ 0064\ 0128...:A_{n+1}-2A_n=0$$ $$\frac1{997999}=0.\ 001\ 002\ 005\ 012\ 029\ 070\ 169\ 408...:A_{n+2}-2A_{n+1}-A_n=0$$

Solution 4:

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 ...