Why does $\frac{1 }{ 99989999}$ generate the Fibonacci sequence?

Question

$\frac{1}{99989999} = 1.00010002000300050008001300210034005500890144... \times 10^{-8}$

(Link), which includes the Fibonacci sequence $(1\ 1\ 2\ 3\ 5\ 8\ 13\ 21\ 55\ 89\ 144\ldots )$.

This is fascinating to me but I can't figure out how this works, and a Google search doesn't seem to reveal anything.

It can't generate the full sequence since it would be non-repeating. — Raskolnikov, Jan 29 '14 at 17:02
This is pretty cool! I wonder what the digits are when the fibonacci numbers exceed 4 digits! — mikhailcazi, Jan 29 '14 at 17:03
@mikhailcazi: they carry. $1/99989999=1.0001000200030005000800130021003400550089014402330377061009871597258441816766094777138661637550371412644978624312217464868661514838098958276817264494622107156936765245892241683090725903497608795855673525909326191712433160440375641967953... × 10^-8$ You can hold off the carries longer with $1/9999899999$ and so on for the reason Henning Makholm gives. — Ross Millikan, Jan 29 '14 at 17:06
@RossMillikan I'll take your word for it; it seems quite a task to actually check! xD — mikhailcazi, Jan 29 '14 at 17:09
also take a look here: http://jm.davalan.org/divers/fibonacci/index-en.html#INV — user127.0.0.1, Jan 29 '14 at 17:14

score 63 · Accepted Answer · edited Apr 13 '17 at 12:20

63

As explained here, the generating function for the Fibonacci sequence is $$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...=\frac{1}{1-(z+z^2)}$$

$10^8/99989999$ is simply the value of this generating function for $z=0.0001$.

The initial segment of the Fibonacci sequence where all numbers have at most 4 digits will appear nice and visible in the decimal expansion. After that, the digits of successive Fibonacci numbers will be added to each other offset by $4$ positions. Eventually this will create a repeating digit sequence somehow.

edited Apr 13 '17 at 12:20

Community

1

answered Jan 29 '14 at 17:04

hmakholm left over Monica

286,031

This is basically a GP, right? $a = 1$, $r = (z+z^2)$ and $n\rightarrow\infty$? – mikhailcazi Feb 17 '14 at 06:33
@mikhailcazi: I don't follow you. – hmakholm left over Monica Feb 17 '14 at 09:50
$1+z+2z^2+3z^3+5z^4+8z^5+13z^6+...$ is an infinite geometric progression, right? The summation being the RHS? – mikhailcazi Feb 17 '14 at 15:15
2

@mikhailcazi: No, it's not a geometric progression. In a geometric progression, the ratio between successive terms is always the same; here they are all different (although converging towards $\phi z$). – hmakholm left over Monica Feb 17 '14 at 15:20
Oh, okay. I tried creating a geometric progression with the starting term = 1 and the ratio = $z+z^2$, and the first few terms turned out to be the same, so I thought it was a GP. – mikhailcazi Feb 17 '14 at 15:23
2

@mikhailcazi: Yes, if you rearrange the result to collect terms with the same power of $z$. But once this rearrangement is done, I don't think the result can be called a geometric progression, because the progression is not there anymore. – hmakholm left over Monica Feb 17 '14 at 15:28
@mikhailcazi You can use the technique of partial fractions to find that $$\frac{1}{1-(z+z^2)} = \frac{1}{\sqrt{5}}\left(\frac{1}{1-z\phi} - \frac{1}{1 - z\Phi}\right)$$, which shows the Fibonacci sequence is equivalent to the sum of two geometric sequences, giving the classic closed-form representation of $$F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n\right)$$ – NovaDenizen Jul 06 '15 at 05:45

score 7 · Answer 2 · answered Jul 06 '15 at 05:43

This answer is meant to pick up where Henning Makholm's answer left off and addresses the

Eventually this will create a repeating digit sequence somehow.

part. The explanation that follows is due to Allen Schwenk and the last part of his article in Math Horizons titled An Unanticipated Decimal Expansion. All the credit goes to him.

Spoiler: The length of the period for the decimal expansion is $496620$ digits.

The length of the period for the expansion: Any fraction of the form $1/D$ may be examined, with period of length $p$ and an irregular lead in sequence of $k$ digits. It must look like this, where we have displayed the irregular lead-in and the first two periods $$ \frac{1}{D}=0.a_1a_2\ldots a_ka_{k+1}a_{k+2}\ldots a_{k+p}a_{k+1}a_{k+2}\ldots a_{k+p}.\tag{1} $$ When we multiply the fraction in $(1)$ by $10^{p+k}$ and subtract the original fraction times $10^k$, everything beyond the decimal point drops out. In other words, we have formed an integer $N$ where $$ \frac{1}{D}10^{p+k}-\frac{1}{D}10^k=\frac{10^{p+k}-10^k}{D}=N. $$ This leads to the integer equation $$ (10^p-1)\cdot 10^k=ND. $$ Now focusing on $D=998999$, we see that it factors into two primes $$ D=998999=179\cdot 5581. $$ Since neither prime shares a factor with $10^k$, we are forced to agree that $N=10^k M$ for another integer $M$. The equation may be rewritten as $$ 10^p-1=MD, $$ implying that the solution has $k=0$; that is, there is no lead-in sequence. We are forced to conclude that $10^p\equiv1\bmod 179$ and $10^p\equiv1\bmod 5581$. This means that $d_1$, the order of $10\mod 179$, must be a divisor of $178$ (by Euler's theorem on finite groups) and also of the period $p$. Likewise, $d_2$, the order of $10\mod 5581$, must be a divisor of $5580$ and also of $p$. In fact, we require $p=\operatorname{lcm}(d_1,d_2)$. Moreover, $178=2\cdot 89$, so the only candidates for the order $d_1$ are $2, 89$, and $178$. Clearly $10^2\not\equiv 1\bmod 179$. What about $10^{89}$? Quick inspection shows that $$ 10^{11} \equiv 157\bmod 179\\[0.5em] 10^{22} \equiv 157^2 \equiv 126\bmod 179\\[0.5em] 10^{44} \equiv 126^2 \equiv 124\bmod 179\\[0.5em] 10^{88} \equiv 124^2 \equiv 161\bmod 179\\[0.5em] 10^{89} \equiv 1610 \equiv -1\bmod 179 $$ The order of $10\bmod 179$ is $d_1=178$.

Similarly, $5580=2^2\cdot 3^2\cdot 5\cdot 31$. This has $36$ possible divisors, and we can check (a tedious task) each of these powers of $10\bmod 5581$. We find that the only power that reduces to $1\bmod 5581$ is $10^{5580}$. So $d_2=5580$, and $p=\operatorname{lcm}(178,5580)=496620$.

Zain Patel · Answer 3 · 2015-07-06T20:40:11.660

Let $F_i$ be the $i$-th Fibonacci number.

We have $$\frac{1}{999,999,999,998,999,999,999,999}$$

represented in decimal form as:

0. 000 000 000 000 
   000 000 000 001 
   000 000 000 001
   000 000 000 002
   000 000 000 003
   000 000 000 005
   000 000 000 008
   000 000 000 013
   000 000 000 021...

This decimal works out to be $$x = 10^{-12}F_1 + 10^{-24}F_2 + 10^{-36}F_3 + \ldots + 10^{-12k}F_k + \ldots \tag{1}$$

where $x$ is our number we let $k$ range over the naturals. Multiply the above by $10^{12}$ to get $$10^{12} x = F_1 + 10^{-12}F_2 + 10^{-24}F_3 +\ldots + 10^{-12k}F_{k+1} +\ldots \tag{2}$$

Multiply the above by $10^12$ once more to get $$10^{24}x = 10^{12}F_1 + F_2 + 10^{-12}F_3 + 10^{-24}F_4 + \ldots + 10^{-12k}F_{k+2} + \ldots \tag{3}$$

Now, we work out the value of $(3) - (2) - (1)$ to get $$(10^{24} – 10^{12} – 1) x = 10^{12}F_1 + (F_2 – F_1) + 10^{-12}(F_3 – F_2 – F_1) + 10^{-24}(F_4 – F_3 – F_2) + \ldots + 10^{-12k} (F_{k+2} – F_{k+1} – F_k) + \ldots $$

Whilst this looks rather tricky to work with, we are quite privileged to be working with the Fibonacci sequence here, since it's defined as $F_n = F_{n-1} + F_{n-2}$ with the initial conditions that $F_1 = 0$ and $F_2 = 1$

Hence you get $$F_{k+2} – F_{k+1} – F_k = \left(F_{k+1} + F_k\right) – F_{k+1} – F_{k} = 0$$

so that our equation simplifies to $$(10^{24} – 10^{12} – 1)x = 1 \iff x = \frac{1}{10^{24} – 10^{12} – 1}$$

And you'll be quite pleased to note that, indeed, $$10^{24} - 10^{12} - 1 = 999,999,999,998,999,999,999,999$$

Why does $\frac{1 }{ 99989999}$ generate the Fibonacci sequence?

3 Answers3

Linked