# Interesting pattern in the decimal expansion of $\frac1{243}$

There appears to be an interesting pattern in the decimal expansion of $\dfrac1{243}$:

$$\frac1{243}=0.\overline{004115226337448559670781893}$$

I was wondering if anyone could clarify how this comes about?

• What do you regard to be interesting about that that doesn't happen for any other rational number? (Note that any rational number $\frac{a}{b}$ has either a terminating or infinite eventually-repeating decimal expansion) Commented Dec 12, 2011 at 4:56
• Perhaps what OP finds interesting is the $\dots11\dots22\dots33\dots44\dots55\dots$ Commented Dec 12, 2011 at 5:00
• the 00, 11, 22, 33, 44, 55 interspersed with 4 5 6 7 8 9, it seems too much to be a coincidence but I don't know the reason for the pattern
– wim
Commented Dec 12, 2011 at 5:01
• The decimal expansion of $1/42$ also has some cute stuff going on. Commented Dec 12, 2011 at 5:05
• A further observation shows that the pattern is $$000 + 4 | 111 + 4 | 222 + 4 | 333 + 4 | 444 + 4 | 555 + 4 | 666 + 4 | 777 + 4 | 888 + 4 + 1.$$ This suggests that $\frac{1}{243}$ is decomposted into fractions of different repeating periods, namely $$\frac{1}{243} = r + \frac{4}{10^{3}-1} + \frac{1}{10^{27}-1}$$, where $$r = \frac{111222333444555666777888}{10^{27}-1} = \frac{111333667111667334112}{1001001001001001001001001}$$ is the rational number with period $000111222333444555666777888$. But I cannot find a particular pattern anymore... Commented Dec 12, 2011 at 5:13

$\frac{1}{243}=\frac{1}{333}+\frac{10}{8991}$

$\frac{1}{333}=.\overline{003}$

$\frac{1}{8991}=.\overline{000111222333444555666777889}=\frac{111}{998001}=\frac{111}{10^6-2\cdot10^3+1}$

• Wow! Now that's what I call a rabbit-out-of-a-hat, how the heck did you know that?! But your hint about 8991 lead me to the following document which explains some things nicely math.sjsu.edu/~goldston/otherpub.pdf
– wim
Commented Dec 12, 2011 at 5:18
• I saw that subtracting off $.\overline{003}=1/333$ made things even prettier. Alpha did the subtraction to get $10/8991$. Then for explanation it seemed useful to divide out the $111$ Commented Dec 12, 2011 at 5:23
• This is a fairly nice observation, but actually we have $\frac{1}{8991} = 0.\overline{000111222333444555666777889}$. Commented Dec 12, 2011 at 5:27
• sos440 is right, the carry mucks things up. so the third line in the answer is not exactly correct. see the pdf i linked in the first comment
– wim
Commented Dec 12, 2011 at 6:48
• @wim: fixed. Thanks. Commented Dec 12, 2011 at 15:07