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3 answers  ·  posted 3y ago by The Amplitwist‭  ·  last activity 3y ago by celtschk‭

#3: Post edited by The Amplitwist‭ · 2020-10-20T17:27:01Z (over 3 years ago)
• Why do the decimal expansions of $1/(10n - 1)$ have this neat pattern?
• Why does the decimal expansion of $1/(10n - 1)$ have this neat pattern?
#2: Post edited by The Amplitwist‭ · 2020-10-19T19:08:19Z (over 3 years ago)
• I was playing around with the reciprocals of some positive integers and found these interesting patterns:
• $$• \frac{1}{19} = 0.\overline{052631578947368421} \tag{(1)} •$$
• Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows:
• \begin{align}
• 1& \\\\
• 21& \\\\
• 421& \\\\
• 8421& \\\\
• {\color{red}{1}}68421 \\\\
• {\color{red}{3}}{\color{green}{3}}68421 \\\\
• \end{align}
• Notice that here I think of the red $\color{red}{1}$ of $16$ and $\color{red}{3}$ of $32$ as being "carried over", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the *left*. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.
• How can I go about making this idea precise?
• -----
• Some more data for this pattern:
• \begin{align}
• {\color{red}{3}}{\color{green}{3}}68421 \\\\
• {\color{red}{6}}{\color{green}{7}}368421 \\\\
• {\color{red}{1}}{\color{green}{34}}7368421 \\\\
• {\color{red}{2}}{\color{green}{69}}47368421 \\\\
• {\color{red}{5}}{\color{green}{38}}947368421 \\\\
• {\color{red}{10}}{\color{green}{77}}8947368421 \\\\
• \end{align}
• In this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.
• I have also observed this to be the case for the decimal expansion of $1/29$:
• $$• \frac{1}{29} = 0.\overline{0344827586206896551724137931} •$$
• In this case, we concatenate the powers of $3$ in a similar fashion.
• Is there any significance to the fact that $19 = {\color{red}{2}}*10 - 1$ and $29 = {\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?
• I also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up:
• $$• \frac{1}{39} = 0.\overline{025641} •$$
• and concatenating the powers of $4$ indeed gives $\dotsc 025641$.
• The last case I checked was $1/49$, and it also obeys this pattern:
• $$• \frac{1}{49} = 0.\overline{020408163265306122448979591836734693877551} •$$
• and concatenating the powers of $5$ gives us this very repeating set of digits.
• Of course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the *forward* direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that
• $$• \frac{1}{49} = \frac{2}{100} + \frac{2^2}{100^2} + \frac{2^3}{100^3} + \dotsb •$$
• which is true by the formula for the sum of an infinite geometric series. So, "forward concatenating" patterns can be explained in this manner, but I'm stumped regarding how to explain the "backward concatenating" ones.
• If anyone can throw some light on this, it would be greatly appreciated!
• I was playing around with the reciprocals of some positive integers and found these interesting patterns:
• $$• \frac{1}{19} = 0.\overline{052631578947368421} •$$
• Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows:
• \begin{align}
• 1& \\\\
• 21& \\\\
• 421& \\\\
• 8421& \\\\
• {\color{red}{1}}68421 \\\\
• {\color{red}{3}}{\color{green}{3}}68421 \\\\
• \end{align}
• Notice that here I think of the red $\color{red}{1}$ of $16$ and $\color{red}{3}$ of $32$ as being "carried over", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the *left*. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.
• How can I go about making this idea precise?
• -----
• Some more data for this pattern:
• \begin{align}
• {\color{red}{3}}{\color{green}{3}}68421 \\\\
• {\color{red}{6}}{\color{green}{7}}368421 \\\\
• {\color{red}{1}}{\color{green}{34}}7368421 \\\\
• {\color{red}{2}}{\color{green}{69}}47368421 \\\\
• {\color{red}{5}}{\color{green}{38}}947368421 \\\\
• {\color{red}{10}}{\color{green}{77}}8947368421 \\\\
• \end{align}
• In this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.
• I have also observed this to be the case for the decimal expansion of $1/29$:
• $$• \frac{1}{29} = 0.\overline{0344827586206896551724137931} •$$
• In this case, we concatenate the powers of $3$ in a similar fashion.
• Is there any significance to the fact that $19 = {\color{red}{2}}*10 - 1$ and $29 = {\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?
• I also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up:
• $$• \frac{1}{39} = 0.\overline{025641} •$$
• and concatenating the powers of $4$ indeed gives $\dotsc 025641$.
• The last case I checked was $1/49$, and it also obeys this pattern:
• $$• \frac{1}{49} = 0.\overline{020408163265306122448979591836734693877551} •$$
• and concatenating the powers of $5$ gives us this very repeating set of digits.
• Of course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the *forward* direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that
• $$• \frac{1}{49} = \frac{2}{100} + \frac{2^2}{100^2} + \frac{2^3}{100^3} + \dotsb •$$
• which is true by the formula for the sum of an infinite geometric series. So, "forward concatenating" patterns can be explained in this manner, but I'm stumped regarding how to explain the "backward concatenating" ones.
• If anyone can throw some light on this, it would be greatly appreciated!
#1: Initial revision by The Amplitwist‭ · 2020-10-19T19:07:53Z (over 3 years ago)
Why do the decimal expansions of $1/(10n - 1)$ have this neat pattern?
I was playing around with the reciprocals of some positive integers and found these interesting patterns:

$$\frac{1}{19} = 0.\overline{052631578947368421} \tag{(1)}$$

Now, this repeating decimal can also be obtained by "concatenating" the powers of $2$ successively to the left as follows:

\begin{align}
1& \\\\
21& \\\\
421& \\\\
8421& \\\\
{\color{red}{1}}68421 \\\\
{\color{red}{3}}{\color{green}{3}}68421 \\\\
\end{align}
Notice that here I think of the red $\color{red}{1}$ of $16$ and $\color{red}{3}$ of $32$ as being "carried over", so the units digit of $32$ and the tens digit of $16$ add to give me the green $\color{green}{3}$. The sequence gets increasingly complicated as more terms are concatenated in this manner, but it definitely looks like it's converging to $\dotsc 052631578947368421$, where the ellipsis indicates that the pattern repeats forever towards the *left*. This is a bit surprising because (informally) this would mean that if I perform this pattern infinitely far out to the right of the decimal point, then I get the decimal expansion of $1/19$.

How can I go about making this idea precise?

-----

Some more data for this pattern:
\begin{align}
{\color{red}{3}}{\color{green}{3}}68421 \\\\
{\color{red}{6}}{\color{green}{7}}368421 \\\\
{\color{red}{1}}{\color{green}{34}}7368421 \\\\
{\color{red}{2}}{\color{green}{69}}47368421 \\\\
{\color{red}{5}}{\color{green}{38}}947368421 \\\\
{\color{red}{10}}{\color{green}{77}}8947368421 \\\\
\end{align}

In this pattern, one can see that at each stage one new digit from the required sequence is correctly added, despite all the carrying-over.

I have also observed this to be the case for the decimal expansion of $1/29$:
$$\frac{1}{29} = 0.\overline{0344827586206896551724137931}$$
In this case, we concatenate the powers of $3$ in a similar fashion.

Is there any significance to the fact that $19 = {\color{red}{2}}*10 - 1$ and $29 = {\color{red}{3}}*10 - 1$ to the use of powers of $2$ and $3$, respectively, in these computations?

I also checked for $1/39$, and though the repeating part of its decimal expansion is not as impressively long as for the other two cases, the pattern still holds up:
$$\frac{1}{39} = 0.\overline{025641}$$
and concatenating the powers of $4$ indeed gives $\dotsc 025641$.

The last case I checked was $1/49$, and it also obeys this pattern:
$$\frac{1}{49} = 0.\overline{020408163265306122448979591836734693877551}$$
and concatenating the powers of $5$ gives us this very repeating set of digits.

Of course, the pattern for $1/49$ makes it obvious that there is a concatenation that works in the *forward* direction: namely, the powers of $2$ concatenated in bunches of two digits. But this just means that
$$\frac{1}{49} = \frac{2}{100} + \frac{2^2}{100^2} + \frac{2^3}{100^3} + \dotsb$$
which is true by the formula for the sum of an infinite geometric series. So, "forward concatenating" patterns can be explained in this manner, but I'm stumped regarding how to explain the "backward concatenating" ones.

If anyone can throw some light on this, it would be greatly appreciated!