Let $N\in\mathbb{Z}_+$ represent as $N = m!\cdot r_m+(m-1)!\cdot r_{m-1}+\cdots+2!\cdot r_2+1!\cdot r_1$ Where $0\le r_i\le i$ for $1\le i\le m$
Algorithm
$$\begin{split}
\frac{N-r_1}{2} &= q_1 \\
\frac{q_1-r_2}{3} &= q_2 \\
\frac{q_2-r_3}{4} &= q_3 \\
\frac{q_3-r_4}{5} &= q_4 \\
\vdots &= \vdots \\
\frac{q_{m-1}-r_m}{m+1} &= q_{m}=0
\end{split}$$
Well $$\begin{split}N &= m!\cdot r_m+(m-1)!\cdot r_{m-1}+\cdots+2!\cdot r_2+1!\cdot r_1\\ &=(r_m,r_{m-1},\dots,r_3,r_2,r_1)_! \end{split}$$
Example: $21=(3,1,1)_!=3!\cdot 3+2!\cdot 1+1!\cdot 1=18+2+1$
Let above conversion may call as, $N$ convert in base(!).
See, power numbers $(a^x)$ have a repeating pattern for digits $(r_i)$. observe below table
Table1: $x=2$
$$\begin{split}
1^2 =1&=(000\color{green}{0}\color{blue}{1})_!
\\2^2=4&=(000\color{green}{2}\color{blue}{0})_!
\\3^2=9&=(001\color{green}{1}1)_!
\\4^2=16&=(002\color{green}{2}0)_!
\\5^2=25&=(010\color{green}{0}\color{blue}{1})_!
\\6^2=36&=(012\color{green}{0}\color{blue}{0})_!
\\7^2=49&=(02001)_!
\\8^2=64&=(02220)_!
\\9^2=81&=(0311\color{blue}{1})_!
\\10^2=100&=(0402\color{blue}{0})_!
\\11^2=121&=(10001)_!
\\12^2=144&=(11000)_!
\\13^2=169&=(120\color{green}{0}\color{blue}{1})_!
\\ \vdots&
\end{split}$$
Table2: $x=3$
$$\begin{split}
1^3 =1&=(0000\color{green}{0}\color{blue}{1})_!
\\2^3=8&=(0001\color{green}{1}\color{blue}{0})_!
\\3^3=27&=(0010\color{green}{1}1)_!
\\4^3=64&=(0022\color{green}{2}0)_!
\\5^3=125&=(0100\color{green}{2}\color{blue}{1})_!
\\6^3=216&=(0140\color{green}{0}\color{blue}{0})_!
\\7^3=343&=(024101)_!
\\8^3=512&=(041110)_!
\\9^3=729&=(10011\color{blue}{1})_!
\\10^3=1000&=(12122\color{blue}{0})_!
\\11^3=1331&=(150121)_!
\\12^3=1728&=(222000)_!
\\13^3=2197&=(3012\color{green}{0}\color{blue}{1})_!
\\ \vdots&
\end{split}$$
Observation in table 1:
For the first column from the right $(r_1)$, the repeating pattern is "$0,1$" and length is $2$
For the second column from the right $(r_2)$, the repeating pattern is "$0,2,1,2,0,0$" and length is $6=3!$
I think for third column from the right $(r_3)$, may have repeating pattern with length $24=4!$
In general, for $n^{th}$ column from right side $(r_n)$, may have repeating pattern with length $(n+1)!$
Similar observation in table 2
Claim:
For every power$(x\in\mathbb{Z}_+)$ each $n^{th}$ columns $(r_n)$ have repeating pattern with length $(n+1)!$ ?
I think representation of base(!) is interesting, if we go deeper it might help to understand something else.
I did not create a program for the above observation (Actually I am not to good at programming nor at English) . Above observation is like similar observation from this post.
Thanks in advance for your comments, ideas and answer.
Best Answer
I’m sorry to disappoint you – it’s great that you’re looking for such patterns (and finding them), but this is unfortunately just a relatively trivial consequence of modular arithmetic.
The digit in the $n$-th column from the right is determined by the remainder of the number modulo $(n+1)!$, and for a power $m^k$, this remainder is in turn determined by the remainder of $m$ modulo $(n+1)!$, so these digits are bound to repeat with a period of (at most) $(n+1)!$, as the remainder of $m$ modulo $(n+1)!$ does.