Descriptive Numbers: Find 10-Digit Number with f Function Cycle of Length 3+

[magneto1]

I ran into an interesting problem while working on a problem set. Given a $10$-digit number $n$ (for our purpose, we will allow $n$ to have leading $0$'s, so you can treat this as a string of $10$-digits). Define a function $f$ that maps $n$ to another $10$-digit number $m$ written in its decimal form $\overline{m_0m_1\dotso m_9}$ where the digit $m_i$ is the number of digit $i$ in $n$.

E.g. $n = 1,890,091,000$. This number has $5$ zeroes, $2$ ones,
$1$ eight, and $1$ nine. So, $f(n) = 5,200,000,012$.

Suppose we iterate this indefinitely (and we can do that quickly using some Mathematica code). From some initial investigation, two possibilities occur.

(a) We will come to a $t$ where $f(t) = t$. This happens when $t = 6,210,001,000$.

or (b) We will come to a $t$ where $f(t) = f(f(t))$. This happens when $t = 6,300,000,100$ or $t=7,101,001,000$.

Since the number of $10$-digit integers is finite, $f \circ f \circ \dotso$ has to form a cycle. From above, there is a cycle of length $1$ and length $2$. Is there any $10$-digit $n$ where repeatedly applying $f$ will lead to a cycle whose length $\geq 3$?

 

[jvicens]

Thank you for sharing your interesting problem with us. I find this problem intriguing and I would like to offer some insights and explanations.

Firstly, it is important to note that the function $f$ you have defined is a type of permutation function, where the digits of the input number are rearranged according to their frequency. This type of function is commonly studied in mathematics and has many interesting properties.

Now, let us consider the two possibilities you have mentioned. In case (a), where we eventually reach a fixed point $t$ such that $f(t)=t$, this means that the digits of $t$ are already in their correct frequency, so applying $f$ again will not change the number. This is why we get a cycle of length $1$.

In case (b), where we eventually reach a number $t$ such that $f(t)=f(f(t))$, this means that $f(t)$ is the same as $t$ but with its digits rearranged. So, applying $f$ again will give us the same number $t$. This is why we get a cycle of length $2$.

Now, to answer your question about whether there exists a $10$-digit number $n$ where repeatedly applying $f$ will lead to a cycle of length $\geq 3$, the answer is no. This is because the maximum possible length of a cycle in this scenario is $10$, and we have already exhausted all the possible cycles of length $1$ and $2$.

To understand this further, let us consider the case where $n$ is a $10$-digit number with all distinct digits. In this case, the function $f$ will simply rearrange the digits of $n$, and it is easy to see that we will eventually reach a cycle of length $10$. This means that for any other $10$-digit number $n$, we will either reach a cycle of length $1$ or $2$, or we will eventually reach a cycle of length $10$.

In conclusion, while it is possible to have cycles of length $1$ or $2$ when repeatedly applying the function $f$, it is not possible to have a cycle of length $\geq 3$ for any $10$-digit number $n$. I hope this helps to clarify your doubts.