Repeated Sum of Cubes of Digits of Multiple of 3

Theorem
Let $k \in \Z_{>0}$ be a positive integer.

Let $f: \Z_{>0} \to \Z_{>0}$ be the mapping defined as:


 * $\forall m \in \Z_{>0}: \map f m = $ the sum of the cubes of the digits of $n$.

Let $n_0 \in \Z_{>0}$ be a (strictly) positive integer which is a multiple of $3$.

Consider the sequence:
 * $s_n = \begin{cases} n_0 & : n = 0 \\

\map f {s_{n - 1} } & : n > 0 \end{cases}$

Then:
 * $\exists r \in \N_{>0}: s_r = 153$

That is, by performing $f$ repeatedly on a multiple of $3$ eventually results in the pluperfect digital invariant $153$.

Proof
We verify by brute force:


 * Starting on $n_0 \le 2916 = 4 \times 9^3$, we will end on $153$.

First we prove that if $3 \divides n_0$, then $3 \divides \map f {n_0}$.

From Divisibility by 3:

The sum of digits of $n_0$ is divisible by $3$.

By Fermat's Little Theorem:


 * $\forall x \in \Z: x^3 \equiv x \pmod 3$

Therefore sum of the cubes of the digits of $n_0$ is also divisible by $3$.

That is the definition of $\map f {n_0}$.

Thus $3 \divides \map f {n_0}$.

It remains to be shown that for every $n_0 > 2916$, $\map f {n_0} < n_0$.

For $n_0 \le 9999$:
 * $\map f {n_0} \le 9^3 + 9^3 + 9^3 + 9^3 = 2916 < n_0$

Now suppose $n_0$ is a $k$-digit number, where $k \ge 5$.

Then:

This shows that $\map f {n_0} < n_0$ for all $n_0 > 2916$.

Thus for any $n_0 > 2916$, $s_n$ is eventually less than $2916$.

Therefore we will eventually reach $153$.