Properties of the recurrent sum of cubes of the digits of integers

Question

While answering this question on Mathematica.SE, I started wondering about the properties of the sequence asked about.

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Consider $a_1=12$, and perform

$a_1 = 12 \rightarrow (1^3 + 2^3) = 9 = a_2$.

Then again

$a_2 = 9 \rightarrow 9^3 = 729 = a_3$,

and

$a_3 = 729 \rightarrow (7^3 + 2^3 + 9^3) = 1080 = a_4$,

and so on. We arrive at the sequence $(12, 9, 729, 1080, 513, 153, 153, 153, 153, ...)$, with the number $153$ repeating on and on. Let's examine how the sequence looks like for different starting values $a_1$:

It turns out, that for $a_1\in\{1, 2, 3, \ldots, 99, 100\}$ the sequence either ends up with a repeating single number, or with two/three alternating values.

So let's check these multiplicities for $a_1\leq 10^4$:

Multiplicity $1$ occurs 8543 times, $2$ --- 501 times, and $3$ --- 956 times.

I haven't noticed any obvious pattern in the $a_1$s that lead to different multiplicities of the limitting sequence. Questions that arise:

1. Are the numbers $1,2,3$ the only multiplicities possible for all $a_1\in\mathbb{N}$?
2. What governs the value of the multiplicity in dependence on $a_1$?
3. How universal this behaviour is when sums of squares, quartics, etc. are considered?
4. Do other types of functions of the IntegerDigits have such interesting/universal properties?
5. What are other interesting questions to ask about?

Answer 1

1. Yes, 1,2,3 are the only possible multiplicities. We can prove this as follows: Firstly, we prove that the sequence must eventually loop on numbers below 10000 (we only need 10000, but this bound can easily be improved below 3000). For an n+1-digit integer where n+1>4 in decimal notation: $$\sum_{k=0}^n{10^k b_k}$$ we consider the difference between it and the next number in the sequence: $$\sum_{k=0}^n{10^k b_k} - \sum_{k=0}^n{b_k^3} = \sum_{k=0}^3{b_k(10^k-b_k^2)}+\sum_{k=4}^n{b_k(10^k-b_k^2)}$$ Now it is clear that $$b_k \neq 0\implies b_k(10^k-b_k^2)>b_k(10000-b_k^2)\ge9999$$for all k at least 4, (if necessary you can verify with a calculator, for 1 to 9), and clearly for some k at least 4, we have $$b_k \neq 0$$ as the number is at least 5 digits. However, we also have $$\sum_{k=0}^3{b_k(10^k-b_k^2)}\ge\sum_{k=0}^3{-b_k^3}\ge 4(-9)^3 = -2916$$ So now we have $$\sum_{k=0}^3{b_k(10^k-b_k^2)}+\sum_{k=4}^n{b_k(10^k-b_k^2)}>9999-2916>0$$ So therefore from any 5 or more digit number the next number is lower, implying we reach a 4 digit number at some point. Furthermore, from any 4 digit number, the next number in the sequence is at most $$4(9)^3 = 2916<10000$$ So this means that after some point, the sequence must be bounded below 10000 and so must loop below this number. But you have already checked all multiplicities below 10000, and clearly they are all 1,2 or 3. So 1,2,3 are the only possibilities.
2. The multiplicity is dependent on what 4 digit number your starting number eventually reduces to, and I do not think there is a simple explicit formula for this.
3. A similar method can be used for sums of other powers to bound and then check, although I suspect an answer for general n will be harder to find.