Squares of sum of digits of numbers part 3

Question

This is the third part of my first post about properties of sum of squares of a number. Anyone who is new may refer to the link below.

Squares of sum of digits of a number

Here is a quick summary:

Take a number then add up it digits and square the result. Repeat the process to the number obtained and keep doing so. I the end we will get either $1,81,169,256$. This was proved in the first post.

Now suppose we do this to a number until we get $1,81,169,256$. The moment we get anyone number out of the $4$ we stop the sequence. What is the probability that the sequence ends in any one of $1,81,169,256$.

Answer 1

Here's a partial answer, under the assumption that you want to take a uniform random starting number from $\{1, \dots, N\}$ and then look at the limiting behavior as $N \to \infty$.

Let $f(n)$ denote the first number from $\{1, 81, 169, 256\}$ that we land on starting at $n$. If $s(n)$ is the sum of the digits of $n$, then $s(n) \equiv n \pmod 9$. This is well-known and easy to show. Then $s(n)^2 \equiv n^2 \pmod 9$.

Let's look at numbers and their squares modulo $9$. I'll split them into three classes: $A = \{0, 3, 6\}$, $B = \{1, 8\}$, and $C = \{2, 4, 5, 7\}$. Every number in each class squares to another number in the same class. Therefore, if our starting number $n$ is in a certain class, then $f(n)$ is in the same class. If $n \in A \pmod 3$ then $f(n) = 81$, and if $n \in B \pmod 9$ then $f(n) = 1$. As $N \to \infty$, the probability that a uniform random $n$ falls into class $A$ approaches $1 / 3$. And similarly for class $B$ we get $2 / 9$.

Now, how is the remaining $4 / 9$ probability split between $169$ and $256$? This I can't answer. All numbers in class $C$ alternate between $4$ and $7$ modulo $9$, so I would have guessed that about half would land on $169 \equiv 7 \pmod 9$ and the other half on $256 \equiv 4 \pmod 9$. However, running a computer simulation seems to indicate that very large numbers are about twice as likely (or more) to land on $169$ than $256$.

Answer 2

I have performed a computer simulation that calculates the exact probabilities of ending at $\{0, 1, 81, 169, 256\}$ from a starting number in $\{0, 1, \dots, 10^d\}$ for $1 \leq d < 5000$. Note that ending at $0$ is only possible if the starting number equals $0$, so this probability will vanish as $d \to \infty$.

I have plotted the result, which can be seen below. In the plot, the value of $d$ is listed on the x-axis and the probability of ending at each of $\{0, 1, 81, 169, 256\}$ is listed on the y-axis.

Note that this answer does not answer the question, it is merely an interesting result, which did not fit in a comment.

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Update: I improved my algorithm, which can now approximate (to unknown accuracy) the probabilities of ending at $\{0, 1, 81, 169, 256\}$ from a starting number in $\{0, 1, \dots, 10^{2^p}\}$ for $1 \leq p \leq 26$.

Listing the probability of ending at $169$ and $256$, respectively, for some $p$'s we get:

20: [0.3229072415185169, 0.1215372029259365]
21: [0.3525997163737325, 0.0918447280707314]
22: [0.3565506315236761, 0.0878938129207874]
23: [0.3553829797083236, 0.0890614647361787]
24: [0.3092395404207834, 0.1352049040237175]
25: [0.3217452008711443, 0.1226992435733908]
26: [0.3294311770014319, 0.1150132674431534]

which give lower probabilities of ending at $256$ than the values I computed before. This may signify that as $p \to \infty$ we may have $0$ probability of ending at $256$, but that is just a guess.