Additional Examples Of Calculating Aliquot Sequences

Recall that the recursive formula for an aliquot sequence is as follows:

 Definition: The aliquot sequence for a number n is defined recursively with the first element in the sequence being $s_0 = n$, and the kth element being $s_k = \sigma (s_{k-1}) - s_{k-1}$.

Also recall that $\sigma (p^n) = \frac{p^{n+1} - 1}{p - 1}$.

We will now go through some more aliquot sequence examples.

# Example 1

Calculate the terms $s_1, s_2, s_3$ and $s_4$ of the aliquot sequence of 425.

We start with $s_0 = 425$.

(1)
\begin{align} s_1 = \sigma (425) - 425 \\ s_1 = \sigma (5^2) \sigma (17) - 425 \\ s_1 = \frac{5^{2+1} - 1}{5 - 1} \cdot (18) - 425 \\ s_1 = \frac{5^{3} - 1}{4} \cdot (18) - 425\\ s_1 = (31)(18) -425 \\ s_1 = 558 - 425 \\ s_1 = 133 \end{align}
(2)
\begin{align} s_2 = \sigma (133) - 133 \\ s_2 = \sigma (7) \sigma (19) - 133 \\ s_2 = (8)(20) - 133 \\ s_2 = 160 - 133 \\ s_2 = 27 \end{align}
(3)
\begin{align} s_3 = \sigma (27) - 27 \\ s_3 = \sigma (3^3) - 27 \\ s_3 = \frac{3^{3+1} - 1}{3 - 1} - 27 \\ s_3 = \frac{3^{4} - 1}{2} - 27 \\ s_3 = 40 - 27 \\ s_3 = 13 \end{align}
(4)
\begin{align} s_4 = \sigma (13) - 13 \\ s_4 = (14) - 13 \\ s_4 = 1 \\ \end{align}

## Example 2

Let $s_i$ denote the elements in the aliquot sequence of 600, and let $t_i$ denote the elements in the aliquot sequence of 602. For which 1 ≤ i ≤ 5 is $s_i ≥ t_i$?

To solve this question we must first calculate the s_1 to s_5 and t_1 to t_5 terms of the aliquot sequences for both 600 and 602. Let's start with 600 that has $s_0 = 600$.

(5)
\begin{align} s_1 = \sigma (600) - 600 \\ s_1 = 1860 - 600 \\ s_1 = 1260 \end{align}
(6)
\begin{align} s_2 = \sigma (1260) - 1260 \\ s_2 = 4368 - 1260 \\ s_2 = 3108 \end{align}
(7)
\begin{align} s_3 = \sigma (3108) - 3108 \\ s_3 = 8512 - 3108 \\ s_3 = 5404 \\ \end{align}
(8)
\begin{align} s_4 = \sigma (5404) - 5404 \\ s_4 = 10864 - 5404 \\ s_4 = 5460 \end{align}
(9)
\begin{align} s_5 = \sigma(5460) - 5460 \\ s_5 = 18816 - 5460 \\ s_5 = 13356 \end{align}

Now let's calculate some values of the aliquot sequence of 602 with $t_0 = 602$.

(10)
\begin{align} t_1 = \sigma (602) - 602 \\ t_1 = 1056 - 602 \\ t_1 = 454 \end{align}
(11)
\begin{align} t_2 = \sigma (454) - 454 \\ t_2 = 684 - 454 \\ t_2 = 230 \end{align}
(12)
\begin{align} t_3 = \sigma (230) - 230 \\ t_3 = 432 - 230 \\ t_3 = 202 \end{align}
(13)
\begin{align} t_4 = \sigma (202) - 202 \\ t_4 = 306 - 202 \\ t_4 = 104 \end{align}
(14)
\begin{align} t_5 = \sigma (104) - 104 \\ t_5 = 210 - 104 \\ t_5 = 106 \end{align}

So for all integers 1 ≤ i ≤ 5, $s_i ≥ t_i$.

## Example 3

Suppose that $\sigma (s_0) = 1092$ and $s_1 = 592$. What is one possible value of $s_0$?

We know that the recursive formula for aliquot sequences is $s_k = \sigma (s_{k-1}) - s_{k-1}$. Hence we know that:

(15)
\begin{align} s_1 = \sigma (s_{0}) - s_{0} \\ 592 = 1092 - s_{0} \\ s_0 = 1092 - 592 \\ s_0 = 500 \end{align}