Aliquot Sequences, Amicable Pairs, and Sociable Numbers Examples 1

# Aliquot Sequences, Amicable Pairs, and Sociable Numbers Examples 1

Recall from the Aliquot Sequences, Amicable Pairs, and Sociable Numbers page that if $n \in \mathbb{N}$ then the corresponding aliquot sequence of $n$ is defined recursively as:

(1)
\begin{align} \quad s_0 & = n \\ \quad s_k & = \sigma(s_{k-1}) - s_{k-1} \quad \forall k \in \mathbb{N} \end{align}

We said that two natural numbers $m, n \in \mathbb{N}$ are an amicable pair if $\sigma(m) - m = n$ and $\sigma(n) - n = m$.

We also said that the collection of natural numbers $\{ n_1, n_2, ..., n_k \}$ are sociable if $\sigma(n_1) - n_1 = n_2$, $\sigma(n_2) - n_2 = n_3$, …, $\sigma(n_k) - n_k = n_1$.

We will now look at some problems regarding this material.

# Example 1

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

We start with $s_0 = 425$.

(2)
\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}
(3)
\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}
(4)
\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}
(5)
\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 $s_0 = 600$.

(6)
\begin{align} s_1 &= \sigma (600) - 600 \\ s_1 &= 1860 - 600 \\ s_1 &= 1260 \end{align}
(7)
\begin{align} s_2 &= \sigma (1260) - 1260 \\ s_2 &= 4368 - 1260 \\ s_2 &= 3108 \end{align}
(8)
\begin{align} s_3 &= \sigma (3108) - 3108 \\ s_3 &= 8512 - 3108 \\ s_3 &= 5404 \\ \end{align}
(9)
\begin{align} s_4 &= \sigma (5404) - 5404 \\ s_4 &= 10864 - 5404 \\ s_4 &= 5460 \end{align}
(10)
\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 $t_0 = 602$.

(11)
\begin{align} t_1 &= \sigma (602) - 602 \\ t_1 &= 1056 - 602 \\ t_1 &= 454 \end{align}
(12)
\begin{align} t_2 &= \sigma (454) - 454 \\ t_2 &= 684 - 454 \\ t_2 &= 230 \end{align}
(13)
\begin{align} t_3 &= \sigma (230) - 230 \\ t_3 &= 432 - 230 \\ t_3 &= 202 \end{align}
(14)
\begin{align} t_4 &= \sigma (202) - 202 \\ t_4 &= 306 - 202 \\ t_4 &= 104 \end{align}
(15)
\begin{align} t_5 &= \sigma (104) - 104 \\ t_5 &= 210 - 104 \\ t_5 &= 106 \end{align}

So for all integers $i \in \{ 1, 2, 3, 4, 5 \}$ we have $s_i ≥ t_i$.

## 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:

(16)
\begin{align} s_1 &= \sigma (s_{0}) - s_{0} \\ 592 &= 1092 - s_{0} \\ s_0 &= 1092 - 592 \\ s_0 &= 500 \end{align}