Theorem: Existence of Arbitrarily Long Aliquot Sequences

It is possible to construct an aliquot sequence of arbitrary length which is monotonically increasing.

Proof:
Let $p_i$ be the $i$th prime.
Construct the sequence $\sequence {t_i}_{t \mathop = 1}^\infty$ of natural numbers $t_i$ recursively as follows:
:$t_1 = 2$
:$t_{i + 1} = \map \phi {p_i^{t_i + 1} } - 1 = p_i^{t_i} \paren {p_i - 1} - 1$ for $i \ge 1$
We claim that this sequence has the property:
:$(1): \quad p_i^{t_i + 1} \divides \map {\sigma_1} {p_{i + 1}^{t_{i + 1} } }$ for $i \ge 1$
For $i = 1$, $p_1 = 2$ and $t_1 = 2$.
:$t_2 = 2^2 \paren {2 - 1} - 1 = 3$
:$\map {\sigma_1} {p_2^{t_2} } = \map {\sigma_1} {3^3} = 40$
which is a multiple of $2^3 = 8$.
For $i > 1$:
{{begin-eqn}}
{{eqn | l = \paren {p_{i + 1} - 1} \map {\sigma_1} {p_{i + 1}^{t_{i + 1} } }
      | r = p_{i + 1}^{t_{i + 1} + 1} - 1
      | c = Divisor Sum of Power of Prime
}}
{{eqn | r = p_{i + 1}^{\map \phi {p_i^{t_i + 1} } } - 1
}}
{{eqn | r = 0
      | rr = \pmod {p_i^{t_i + 1} }
      | c = Euler's Theorem (Number Theory)
}}
{{end-eqn}}
For $i > 1$, $p_{i + 1} - 1 > 2$.
Hence $p_{i + 1} - 1$ is even and composite.
Since $p_i$ is odd and $p_i > \dfrac {p_{i + 1} - 1} 2$ by Bertrand-Chebyshev Theorem:
:$p_{i + 1} - 1$ and $p_i^{t_i + 1}$ are coprime.
By Euclid's Lemma, $p_i^{t_i + 1} \divides \map {\sigma_1} {p_{i + 1}^{t_{i + 1} } }$.
We define for $l \ge 1$: $A_l = \set {m \in \N: p_i^{t_i} \divides m, p_i^{t_i + 1} \nmid m, \forall i: 1 \le i \le l}$.
Denote $\map s n$ the aliquot sum of $n$.
Then for $m \in A_l$ where $l \ge 2$, we claim:
{{begin-eqn}}
{{eqn | n = 2
      | l = \map s m
      | o = >
      | r = m
}}
{{eqn | n = 3
      | l = \map s m
      | o = \in
      | r = A_{l - 1}
}}
{{end-eqn}}
Let $m \in A_l$ where $l \ge 2$.
From definition we have:
{{begin-eqn}}
{{eqn | l = p_1^{t_1}
      | r = 4 \divides m
}}
{{eqn | l = p_2^{t_2}
      | r = 27 \divides m
}}
{{end-eqn}}
Hence $m$ is a multiple of $27 \times 4 = 108$.
Since $108$ is abundant, $m$ is also abundant by Multiple of Abundant Number is Abundant.
$\map s m > m$ follows from definition of abundant numbers.
Write $m = p_1^{t_1} \cdots p_l^{t_l} k$, where $k$ is not divisible by $p_1, \dots, p_l$.
Then:
{{begin-eqn}}
{{eqn | l = \map s m
      | r = \map {\sigma_1} m - m
}}
{{eqn | r = \map {\sigma_1} {p_1^{t_1} } \map {\sigma_1} {p_2^{t_2} } \cdots \map {\sigma_1} {p_l^{t_l} } \map {\sigma_1} k - m
      | c = Divisor Sum Function is Multiplicative
}}
{{eqn | r = K p_1^{t_1 + 1} \cdots p_{l - 1}^{t_{l - 1} + 1} - p_1^{t_1} \cdots p_l^{t_l} k
      | c = for some $K \in \N$, by $(1)$
}}
{{eqn | r = p_1^{t_1} \cdots p_{l - 1}^{t_{l - 1} } \paren {K p_1 \cdots p_{l - 1} - p_l^{t_l} k}
}}
{{end-eqn}}
Since $K p_1 \cdots p_{l - 1}$ is divisible by $p_1, \dots, p_{l - 1}$ while $p_l^{t_l} k$ isn't:
:$p_1, \dots, p_{l - 1} \nmid \paren {K p_1 \cdots p_{l - 1} - p_l^{t_l} k}$
Hence for each $i$ where $1 \le i \le l - 1$:
:$p_i^{t_i} \divides \map s m, p_i^{t_i + 1} \nmid \map s m$
This shows that $\map s m \in A_{l - 1}$.
Hence to create an aliquot sequence of length $l$, one picks an element $m$ from $A_l$.
Applying $(2)$ and $(3)$ repeatedly, we see that:
:$m, \map s m, \map s {\map s m}, \dots, \map {s^l} m$
is a monotonically increasing aliquot sequence of length $l$.
{{qed}}