<s>[INST] Abundant products of iterations of Euler's totient function Let $a_0(n) = n$ and $a_{i+1}(n) = \varphi(a_i(n))$ for $i\geq 0$, where $\varphi(n)$ is Euler's totient function (the number of positive integers less than or equal to $n$ and coprime with $n$). Denote $f(n) = \prod_{k=0}^{\infty}a_k(n)$.
Determine all positive integers $n$ such that the sum of positive divisors of $f(n)$ is strictly
greater than $2f(n)$.
 [/INST] Let's observe that:
\begin{align*}
a_0(n) &= n \,, \\
a_1(n) &= \varphi(n) \,, \\
a_2(n) &= \varphi(a_1(n)) = \varphi(\varphi(n)) = \varphi^{(2)}(n)\,, \\
... \\
a_k(n) &= \varphi^{(k)}(n)\,,
\end{align*}
where $\varphi^{(k)}(n)$ is $k$ times composition of the totient function. This gives us the function $f(n)$ as
\begin{align*}
f(n) = n\prod_{k=1}^{\infty}\varphi^{(k)}(n) \,.
\end{align*}
Now it is intuitively clear why $\varphi^{(k+1)}(n)\leq\varphi^{(k)}(n)$; that is, we expect the totient function to decrease with each iteration (until it becomes $1$). This means that for a finite number $n$, we have a finite number of elements in the product of $f(n)$.
The problem asks us to find all numbers $n$ that give us $2f(n)<\sigma(f(n))$, where
\begin{align*}
\sigma(f(n)) = \sum_{d|f(n)}d\,,
\end{align*}
is called the divisor function and is the sum of all divisors of $f(n)$.
Numbers that satisfy $2f(n)<\sigma(f(n))$ are called abundant numbers. However, we are going to look for numbers $2f(n)>\sigma(f(n))$, where $n$ is referred to as deficient numbers. Deficient numbers contain all odd numbers with distinct primes and all even numbers that are powers of $2$. Why we choose to look at this case will become apparent soon.
The totient function can be written as
\begin{align*}
\varphi(n) = n\prod_{p|n}\bigg(\frac{p-1}{p}\bigg)\,,
\end{align*}
where $p|n$ means all primes that divide $n$.
The only number that is even and prime is $2$. This means that numbers that are made up of primes greater than $2$ will always give $\varphi(n)$ as an even number. This means that for all $n>2$ we have that $f(n)$ will always be an even number so we do not need to worry about odd deficient numbers.
However, for $n = 2^{m}$ we have
\begin{align*}
\varphi(2^{m}) = 2^{m}\bigg(\frac{2-1}{2}\bigg) = 2^{m-1}\,.
\end{align*}
Thus $f(2^{m})$ is
\begin{align*}
f(2^{m}) = 2^{m+(m-1)+(m-2)+...+1} = 2^{m(m+1)/2}
\end{align*}
and will be a deficient number because $f(2^{m})$ is a power of $2$ (as stated before).
We need to also exclude perfect numbers that satisfy $\sigma(f(n)) = 2f(n)$. We already worked out that $f(n)$ is always even and we do not need to worry about odd perfect numbers (still an open problem in number theory). Even perfect numbers are of the form $2^{p-1}(2^{p}-1)$, where $p$ is a prime and $2^{p}-1$ is also a prime (Mersenne prime). We need $n = 2^{p} - 1$ because this is the only term that is odd. Now if we apply the totient function $\varphi(n) = n - 1 = 2^{p}-2$ (because $n$ is prime and will be always coprime with all numbers up to itself), we expect $\varphi(n) = 2^{p-1}$ and this gives us the equation
\begin{align*}
2^{p} - 2 &= 2^{p-1} \\
2^{p-1} - 1 &= 2^{p-2} \,.
\end{align*}
The LHS is odd while RHS is even. So only $p = 2$ is valid and we get $n = 3$ and $f(3)$ will be a perfect number.
In conclusion, the numbers that do not satisfy the inequality are $3$ and $2^{m}:m\in\mathbb{N}$. Any other number satisfies the inequality.
</s>