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 $$2f(n)$$.

• No, the definition implies $\varphi(1) = 1$. Apr 2 '21 at 17:52
• My first analysis of the question is that these $f(n)$ numbers are highly composite and most of them should be abundant numbers. I think the proof should just focus on the tail of the totient chain which should end with a series of powers of $2$ followed by "the product of some Fermat primes and a power of two ". If the Fermat primes are $3$ or $5$ then $f(n)$ should be divisible by $6$ or $20$ and this makes the number abundant. it remains to check other general Fermat primes and this probably would give proof that only powers of $2$ do not satisfy the abundance criteria Apr 4 '21 at 16:30

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.

• +1, seems correct. But please do make it concise. You've dragged out the answer really long, a lot of trivial stuff can be omitted. Apr 2 '21 at 4:18
• Thanks for the feedback, new around here and will keep it in mind. Apr 2 '21 at 10:27
• @Schwauss how do we know that for the case of even deficient numbers we need to consider only powers of 2? I was not able to find a corresponding reference. Apr 2 '21 at 17:58
• @Schwauss But it says that all powers of two are deficient but not that all even deficient numbers are powers of two? Apr 3 '21 at 12:12
• I don't think this answers the question Apr 4 '21 at 16:19