# A problem inspired in the definition of tau numbers and a divisibility relationship related to powers of two

It is (I assume that this easy fact is well-known) obvious that an integer $$n>1$$ is a power of two $$n=2^{\alpha}$$, where $$\alpha\geq 1$$ is integer, if an only if $$n$$ satisfies the divisibility relationship $$(\text{rad}(n)\cdot\varphi(n))\mid n\tag{1}$$ where $$\text{rad}(n)=\prod_{p\mid n}p$$ denotes the product of distinct prime numbers dividing $$n$$ and $$\varphi(n)$$ is the Euler's totient function.

(A draft of the proof is that for integers $$n>1$$ one can to write $$\text{rad}(n)\varphi(n)=n\prod_{p\mid n}(p-1)$$ thus our integer $$n$$ satistying $$(1)$$ has $$\omega(n)=1$$ distinct prime factor in its factorization, this arithmetic function $$\omega(x)$$ counts the number of distinct prime factors, and one has $$p-1=1$$ implies thus that our integer has the form $$n=2^{\alpha}$$, where $$\alpha\geq 1$$ is integer.)

Remarks. A) One knows that the following series is convergent $$\sum_{n=1}^\infty\frac{1}{2^n}$$, thus the sum of the reciprocals of the solutions of $$(1)$$ is convergent (really $$n=1$$ also will be a solution and thus the previous series has value $$2$$ when we consider that $$1$$ also is a power of two $$1=2^{0}$$ and solution of $$(1)$$). B) We notice that the Euler's totient function is a counting function, because counts the positive integers $$k$$ up to $$n$$ that satisfy $$(n,k)=1$$.

This week I've known the nice article [1] in which were studied problems concerning the known as tau numbers (see introductory section of [1], this journal is a free access electronic journal, in the first paragraph the author recalls the definition, or see the linked Wikipedia). From the definition of tau numbers, (look at if you want the Wikipedia Refactorable number) I was inspired to define the following variation of tau numbers, or if you prefer a variation of the divisibility relationship $$(1)$$ since $$\tau(n)=\sigma_0(n)=\sum_{1\leq d\mid n}1$$ also is a counting function.

Definition. For integers $$n>1$$, I say that $$n$$ is a rad-refactorable number if and only if satisfies $$(\text{rad}(n)\cdot\tau(n))\mid n,\tag{2}$$ where $$\tau(n)$$ denotes the number-of-divisors function.

The first few terms of this sequence are the following where I've considered thus add $$n=1$$ as a solution $$(2)$$

$$a_1=1, a_2=8,a_3=9,a_4=72,a_5=128, a_6=384,a_7=625,\ldots$$

Thus we've redefined the solutions as those positive integers $$n\geq 1$$ satisfying $$(2)$$ with the purpose to include $$a_1=1$$ as a legitimate rad-refactorable number.

Question. Can you to prove or refute that $$\sum_{\substack{\text{positive integers } a_m\\ \text{that are rad-refactorable}}}\frac{1}{a_m}$$ is convergent? Many thanks.

I've computed more terms of the sequence, my thought is that isn't easy to bound the terms $$\frac{1}{a_m}$$ using a geometric series, and I don't know if there are infinitely many rad-refactorable numbers.

## References:

[1] Joshua Zelinsky, Tau Numbers: A Partial Proof of a Conjecture and Other Results, Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.8.

• To avoid being pedantic, I have not written in the body of the post that one can also do the comparison $1\cdot\tau(n)$ versus $\text{rad}(n)\cdot\tau(n)$ to show that the functions $f(n)=1$ and $\text{rad}(n)$ are both multiplicative functions. – user142929 Oct 8 '19 at 9:45

If I'm understanding your problem correctly, your sum diverges and does so very rapidly. Every positive integer of the form $$2^7p$$ for an odd prime $$p$$ is a rad-refactorable number, so your sum is bounded below by $$\sum \frac{1}{2^7p} = \frac{1}{2^7}\sum \frac{1}{p}.$$ Here the sums are over all odd primes. Since the sum of the sum of the reciprocals of the primes diverges, so does this sum.

Let $$A$$ be the set of tau numbers and let $$B$$ be the set of rad-refactorable numbers. More generally then the set of numbers of the form $$2^{2^k-1}p_1p_2 \cdots p_s$$ is in $$B$$ (and hence in $$A$$) as long as $$s$$ is only a little smaller than $$2^k$$, where $$p_1 \cdots p_s$$ are allowed to range over distinct primes. Set

$$f(x)=\sum_{n \leq x, n \in A} \frac{1}{n},$$

and set

$$g(x)=\sum_{n \leq x, n \in B} \frac{1}{n}.$$

Then trivially $$f(x) \geq g(x)$$ and $$g(x)$$ grows faster than the sum of the reciprocals of products of $$m$$ distinct primes for any $$m$$, as long as one goes out far enough, due to the construction above. I don't know how much faster $$f(x)$$ grows compared to $$g(x)$$. My guess is that their overall growth rates are will be similar since it isn't that hard to write down lots of other sequences of rad-refactorable numbers. I don't know anything about the asymptotics for rad-refactorable numbers themselves. There is a paper by Claudia Spiro which gives a strong bound on the asymptotics for the tau numbers, and that might be worth checking out; I suspect one can use her techniques to get a slightly tighter upper bound on the rad-refactorable numbers.

• Many thanks for your great answer, it is incredible. I am going to study it this week, but I will accept it as answer in the next few hours. – user142929 Oct 8 '19 at 12:51