We denote the radical of the integer $n> 1$ as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\ p\text{ prime}}}p,$$ taking $\operatorname{rad}(1)=1$ that is this definition from Wikipedia.

In this post we do a comparison with the so-called Faulhaber polynomials (their factorization), see this Wikipedia in the way that I am going to define sequences using the arithmetic function $a(k)=\operatorname{rad}(n)$ instead of the arithmetic function $a(k)=k$ inside the finite sums $$\sum_{k=1}^n (a(k))^p$$ where $p\geq 1$ is a fixed integer. The comparison that we evoke is ask us when (for a fixed integer $p\geq 1$) the sequence $$\sum_{k=1}^n (\operatorname{rad}(k))^p\tag{1}$$ reaches/contains infinitely many prime numbers, this is that our sequence $(1)$ has infinitely many terms being prime numbers for a fixed choice of $p$.

I think that this conjecture is wild, since I did few experiments (and I hope that these were rights) and by the mentioned comparison. Any case I would like to know feedback about our question from this MSE.

Question. Can you find an integer (or what do you think about it) $p\geq 1$ such that the sequence $$\sum_{k=1}^n (\operatorname{rad}(k))^p$$ contains only a finite number of prime numbers in its terms as $n$ runs over the positive integers $\geq 1$? Or well, what are your heuristic or reasoning to elucidate if this kind of sequences $(1)$ should have finitely many or infinitely many prime numbers for a choice of $p$, when $n\geq 1$? Many thanks.

I think that this question can be very difficult. I should to choice some answer that provide good feedback about it, thus isn't required solve all minor questions.

I did this variation since the radical of an integer $\operatorname{rad}(n)$ is an important arithmetic function (related to the so-called abc conjecture) in analytic number theory, is a multiplicative function and it's like a substitute of the function $a(n)=n$.

  • $\begingroup$ Many thanks for previous upvote. $\endgroup$ – user243301 Apr 8 '18 at 13:56
  • $\begingroup$ Why the prime factors of $\sum_{k=1}^{n}\text{rad}(k)^p$ should somehow relevant? $\endgroup$ – Jack D'Aurizio Apr 8 '18 at 18:15
  • $\begingroup$ I am agree that there is no a special reason why to study the prime numbers of the form $\sum_{k=1}^{n}\text{rad}(k)^p$. It doesn't seem relevant, just was that it seems to me a curious phenomenon that the few sequences that I've calculated seem to have many primes. Many thanks @JackD'Aurizio $\endgroup$ – user243301 Apr 8 '18 at 18:45
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    $\begingroup$ How many primes do you mean by "many primes"? I computed the first 100,000 terms for the case $p = 2$ and observed that around 3% of the terms were primes. $\endgroup$ – rwbogl Apr 16 '18 at 1:43
  • $\begingroup$ Many thanks in the question I am asking if we can find that for some $p\geq 1$ there are finitely many or infinitely many prime numbers. Many thanks for your computations and help @rwbogl Feel free to add your remarks about the case $p=2$ or /and computational evidence or reasonings about other cases in a table as a (partial) answer. $\endgroup$ – user243301 Apr 16 '18 at 6:41

This is far from a complete answer to your question, but since you have the experimental-mathematics tag you might appreciate the computations.

Below is a graph counting the number of primes in the sequence $a_p(n) = \sum_{k = 1}^n (\operatorname{rad}(k))^p$ from $n = 1$ to $n = 10,000$, and for $p$ from $1$ to $60$. Note that the sequence is strictly increasing, so each prime counted is actually unique.

enter image description here

The count seems to decay quickly for large $p$, but this isn't proof that there are only finitely many primes in $\{a_p(n)\}_n$ for any $p$.

Below is a graph showing how the count of primes in $\{a_2(n)\}_{n = 1}^u$ develops as $u$ grows. That is, it shows what proportion of the numbers $a_p(n)$ are prime from $n = 1$ to $n = u$ as $u$ varies for $p = 2$.

Proportion of primes in partial rad-sum sequence.

It is interesting to note that this graph seems to level off. Perhaps the proportion approaches some non-zero value.

  • $\begingroup$ I appreciate your answer because is very nice and has a high quality. Very good work, many thanks. $\endgroup$ – user243301 Apr 16 '18 at 15:50

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