Gaps between consecutive prime powers

Question

After I've read and I've understand [1] (I have no free access), I would like to know if is it possible to prove or discard that the sequence A025528, that counts the number of prime powers less than or equal to a fixed integer $n\geq 1$ in OEIS has gaps between their terms arbitraly large, as same as than occurs in the sequence of prime numbers.

With it, and calculations and reasoning of the author should be easy to deduce, in case that the sequence of gaps is unbounded, the irrationality of a real number following the proof in the article.

Question. Are the gaps between consecutive terms in the sequence of prime powers A000961 arbitrarly large? I am asking if such sequence is unbounded. I think that should be well known. Thanks in advance.

As a detail, and as comparison with a reasoning in the article I know that, for instance, in the sequence $4!+1,4!+2,4!+3,4!+4$ the first term is a prime power, $25$.

References:

[1] Cilleruelo, Una serie que converge a un número irracional, La Gaceta de la RSME, Vol. 18 (2015), No. 3, page 568.

It would be better to refer to the OEIS sequence https://oeis.org/A000961 -- see also https://oeis.org/A057820 — Barry Cipra, Apr 02 '17 at 20:02
See also http://math.stackexchange.com/questions/288439/is-the-asymptotic-density-of-powers-of-primes-zero — Barry Cipra, Apr 02 '17 at 20:16

score 2 · Accepted Answer · answered Apr 27 '17 at 04:05

2

In On the Densities of some subsets of integers (Missouri J. Math. Sci. Volume 19, Issue 3 (2007), 167-170), Florian Luca proves that the set $SP$ of sigma-primes (positive integers $n$ for which $n$ and $\sigma(n)$ are coprime) has asymptotic density zero. Since the set of prime powers is a subset of $SP$, the set of prime powers has asymptotic density zero. As a result, there must exist arbitrarily large gaps between consecutive prime powers.

(Added: there are easier ways to prove the density of the prime powers is zero, but this is one official/credible source.)

answered Apr 27 '17 at 04:05

Matthew Conroy

11,234

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If the gap of a strictly increasing sequence $a_n \in \mathbb{Z}{\ge 1}$ is bounded then $\lim \inf{s \to 1^+} \frac{\sum_n a_n^{-s}}{\zeta(s) } >0$. Here $\sum_n a_n^{-s} = \sum_{p^k} p^{-sk} = \log \zeta(s)+O(1)$ so that $\lim \inf_{s \to 1^+} \frac{\sum_n a_n^{-s}}{\zeta(s) } = 0$ – reuns Apr 27 '17 at 04:12
I will read when I can Florian's article, is a very good mathematician. And many thanks for your reference, when I can I choose it as an answer. – Apr 27 '17 at 08:52
Merci @user1952009 for your remarks, feel free to add your remarks as an answer. – Apr 27 '17 at 08:53

score 2 · Answer 2 · answered Apr 27 '17 at 06:16

Matthew Conroy's answer shows that this follows easily from well-known facts about the density of prime powers. Here is a more elementary approach: the OP seems to be aware of the standard construction that for $n\ge 1$ the set of numbers $$n! + 2, n! + 3, \ldots, n!+n$$ forms a string of $n-1$ consecutive non-primes. In fact it's easy to adjust this construction to ensure that none of them are prime powers: just replace $n!$ by a larger function $F(n)$ with the property that for any prime power $p^r \le n$, $F(n)$ is a multiple of $p^{r+1}$. Then $$F(n) + 2, F(n) + 3, \ldots, F(n)+n$$ contains no prime powers, as $F(n) + p^r$ will be divisible by $p^r$ and clearly exceeds $p^r$, but is not divisible by $p^{r+1}$. (And $F(n) + k$, being divisible by $k$, is clearly not a prime power when $k$ is not itself a prime power).

One could crudely take $F(n)$ to be $(n!)^{r+1}$ where $2^r$ is the largest power of $2$ not exceeding $n$.

Many thanks, your are generous with this nice answer. I am going to take notes also of this in my notebook. — , Apr 27 '17 at 09:00

Gaps between consecutive prime powers

References:

2 Answers2