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If $S$ is a subset of the set of the positive integers $\mathbb N$, we may consider the set $S^*$ of all products of elements of $S$, allowing for repeated factors —this is a multiset, really, in general. For example, if $S$ is the set of all prime numbers, then $S^*$ is simply $\mathbb N$, with all elements of multiplicity one —this is the fundamental theorem of arithmetic— and in fact this set $S$ is the only one with that property, so that this characterizes the set of prime numbers.

If $S$ is not the set of prime numbers, $S^*$ will have holes and there will probably be regions of $\mathbb N$ with extra density.

Are there standard useful ways to quantify how far from being the set of primes a set $S$ is in this sense, that is, how irregularly distributed is the set $S^*$ in $\mathbb N$?

There are sets $S$ with extremely irregular $S^*$. An example is the set of powers of a number. Once one has a qualitative measure of irregularity, one can make sense of the question:

If we pick a random set $S$ (for some sense of «random»), what is the expected irregularity?

From $S$ one can construct an "Euler product" $\zeta_S(s)=\prod\limits_{p\in S}\frac{1}{1-\frac{1}{p^s}}$ which is the Riemann zeta function if $S$ is the set of prime numbers. The function $\zeta_S$ is the Dirichlet enumerating function for the multiset $S^*$.

Are there analytic properties of the function $\zeta_S$ that characterize the set of primes among the other candidates for $S$?

Of course, $\zeta_S=\zeta$ iff $S$ is the set of primes, but I imagine there are much coarser conditions that one can put on $S$ that already imply the same conclusion.

The sort of thing one could imagine involves the distributions of values, of zeros, growth conditions, and so on. For example (tweaked after the comments of Steve and Greg below):

Does the Riemann Hypothesis hold for $\zeta_S$ iff $S$ is the set of primes up to a finite set?

GH from MO
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Mariano Suárez-Álvarez
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  • What do you mean by the set of elements in $S^\ast$ all have multiplicity one if $S$ is the set of primes? Not all numbers are square-free... – Stanley Yao Xiao Oct 26 '17 at 21:30
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    Every positive integer is a product of primes in exactly one way. That claim is exactly the same as «$\zeta_S=\zeta$ iff $S$ is the set of prime numbers». – Mariano Suárez-Álvarez Oct 26 '17 at 21:31
  • The Euler product for $\zeta(s)$ really characterizes the primes in the simplest possible way : $\exp(-\sum_p \log(1-p^{-s})) = \sum_n n^{-s}$, unfortunately the $\log,\exp$ operators on Dirichlet series are very complicated. I had this old question of mine. Here the natural way to measure how $S$ differs from $\mathcal{P}$ is to look at a distance function between the logarithm Dirichlet series $\sum_{q \in S, k \ge 1} \frac{q^{-sk}}{k}$, or to measure how $\prod_{q \in S} (1-q^{-s})^{-1}$ fails to be a "periodic" distribution. – reuns Oct 26 '17 at 22:10
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    One trivial-ish observation; if $S$ is the odd primes, then $S^=\mathbb{N}-2\mathbb{N}$ is just the odd naturals; more broadly, if $S$ is the primes minus a finite set, then $S^$ will be the naturals minus a very regular set (with multiplicity 1). Obviously all of these sets are very 'close to' the primes and all of the suggested measures in the Q seem to reflect that closeness well in this case. OTOH, I believe it implies a trivially false answer to the last question as currently posed, as I'd think the RH for primes minus a finite set should be exactly RH. – Steven Stadnicki Oct 26 '17 at 22:40
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    You should search on the term "Beurling primes" or "Beurling numbers", where much of these ideas have already been anticipated, including looking at the properties of the corresponding zeta function. (And yes, Steven Stadnicki is correct about RH holding for $S$ being all but finitely many primes, or, for that matter, the union of all but finitely many primes with any other finite multiset.) – Greg Martin Oct 26 '17 at 22:54
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    @Greg, is the set of primes characterised up to finite sets by RH, then? :-) – Mariano Suárez-Álvarez Oct 26 '17 at 23:15
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    Good question, which I suggest editing into the OP. – Greg Martin Oct 26 '17 at 23:43
  • "we may consider the set $S^∗$ of all products of elements of $S,$ allowing for repeated factors —this is a multiset, really" At first I thought you just meant $S^*$ is the closure of $S$ under multiplication, but calling it a multiset makes me think you mean that if there is more than one way to express a number as a product of members of $S$ then the product has multiplicity $>1.$ Thus if $S= {2,3,4,6},$ then the multiplicity of $12$ is $3,$ since you have $12=2\times2\times3 = 2\times6 = 3\times 4.$ Is that what you meant? $\qquad$ – Michael Hardy Jan 22 '19 at 04:20

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As Greg Martin has pointed out this topic of Beurling numbers has been extensively investigated. Usually the Beurling numbers are studied as semigroups contained in ${\Bbb R}$. In the particular context of subsets of the natural numbers, you may wish to look at the work of Lagarias who considers such sets satisfying the "Delone property" that the gaps between consecutive elements in $S^*$ are bounded above and below. One of his results then states that the set $S$ consists of all but finitely many primes plus finitely many composite numbers. Lagarias's paper will also give you other references. Since you mentioned irregularities and the Riemann hypothesis, you may also look at my answer to Why does the Riemann zeta function have non-trivial zeros?

Lucia
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