Timeline for Picking a new set of primes
Current License: CC BY-SA 3.0
13 events
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| Jan 22, 2019 at 4:20 | comment | added | Michael Hardy | "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? $\qquad$ | |
| Oct 27, 2017 at 3:30 | history | edited | GH from MO |
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| Oct 27, 2017 at 2:45 | answer | added | Lucia | timeline score: 5 | |
| Oct 27, 2017 at 2:23 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Oct 26, 2017 at 23:43 | comment | added | Greg Martin | Good question, which I suggest editing into the OP. | |
| Oct 26, 2017 at 23:15 | comment | added | Mariano Suárez-Álvarez | @Greg, is the set of primes characterised up to finite sets by RH, then? :-) | |
| Oct 26, 2017 at 22:54 | comment | added | Greg Martin | 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.) | |
| Oct 26, 2017 at 22:40 | comment | added | Steven Stadnicki | 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. | |
| Oct 26, 2017 at 22:29 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 3.0 |
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| Oct 26, 2017 at 22:10 | comment | added | reuns | 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. | |
| Oct 26, 2017 at 21:31 | comment | added | Mariano Suárez-Álvarez | 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». | |
| Oct 26, 2017 at 21:30 | comment | added | Stanley Yao Xiao | 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... | |
| Oct 26, 2017 at 21:23 | history | asked | Mariano Suárez-Álvarez | CC BY-SA 3.0 |