Timeline for What is the natural density of hyper prime numbers?
Current License: CC BY-SA 4.0
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Feb 19, 2019 at 20:44 | history | edited | kodlu | CC BY-SA 4.0 |
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| Feb 19, 2019 at 20:06 | answer | added | Renny Barrett | timeline score: 1 | |
| May 27, 2018 at 20:00 | vote | accept | Morteza Azad | ||
| Apr 28, 2018 at 19:12 | answer | added | MTyson | timeline score: 4 | |
| Apr 28, 2018 at 16:17 | comment | added | Morteza Azad | @AliCaglayan By the way, the hyperoperators aren't useless or pure generalizations at all. For some applications of hyperoperators you may take a look at the Ackermann function in computability, Graham's number and Shelah's primitive recursive bound for van Der Warden numbers in Ramsey theory or their potential use in set theory and cardinal arithmetic. | |
| Apr 28, 2018 at 16:06 | comment | added | Morteza Azad | @AliCaglayan Well, in principle the commutative nature of an arithmetic operator actually contributes to the increase in the number of its corresponding primes because it reduces the number of composite numbers by making combinations such as $i*j$ and $j*i$ equal. In a non-commutative operator $i, j$ are likely to produce two different composite numbers. Also, as it is mentioned in the question 3, it is not immediately clear if the hyperoperators satisfy a variant of unique factorization or not. I think it is actually the high growth speed of hyperoperators which makes them have many primes. | |
| Apr 28, 2018 at 15:52 | comment | added | Morteza Azad | @MTyson (+1) That is interesting! I think your comment could be posted as an answer. Could you please explain how you obtained this estimation? | |
| Apr 28, 2018 at 13:42 | comment | added | Ali Caglayan | The reason we have so few "2-primes" is because the operation of multiplication is very relaxed. It is commutative and satisfies unique factorization. The higher operations would be very restrictive meaning that almost all numbers are hyperprime. I believe it would be very difficult to say something about all hyperoperators. Take the difference between addition and multiplication, "1-primes" don't really relate to "2-primes". So it is very unlikely any hyperoperators are related. Each hyperoperation would be interesting in its own right, although imo past tetration its pretty boring. | |
| Apr 28, 2018 at 2:42 | comment | added | MTyson | The $3$-hypercomposites already have density $0$. From $1$ to $n$ there are at most (roughly) $n^{1/2}+\dots+n^{1/\log_2(n)}$. | |
| Apr 28, 2018 at 1:03 | history | edited | Morteza Azad |
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| Apr 28, 2018 at 0:45 | history | edited | Morteza Azad | CC BY-SA 3.0 |
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| Apr 28, 2018 at 0:39 | history | edited | Morteza Azad | CC BY-SA 3.0 |
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| Apr 28, 2018 at 0:12 | history | asked | Morteza Azad | CC BY-SA 3.0 |