What do we mean by *hyper* prime numbers? Well, roughly speaking they are natural numbers which are *prime* with respect to [hyperoperators](https://en.wikipedia.org/wiki/Hyperoperation) in arithmetic such as exponentiation, [tetration](https://en.wikipedia.org/wiki/Tetration), [pentation](https://en.wikipedia.org/wiki/Pentation), et cetera.


The common soul of these generalized prime numbers is their *irreducibility* to the smaller numbers with respect to the given hyperoperator; the same property that distinguishes usual primes from other natural numbers. In fact, such numbers don't arise as a non-trivial *combination* of smaller building blocks. The underlying *combination method* could be multiplication, exponentiation or any other hyperoperation.



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> **Definition 1.** We call addition, multiplication, exponentiation, tetration,... , the $1$-hyperoperator, $2$-hyperoperator, $3$-hyperoperator, $4$-hyperoperator, ... respectively and use the $\*\_n$ notation to denote the $n$-hyperoperator for $n\geq 1$.
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> **Definition 2.** For any $n\geq 1$, the natural number $p>1$ is called a $n$-hyperprime number if there are no $i,j<p$ such that $p=i\*\_{n}j$.
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**Remark 1.** Note that according to the definition, there are no $1$-hyperprime numbers, simply because every natural number greater than $1$ arises as the addition of two smaller natural numbers. $2$-hyperprime numbers are exactly the usual prime numbers and the set of $3$-hyperprime numbers consists of usual primes and numbers like $12$ which don't arise as the exponentiation of two smaller numbers. In general, if $p>1$ is a $m$-hyperprime number for some $m\geq 1$ then it is $n$-hyperprime for any $n>m$.


**Remark 2.** It is intuitively clear that *almost all* natural numbers are hyperprime in some sense, simply because the hyperoperator $\*\_n$ gets violently powerful when $n$ grows. Thus for almost any natural number $p>1$ one may find sufficiently large $n$ such that even the smallest combinations of natural numbers like $2\*\_{n}3$ and $3\*\_{n}2$ exceed $p$. So based on the increasing nature of hyperoperators, $p$ could not arise as the $\*\_n$ combination of any smaller numbers and is a $n$-hyperprime number. The only (magical?) exception is the *eternally composite* number $4$ which is NOT $n$-hyperprime for any $n\geq 1$ because $\forall n\geq 1~~~4=2\*\_{n}2$.


Now the first question is: "*How many hyperprimes of any type are there?*" We know that there are no $1$-hyperprimes. Also, there are *very few* $2$-hyperprime numbers in the sense that the [natural density](https://en.wikipedia.org/wiki/Natural_density) of the set of prime numbers in $\mathbb{N}$ is $0$. Roughly speaking, the usual prime numbers are so rare among natural numbers that one may assume that there is almost no such number!


But as stated, there is a promising increasing trend among hyperprime numbers of higher order. When $n$ gets larger and larger, the $n$-hyperoperator skips more and more numbers in its range (while receiving non-trivial input) leaving them as $n$-hyperprimes. Consequently, the corresponding set of $n$-hyperprime numbers gets rapidly larger by varying $n$. So one may cover almost the entire set of natural numbers this way eventually. Thus it is natural to ask:



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> **Question 1.** Where is the first place that almost all natural numbers are prime with respect to a hyperoperator?
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> Precisely, what is the least natural number $n$ such that the set of all $n$-hyperprime numbers is of natural density $1$ in $\mathbb{N}$?
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> What about the other values in between? Is there a natural number $n$ such that the natural density of the set of all $n$-hyperprime numbers lies in the interval $(0,1)$? (And so we have a *balanced* number of prime numbers, neither too few nor too many!)
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In order to answer the above question one may need to figure out the distribution of hyperprime numbers in $\mathbb{N}$ which in the case of usual prime numbers is given by [Prime Number Theorem](https://en.wikipedia.org/wiki/Prime_number_theorem) that provides the $\frac{k}{\ln(k)}$ estimation for the number of $2$-hyperprime numbers in the interval $[1, k]$.



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> **Question 2.** What is the analogy of Prime Number Theorem for $n$-hyperprimes in general? In other words, what is the growth rate of the [$n$-hyperprime counting function](https://en.wikipedia.org/wiki/Prime-counting_function) $\pi\_{n}(x)$ which assigns the number of $n$-hyperprime numbers $\leq x$ to the real number $x$? For $n=2$ it is known to be $\frac{x}{\ln (x)}$.
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Finally, one may ask about the analogy of [Fundamental Theorem of Arithmetic](https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic) in the case of hyperprimes. Although, we are not dealing with a commutative hyperoperator for $n\geq 3$ but it is still meaningful and interesting to wonder whether every natural number has a unique representation in terms of $n$-hyper combination of $n$-hyperprime numbers or not? For instance, $8^{9^{10}}$ is a natural number. While $8=2^3$ and $9=3^2$ aren't $3$-hyperprime but this number has a representation $2^{3^{21}}$ via $3$-hyperprime numbers $2, 3, 21$. Does such a representation exist for every other number? Is it unique?



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> **Question 3.** Does a variant of Fundamental Theorem of Arithmetic hold for any $n$-hyperoperator using the mentioned notion of $n$-hyperprimeness?
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> Precisely, is it true that for every natural numbers $n>1$ and $k>1$, there is $s\geq 1$ and a unique sequence of $n$-hyperprime numbers $p\_1, \cdots, p\_{s}$ such that $k=p\_s \*\_{n}(\cdots (p\_3 \*\_{n}(p\_2 \*\_{n} p\_1)))$?
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What is the natural density of hyper prime numbers?
The answer to Question 1 is that the $3$-hyperprimes have density $1$.


For any $k\ge 2$ there are only $\lfloor n^{1/k}\rfloor$ numbers are of the form $a^k$ between $1$ and $n$. When $k>\log\_2(n)$ (i.e. $2^k>n$) only $1$ is of that form. Hence there are at most $n^{1/2}+\cdots+n^{1/\log\_2(n)}=O(n^{1/2}\log(n))$ $3$-hypercomposites from $1$ to $n$.


