A usual prime number is a number greater than $1$ which is not in the form of multiplication of two numbers greater than $1$. We may consider the following natural generalizations:

$p>1$ is $+$ - prime iff there are no $m,n>1$ such that $p=m+n$

$p>1$ is $\times$ - prime iff there are no $m,n>1$ such that $p=m\times n$

$p>1$ is $\hat{}$ - prime (or exponentially prime) iff there are no $m,n>1$ such that $p=m^n$

Remark 1: $2,3$ are the only $+$ - prime numbers and $\times$ - prime numbers are the ordinary primes. Also each $+$ - prime is $\times$ - prime and each $\times$ - prime is $\hat{}$ - prime. Furthermore there are many numbers like $6$, $10$ that are $\hat{}$ - prime but not $\times$ - prime.

Question 1: How many $\hat{}$ - prime numbers are there in the interval $[2,n]$ for a given natural number $n>1$? In the other words, is there any formula to compute this as well as $\times$ - primes?

Question 2: Is there any representation theorem for natural numbers in terms of exponentially prime numbers as well as factorization theorem for $\times$ - primes?

Remark 2: Regarding question 2, note that there are too many exponentially prime numbers so we may expect that there is a simpler representation theorem for all natural numbers with respect to this type of primes. More precisely consider the following form of question 2:

Question 3: Is there any fixed number $n_0\geq 0$ such that for all $m>1$ there exists an exponentially prime number $p$ with $m=p+n_0$ (i.e. $p$ is less than or equal to $m$ and near to $m$ in the sense of $n_0$)? What about $|m-p|=n_0$?

Remark 3: Note that in the case of $m=p+n_0$ in above question, $n_0=0$ doesn't work because we may assume $m$ to be a non-exponentially prime number (e.g. $m=4$). Also $n_0=1$ doesn't work because if we consider $m=9$ then $p$ should be $8$ which is not an exponentially prime number.

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    $\begingroup$ Long ago I found this at ArXiv: arxiv.org/abs/1003.3509v1 Maybe can be interesting. $\endgroup$ – MphLee Nov 22 '14 at 9:03
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    $\begingroup$ @MphLee (+1) Thanks for introducing this interesting reference. $\endgroup$ – user180918 Nov 22 '14 at 10:11
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    $\begingroup$ @MphLee Do you possibly know anything about my unanswered questions here and here? $\endgroup$ – user180918 Nov 22 '14 at 10:22
  • $\begingroup$ I did comment on one of the question with a trivial obsevation. Sad to say that I can't answer to those question because... I was planning to ask the same questions for long. I'm an amateur mathematician so is hard to go that deep for me. Brandenburg's Answer on the second question intrigued me so much.. I was thinking about that for alot. Nowdays, even if is beyond my reach, i'm playng with the idea of the definition of Hyperoperations object. About the Hyperoperation buisness there is alot on the Tetration forum. continue $\endgroup$ – MphLee Nov 22 '14 at 10:39
  • $\begingroup$ continue But there, on the Tetration Forum, few thread was discussed about Hyperoperations in Number Theory, and even less about the relation between H.os. and Category theory. Here is the only one I know on thetration forum (math.eretrandre.org/tetrationforum/… is really poor imho. $\endgroup$ – MphLee Nov 22 '14 at 10:47

These are the non-powers, so there are $x-O(\sqrt x)$ of them up to $x$.

For question 1 this can be computed by inclusion-exclusion. Take $\lfloor x\rfloor$, then subtract the $\lfloor\sqrt x\rfloor$ squares up to $x$, then subtract the $\lfloor\sqrt[3]x\rfloor$ cubes up to $x$, then skip the fourth powers (they're already counted), then subtract the $\lfloor\sqrt[5]x\rfloor$ squares up to $x$, then add in the $\lfloor\sqrt[6]x\rfloor$ squares up to $x$, which were counted as both squares and cubes, etc. Generally numbers with exponent $e$ are given weight $\mu(e).$

There's no obvious answer for question 2 since the numbers are so common. One representation you could choose is as a product of primes, which are all ^-primes.

For question 3 you can have the first variant since it requires that you have only finitely many ^-primes (any power plus $n_0$ would fail). A slightly stronger version of Catalan's conjecture would give you the second version with $n_0=1$ and large enough numbers.


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