Extending the primes

Question

I had an idea and I'd like to find out whether it has a name or has been studied before.

Imagine the natural numbers and the operations of addition and multiplication, but with the following restriction: multiplication can only be carried out $d$ times, and only with prime numbers. Any number that usually has more than $d$ prime factors now becomes 'prime' in the new system. So, if $d=2$ say, then $8$ becomes a new 'prime' because the usual prime decomposition of $8= 2\times2\times2$ takes 'too many' multiplications; similarly, if $d=3$ then $16=2\times2\times2\times2$ becomes 'prime'.

Has this kind of thing ever been studied?

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Edit: I've been asked to clarify this, so here's two other ways of describing the same object.

Geometrically. In a $d$-dimensional Euclidean space, let $\mathscr{D}$ be the smallest set of natural numbers such that you can represent every natural number as the volume of a $d$-dimensional box with edge lengths that are members of $\mathscr{D}$. $\mathscr{D}$ always contains the primes, and if $d<\infty$ it also contains other numbers. Perhaps it'd be interesting to understand those other numbers.

Algorithmically. First, put all the prime numbers in a set $\mathscr{D}$. Then, put every number $n$ with $\Omega(n)>d$ in $\mathscr{D}$ (where $\Omega(n)$ is the number of indistinct prime divisors of $n$). Finally, go through each of the numbers $n$ for which $\Omega(n)>d$ (i.e. the ones which were just added to $\mathscr{D}$) in increasing order, and for each such number $n$ that has not yet been removed from $\mathscr{D}$, remove from $\mathscr{D}$ every number $m$ that is product of $n$ and $d-1$ (possibly identical) numbers $k_1,k_2,...,k_{d-1}$ such that each $k_i\in\mathscr{D}$ and each $k_i \le n$. We can call the remaining numbers in $\mathscr{D}$ '$d$-prime'.

Answer 1

First, as Jyrki suggested, let's find some other term than prime. Maybe $d$-Mariusian is appropriate. We define the set $$\mathscr D_d:=\{n\in\mathbb N\mid n \text{ is $d$-Mariusian}\}$$ Now, forbidding multiplication is somehow difficult, so let's find a more formal definition for Mariusian numbers:

Definition: Let $n\in\mathbb N$. $n$ is called "$d$-Mariusian", if $n\neq 1$ and either $n$ is a prime or $$n=\prod_{i=1}^km_i \;\text{ for }\; m_1,\dots,m_k\in\mathscr D_d \text{ implies } k>d \text{ or } k=1$$

I hope, this agrees with your concept, otherwise skip the following and let me know.

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Note, that $\mathscr D_1=\mathbb N\setminus\{1\}$ and $\mathscr D_{\infty}=\mathbb P$. Now, let's study $2$-Mariusianity.

If $n$ is prime, we have $n\in \mathscr D_2$. If $n=pq$ for primes $p,q$, then $pq\not\in\mathscr D_2$.

Let $n=pqr$ for primes $p,q,r$. Assume $n$ is not $2$-Mariusian, then it is the product of two $2$-Mariusian integers. There are only two (kinds of) ways to write $pqr$ as product of two integers:

• Seperate one prime, i.e: $n=p(qr)=q(pr)=r(pq)$, but then you have a non $2$-Mariusian factor.
• $n=(pqr)\cdot 1$, but $1\not\in\mathscr D_2$ (You can see here, that it is important to the well-definedness, that we defined $1$ to be not-Mariusian)

So $n=pqr$ is $2$-Mariusian. You can go on and see, that $$n=\prod_{i=1}^kp_i^{e_i}\in\mathscr D_2 \Leftrightarrow 2\nmid \sum_{i=1}^ke_i$$ Now you can try to give a characterisation for $d$-Mariusianity for arbtirary $d$'s.

Answer 2

Not an answer, but an attempt to formalize the OP's definition:

Let $M_d(S)$ be the set of all products of up to $d$ elements of $S$.

Define the sets $P_i$ and $C_i$ recursively as follows:

• $P_0$ are the prime numbers;
• $C_i = M_d(P_i) \setminus P_i$;
• $P_{i+1} = M_{d+1}(P_i) \setminus C_i$.

$P_i \subset P_{i+1}$, so we can say that a number is prime* if it is an element of $P_\infty$, i.e. an element of $P_i$ for some $i$.

Taking $d=2$ and looking only at powers of 2, we have that

• $P_0 = \{2\}$
• $C_0 = \{4\}$
• $P_1 = \{2,8\}$
• $C_1 = \{4,16,64\}$
• $P_2 = \{2,8,32,128,512\}$

and so forth. Marius, is this what you have in mind?