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Let $P=(p_j)_{j=1}^\infty$ be an increasing sequence of real numbers with $1<p_1$ and $\lim_{j\to\infty}p_j=\infty$. As mentioned in [1], Beurling proved that if the multiplicative group $N_P$ generated by $(p_j)_{j=1}^\infty$ satisfies that

$$\# \left(N_P\cap[0,x]\right)=Ax+O(x\log(x)^{-\gamma})\qquad(1)$$

for some constants $A>0$ and $\gamma>3/2$, then the conclusion of the prime number theorem is valid for the system $P$.

My question is as follows: If a system $P$ satisfies condition (1) above, what happens with the system $P'=(\lfloor p_j\rfloor)_j$ of integer values of the primes $p_j$ (I guess we should assume $p_1\geq2$)? Do they still satisfy condition (1) in general? And if not, are there known conditions under which condition (1) would be preserved by taking integer values?

I am not an expert in the topic of Beurling primes, so sorry if the question is easy due to some known result.

[1]: Harold G. Diamond, The Prime Number Theorem for Beurling's Generalized numbers.

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  • $\begingroup$ Maybe I'm missing something, but it seems like the counting functions up to $x$ of $N_P$ and $N_{P'}$ should differ by $O(1)$, and so condition (1) and Beurling's result should also apply to $P'$. $\endgroup$
    – Anurag Sahay
    Commented Dec 22, 2023 at 11:15
  • $\begingroup$ I do not see why the counting functions would differ by $O(1)$. I guess you are using something more than the fact that $|p_i-\lfloor p_i\rfloor|<1$ for all $i$? Because for example with the usual primes, it seems if we just change $p=2$ by $1.9$, then that would change the counting function up to $x$ by at least something linear in $x$ $\endgroup$
    – Anon12345
    Commented Dec 22, 2023 at 11:32
  • $\begingroup$ Ah, you're quite right; I was missing something -- I confused the counting function of the Beurling primes with that of the Beurling integers. $\endgroup$
    – Anurag Sahay
    Commented Dec 22, 2023 at 12:26

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