While I am studying the famous article [1], in English this is Andrew Granville and Greg Martin, Prime Number Races, The American Mathematical Monthly, vol. 113, (2006), I wondered what about a race of odd semiprimes.

A semiprime is a positive integer that is the product of two prime numbers (see the Wikipedia Semiprime or the article from the encyclopedia MathWorld).

As in the cited article we consider that our race the odd semiprimes separated in two teams, those semiprimes of the form $s=4n+1$ for some positive integer $m$, this is $s\equiv 1\text{ mod }4$ that are our Team 1


and those semiprimes of the form $s=4n+3$ for some positive integer $m$, that is $s\equiv 3\text{ mod }4$ are our Team 2

$$15 ,35, 39, 51, 55 ,63, 75, 87, 91 ,95, 99\ldots.$$ We denote the corresponding counting functions as $$\#\{\text{semiprimes }4n+1\leq x\}\tag{1}$$ and $$\#\{\text{semiprimes }4n+3\leq x\}.\tag{2}$$

Question. Is it possible to know which team will win the race? I am asking if it is possible to study in a similar way that shows the first section of the mentioned article our race of semiprimes, maybe it is possible to get a similar statement than Hardy's theorem about the difference of previous counting functions $(1)$ and $(2)$, or maybe one can to state a conjecture in the spirit of the conjecture due to Knapowski and Turán but now for the race of semiprimes of the form $4n+1$ versus $4n+3$. Many thanks.

I don't know if this problem is in the literature, feel free to refer the literature and I try to search and read it from the literature to know how does the race end.


[1] Andrew Granville and Greg Martin, Carreras de números primos, La Gaceta de la RSME, Volumen 8, Número 1 (enero-abril, 2005).

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    $\begingroup$ I believe that the same article, and in English, by the authors is on arXiv. I recommend the articles from the journals. $\endgroup$
    – user142929
    Aug 28 '19 at 20:47
  • $\begingroup$ Is Spanish your mother tongue? If I'm not mistaken, you mentioned a source in this language in another question of yours. $\endgroup$ Aug 28 '19 at 20:58
  • $\begingroup$ Yes, @SylvainJULIEN If I'm not mistaken this article was written for The American Mathematical Monthly and also for La Gaceta de la Real Sociedad Matemática Española $\endgroup$
    – user142929
    Aug 28 '19 at 21:09
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    $\begingroup$ Does "real" in Spanish mean "royal"? I can read Spanish with less ease than English. $\endgroup$ Aug 28 '19 at 21:17
  • $\begingroup$ Yes @SylvainJULIEN you can find it in the Spanish Wikipedia Real Sociedad Matemática Española $\endgroup$
    – user142929
    Aug 28 '19 at 21:35

This interesting question has indeed been considered. See the paper by Ford and Sneed: here's a link to the Math Review (I recommend also clicking on the "From References" link there and following up on those four papers). Questions of this sort go back at least as far as Shanks in 1959.

The short answer is that when counting prime factors with multiplicity, as you have done, residue classes that are quadratic residues are favored over quadratic nonresidues (for instance, the $4n+1$ team is favored over the $4n+3$ team), and this is true for any fixed number $k$ of prime factors, not just the semiprime case $k=2$.

When counting primes without multiplicity, the bias switches back and forth depending on the parity of $k$: when $k$ is odd (as in the classical prime number race) quadratic nonresidues have the advantage, while when $k$ is even quadratic residues have the advantage.

  • $\begingroup$ I didn't know it, thank you very much. I am going to see it. $\endgroup$
    – user142929
    Aug 29 '19 at 16:47
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    $\begingroup$ Here is an an imprecise heuristic argument: first, it is not the same to count semiprimes with both factors below $\sqrt(x)$ but let’s consider that. Suppose there are $2k$ odd primes which could be factors. One might expect about $k$ of each form. Then that gives $k^2$ products of the form $4n+3$ and either $k^2+k$ or $k^2-k$ of the form $4n+1$ depending on if you count squares or not. $\endgroup$ Sep 1 '19 at 3:41

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