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*](https://en.wikipedia.org/wiki/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 


$$9,21,25,33,45,49,57,65,69,77,85,93,\ldots,$$


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 $$\#\{\text{semiprimes }4n+3\leq x\}.\tag{2}$$



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> **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.**
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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.


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



Races that involve odd semiprimes: a first statement or conjecture
This interesting question has indeed been considered. See the paper by Ford and Sneed: here's a [link to the Math Review](https://mathscinet.ams.org/mathscinet-getitem?mr=2778652) (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](https://mathscinet.ams.org/mathscinet-getitem?mr=108470).


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.


