Let $A(N)$ denote the number of positive integers $n\le N$ composed of prime numbers $p\equiv 1\pmod 4$ only. Is there an asymptotic formula for $A(N)$ (as $N$ tends to infinity)?
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Yes, $A(X) = cX/\sqrt{\log(X)} + O(X/\log^{3/2}(X))$ for a positive real number $c$ which I think is 1 (edit: This remark on the constant was just a vague recollection which is wrongly remembered as it turns out. See the comments below.). This result holds in much much more generality, e.g. for primes in congruence classes or even for positive-density subsets of primes defined by modular forms. See Theorem 2.8 of Serre's "Divisibilit\'e de certaines fonctions arithmetiques" from 1976 in L'Enseignement Mathematique. You'd replace $\sqrt{\log(X)}$ with $\log^{1-\delta}(X)$ for a more general set of primes of density $\delta$, but the primes congruent to 1 mod 4 and 3 mod 4 respectively make up density $1/2$ subsets of primes. You can also get secondary, tertiary, or as many error terms as you wish.
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1$\begingroup$ $c = \tfrac{1}{\sqrt{2}} \prod_{p \equiv 3 \bmod 4} 1/\sqrt{1-p^{-2}} \approx 0.764$. See mathworld.wolfram.com/Landau-RamanujanConstant.html math.stackexchange.com/questions/264069/… $\endgroup$ Commented Jul 14, 2016 at 0:10
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$\begingroup$ That is the $c$ value for the sums of two squares counting function, which is larger than $A(N)$. $\endgroup$ Commented Jul 14, 2016 at 0:33
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2$\begingroup$ @JeremyRouse I asked on one of the sites for counting primitively represented $x^2 + y^2,$ it was the same as unrestricted but with a smaller constant, surprised me. I guess this question would simply halve that number because we have even and odd about the same. Let me try to find that. math.stackexchange.com/questions/1282550/… $\endgroup$ Commented Jul 14, 2016 at 1:15
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$\begingroup$ You can also look at the Selberg–Delange method, see for example Tenenbaum's "Introduction to Analytic and Probabilistic Number Theory", section II.5 I think. $\endgroup$ Commented Jul 31, 2016 at 17:36
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$\begingroup$ For completeness, the constant $c$ here (based on the answer to the question by @WillJagy above) should be $1/(4K) \approx 0.327129366941$ where $K \approx 0.764223653589$ is the Landau–Ramanujan constant. (Sort of confirmed by experiment -- it converges very slowly -- e.g. $A(10000000) = 814182 \approx 0.3269 \times 10000000/\sqrt{\log 10000000}$.) $\endgroup$ Commented Feb 24, 2020 at 22:31