Density of Sophie Germain $3\bmod 4$ primes

Is there any reason to expect the density of primes $p$ such that $2p+1$ is also a prime where $p=3\bmod4$ holds would be different from case of $p=1\bmod4$?

What if $2p+1$ is replaced by $2p-1$ and what is the conjectured density and what is known so far for both cases $2p+1$ and $2p-1$?

• The first hundred thousand Sophie Germain primes are tabulated at oeis.org/A005384/b005384.txt so you could run a computer over them to see how the counts go. May 27 '16 at 2:19
• If $100000$ would say anything I would do that (in any case this is a good link).
– user76479
May 27 '16 at 2:22
• It wouldn't say anything conclusive, but if you found, say, that in every 10,000, there were twice as many $1\bmod4$ as $3\bmod4$ you'd probably find that very suggestive, at least. May 27 '16 at 2:26
• @GerryMyerson does not seem to be biased.
– user76479
May 27 '16 at 2:48
• For what it's worth, up to $10^8$, there are $664578$ odd primes: $332180$ of the form $4k+1$ and $332398$ of the form $4k+3$; and of these, respectively $27940$ and $28091$ are such that $2p+1$ is also prime, and respectively $27981$ and $28175$ such that $2p-1$ is also prime. Jun 3 '16 at 9:49

As with most questions about finding primes in linear forms, nothing is known. We expect that the number of Sophie Germain primes $\leq x$ equivalent to 3 modulo 4 is $$\sim \frac{2x}{\log^2 x} \prod_{p > 2} (1-2/p) (1-1/p)^{-2},$$ and for the same estimate to hold with 3 replaced by 1. The same should hold true with $2p+1$ replaced by $2p-1$.

The question of finding Sophie Germian primes equivalent to 3 modulo 4 can be formulated as a sieve theory problem. Start with the set $$A : = \{n \leq x : n \equiv 3 \mod 4\}.$$ Fix a parameter $z \leq \sqrt{x}$. We are interested in the $S(A,z)$, the number of $n \in A$ such that $n(2n+1)$ has no prime factors $\leq z$. For instance, when $z = \sqrt{x}$ then $S(A,z)$ in the number of Sophie Germain primes in $(z,x]$ (plus 1 because of the integer 1).

A sieve will tell us that if $z = x^{o(1)}$, then $$S(A,z) \sim \frac{x}{2} \prod_{2 < p \leq z} \left(1- \frac{2}{p}\right).$$ If $z = x^{\delta}$ for some fixed $\delta >0$, then a sieve yields $$S(A,z) \gg \frac{x}{\log^2 x}.$$

Current sieves do not allow one to take $z = \sqrt{x}$ and actually do not predict the correct asymptotic formula. For instance, the number of primes $\leq x$ is $\sim x/ \log x$, while a sieve would predict $$x \prod_{p < \sqrt{x}} (1 - \frac{1}{p}) \sim x 2 e^{- \gamma}/ \log x.$$ This discrepancy is related to the fact that if a prime $p$ divides $n\leq x$ and $p > \sqrt{x}$, then it is impossible that a different prime $q > \sqrt{x}$ also divides $n$.

To formulate a heuristic, we make the assumption that each element of $A$ is independently prime with probability $1 / \log x$, in accordance with the prime number theorem. Then we consider $\mathbb{P}(n , 2n+ 1 \text{ is prime})$. If these events were independent, then the probability would be $\frac{1}{\log^2x}$. But there are some minor local obstructions to independence that have to be accounted for. The probability that $p>2$ divides $n(2n+1)$ is $1-2/p$, while the probability that $p | n$ times the probability that $p | 2n+1$ is $(1-1/p)^2$. Thus we see we should add the correction faction $$\prod_{p>2} (1-2/p) (1-1/p)^{-2}.$$ Note that the above product is convergent. For $p =2$, the correction factor is $(1-1/2)^{-2}$, since we assumed all the elements of $A$ are odd.

Thus we expect the number of Sophie Germain primes equivalent to 3 modulo 4 to be $$\frac{2x}{\log^2 x} \prod_{p > 2} (1-2/p) (1-1/p)^{-2}.$$ This is known as the twin prime constant and is in the same spirit of the singular series arising from the the Hardy-Littlewood prime $k$-tuples conjecture. The only difference is that in this question, we seek primes in linear forms $\{a_i n + b_i\}_{i=1}^k$ where $a_i$ is not required to be 1. State of the art work on finding primes in linear forms can be found here.

• why should the twin prime constant occur in sophie germain prime density?
– user76479
May 27 '16 at 23:16
• $n (n+2) =0$ and $n (2n+1) =0$ have the same number of solutions modulo every prime. May 27 '16 at 23:57
• Is $x \prod_{p < \sqrt{x}} (1 - \frac{1}{p}) \sim x 2 e^{- \gamma}$ correct? I have a different asymptotic $x \prod_{p < \sqrt{x}} (1 - \frac{1}{p}) \sim x 2 e^{- \gamma}/ \log x$.
– user92450
Jun 2 '16 at 23:02
• Yes of course! I made the appropriate change Jun 3 '16 at 1:05