Sequence ${2n}\cdot p_{2n}+1$ for $n=1,2,...20$ is
$7,29,79,3^2 \cdot 17,3 \cdot 97,5\cdot 89,3^2 \cdot 67, 3 \cdot 283, 7 \cdot157,7^2\cdot29,37\cdot47,2137,37\cdot71,3^4 \cdot 37,33917 \cdot 599,29 \cdot 163,5437,3 \cdot 5 \cdot7 \cdot59,3^2 \cdot769$
If we denote with $\operatorname {diff}(n)$ the number of different prime factors of $n$ then we have for the first $20$ members of a sequence ${2n}\cdot p_{2n}+1$
$1,1,1,2,2,2,2,2,2,2,2,1,2,2,1,2,2,1,4,2$
I am interested in what happens for very big $n$´s, more precisely, is it impossible that we have $$\lim_{n \to +\infty} \dfrac {|\{\operatorname {diff}(2k \cdot p_{2k}+1)=1 \text{or}\operatorname {diff}(2k \cdot p_{2k}+1)=2:k=1,...,n \}|}{n}=1?$$
And, is it possible that we have
$$\lim_{n \to +\infty} \dfrac {|\{\operatorname {diff}(2k \cdot p_{2k}+1)=1 \text{or}\operatorname {diff}(2k \cdot p_{2k}+1)=2:k=1,...,n \}|}{n}=0?$$
The expression in numerator is just the cardinality of all the numbers in the set $\{2\cdot p_2+1,...,2n \cdot p_{2n}+1\}$ that have either $1$ or $2$ different prime factors, so this limit measures the density.
If someone of you pushes these calculations much further, let me know.
Peter calculated that $2646448$ times the expression has at most two prime factors upto $n = 10^7$, which is, I think, a very high percent because much of these numbers should be divisible with $3$.