Do 50% of primes give primes of the form $P=15p+4$?

Question

Do 50% of primes give primes of the form $P=15p+4$?

I checked this for primes $p$ form 7 up to 233 and found 21 primes out of 38 produce primes of the form $15p+4$:

$(p, P):(7,109), (13,199),(23,349),(29,439),(41, 619), (47, 709), (61, 919), (67, 1009), (71, 1069), (71, 1069), (81, 1249), (97, 1459), (103, 1549), (107, 1609), (113, 1699), (139, 2089), (149, 2239), (151, 2269), (179, 2689), (181, 2719), (211, 3169), (233, 3499)$

Since all primes end with digits 1, 3, 7 and 9, the last digit of all these generated primes is $9$.We can write $P=3(5p+1)+1$. Due to Dirichlet theorem the number of primes of the form $3k+1$ can be infinite, particularly this set of primes , which can also be written as $P=5(3p)+4$,are a subset of primes of the form $5k+4$ which is proved to be infinitely many.The point is that K has especial form $k=5p+1$ in $P=3k+1$ and $k=3p$ in $P=5k+4$.

My question is that : is this trend i.e. 50% of primes generate primes of the form $15p+4$ continues? It need a powerful computer. Could someone check this please?

Answer 1

Edit: the original answer below has been based on a grievous misreading of the question, and you are well advised to ignore it. The correct density is 0, just like with the twin primes; indeed, Brun's sieve methods work for this as well.

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Dirichlet's theorem for arithmetic progressions is well known, and so is the stronger PNT for arithmetic progressions. The allowable residues mod $15$ for primes are just the positive integers at most and coprime to $15$, of which there are 8: $1, 2, 4, 7, 8, 11, 13, 14$. Primes are equally represented among these eight, so the actual density is not $50\%$, but $1/8 = 12.5 \%$.