Finding huge AP3 of type (3, p, q)

[JeppeSN]

When I look at Caldwell's Top 20 for Arithmetic Progressions of Primes, I see it requires only 7035 (decimal) digits to reach that list.

I wonder if it would be "allowed" (by Caldwell, does anyone know?) to use arithmetic progressions starting with the smallest odd prime 3, followed by two more odd primes, that is:
\[(3,\quad p,\quad q)\]
One would conjecture there are plenty of those (related entries in Sloane's OEIS are A063908 and A092109).

Of course, we can express q in terms of p:
\[(3,\quad p,\quad 2p - 3)\]
so it is similar to Sophie Germain primes (its Top 20 requires 31112 digits), or we can take t = p-1 and write:
\[(3,\quad t + 1,\quad 2t - 1)\]
(The common difference in the progression is t-2.)

This seems attractive because we can take t to be some smooth number and use fast N-1 and N+1 tests to check that t+1 and 2t-1 are prime.

Would it not be relatively easy to find such AP3s (arithmetical progressions of length three), where t is a number with more than 7035 digits? I think you could use a kind of double sieve first.

/JeppeSN

[science_man_88]

from primes greater than 3 being 1 of 5 mod 6 you get t-2 must be 2 or 4 mod 6 so t must be 4 or 0 mod 6 respectively.

[paulunderwood]

More relevant to sizes is: https://primes.utm.edu/top20/sizes.php

[JeppeSN]

Originally Posted by paulunderwood

More relevant to sizes is: https://primes.utm.edu/top20/sizes.php

OK, so the required size is much larger than what I said. But would it not still be easier to use the form \((3,\quad t+1,\quad 2t-1)\) where the first prime, 3, is "free"? Or is it actually just as easy to find triples where all three terms are "huge"?

Say, if we wanted to find an AP3 whose last term was a megaprime, would it be easiest to use the form \((3,\quad t+1,\quad 2t-1)\)?

/JeppeSN

[paulunderwood]

Originally Posted by JeppeSN

OK, so the required size is much larger than what I said. But would it not still be easier to use the form \((3,\quad t+1,\quad 2t-1)\) where the first prime, 3, is "free"? Or is it actually just as easy to find triples where all three terms are "huge"?

Say, if we wanted to find an AP3 whose last term was a megaprime, would it be easiest to use the form \((3,\quad t+1,\quad 2t-1)\)?

/JeppeSN

I am not sure if searching for (3,t+1,2t-1) or 3 megaprimes is easiest. The current record took the might of PrimeGrid, running their SG/twin search, and the ingenuity of David Broadhurst to scan the top5000 for the second term. A mega AP3 seems to be some decades away.

[axn]

Originally Posted by paulunderwood

I am not sure if searching for (3,t+1,2t-1) or 3 megaprimes is easiest.

A search for (3,t+1,2t-1) is the same as the search for SG or twin prime of size t.

Empirically, primegrid's effort on n=1290000 yielded two such pairs, one SG & one Twin. Using the same set of reported primes from that effort, at least 12 APs were found. Therefore, we can conclude that finding APs the straightforward way is easier than searching for (3,t+1,2t-1)

[JeppeSN]

Originally Posted by axn

A search for (3,t+1,2t-1) is the same as the search for SG or twin prime of size t.

Empirically, primegrid's effort on n=1290000 yielded two such pairs, one SG & one Twin. Using the same set of reported primes from that effort, at least 12 APs were found. Therefore, we can conclude that finding APs the straightforward way is easier than searching for (3,t+1,2t-1)

That makes sense. Thanks to you and paulunderwood. /JeppeSN