Message boards : AP26 - AP27 Search : 23#

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Darryl

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Message 127380 - Posted: 24 Feb 2019 | 10:48:06 UTC

Hi All,

I just bought a new crunching rig with a decent graphics card, and decided to put it to work on the AP27 project. Looking at the first few results I found yesterday and today, and comparing with an ancient AP21 I found here at Primegrid 9 years ago (until I switched my new PC on yesterday this AP21 was the only AP I had ever found) I noticed that one common feature they all share is that the prime sequences are constructed by multiplying a given quantity by 23#. For example:

11988702023996197+4465007*23#*n for n=0..20

This made me curious as to whether there is any significance in the choice of 23# here (since it seems not to have changed in 9 years) - are any other sequences, constructed of say 29# or 17#, being tested as well? For numbers of the form a+b*c#*n for some range of n, it occurs to me we could get similar densities of primes and similar AP's as we are testing now, but using a different value than c=23.

Thanks
Darryl
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dukebg
Volunteer tester

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Message 127383 - Posted: 24 Feb 2019 | 12:02:28 UTC

Including a primorial like that helps with the numbers not being divisible by the smallest primes.
Check out the full paper that discusses the tricks to find APs fastest here (link from one of the sticky topics in this subforum).

JeppeSN

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Message 127385 - Posted: 24 Feb 2019 | 12:18:09 UTC

In an arithmetic progression of form An + B, a prime p which does not divide the common difference A, will divide every p-th term of the progression.

For example, in any progression of form 60n + B, no matter what B is, every seventh term is divisible by the prime 7. That is because 7 does not divide 60. So if you want an AP that is greater than 7 in length (or just greater than 6 if no term of the AP equals 7 itself), you want a coefficient A which is divisible by 7.

If you want an AP of length 23, 24, 25, 26, 27, or 28, then the A in the formula An + B must be divisible by all the primes 2, 3, 5, 7, 11, 13, 17, 19, and 23.

This explains the 23# you see.

When, in the future, someone searches for an AP29 (and one not starting from the term 29, so B≠29), they would need to substitute 23# with 29#.

/JeppeSN

Darryl

Joined: 21 Mar 09
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ID: 37210
Credit: 336,896,165
RAC: 222,625

Message 127388 - Posted: 24 Feb 2019 | 13:09:22 UTC

Thanks for both of these answers. It makes much more sense now!

Cheers,
Darryl
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Message boards : AP26 - AP27 Search : 23#