Hi, how can I test my probable prime number?

[R.D. Silverman]

Originally Posted by retina

Yes, perhaps. Often people post to several boards at the same time. They get belittled on one board, ignored on a second board, and nanny coddled on a third. It doesn't take a genius to guess which board they direct their attention to.

Yep.

But I saw none of that here. The OP received honest, reasonable replies.

[jonno]

I did try this program. Got the basic idea then the computer died. Good stress test. Lol.

OP said he had a specific number. Second person said put it in prime95. How do I put a specific number in? I'm looking at playing with 6^n +/- 1

Is there a way to put a certain number of that form or can I manipulate the programs search to leave the 2^n and just use 6^n

Thank you

[LaurV]

Don't need to bother with 6^n-1, that is always divisible by 5. (why?)

[Batalov]

Don't need to bother with 6^n+1, either. It can only be prime when n=2^m (for bonus points, -- why?)

And even then, after the first three primes (7, 37 and 1297), probabilistically, -- never. See http://www.prothsearch.net/GFN06.html



P.S. But you can search for factors of these. You will be in an awesome company, too! Look: H.Riesel, W.Keller, H.Dubner, and many other interesting people. One tool for that is mmff-gfn (on a GPU) and there are other programs.

[Brian-E]

And in general, unless I'm mistaken, prime95 is specifically designed, written, and highly optimised, for testing the primality of numbers of the form 2^n-1 and cannot be used for other numbers.

[axn]

Originally Posted by Brian-E

And in general, unless I'm mistaken, prime95 is specifically designed, written, and highly optimised, for testing the primality of numbers of the form 2^n-1 and cannot be used for other numbers.

Correct, as far as primality tests go. But it can do a PRP-test on a much larger set of numbers, (k*b^n+/-c)/f

[Brian-E]

Originally Posted by axn

Correct, as far as primality tests go. But it can do a PRP-test on a much larger set of numbers, (k*b^n+/-c)/f

Ah, thanks. I never even knew about PRP testing with Prime95. I do now.

[jonno]

Originally Posted by Batalov

Don't need to bother with 6^n+1, either. It can only be prime when n=2^m (for bonus points, -- why?)

And even then, after the first three primes (7, 37 and 1297), probabilistically, -- never. See http://www.prothsearch.net/GFN06.html



P.S. But you can search for factors of these. You will be in an awesome company, too! Look: H.Riesel, W.Keller, H.Dubner, and many other interesting people. One tool for that is mmff-gfn (on a GPU) and there are other programs.

Thank you. I'll look that up.
I know 6^n-1 is always a product of 5 but wanted to play with numbers like
6^n - 6^n-1 - 6^n-2..... -6^0
I've got a working sieve for elimination of all composite numbers and wanted to run tests against it.

[Batalov]

Originally Posted by jonno

Thank you. I'll look that up.
I know 6^n-1 is always a product of 5 but wanted to play with numbers like
6^n - 6^n-1 - 6^n-2..... -6^0
I've got a working sieve for elimination of all composite numbers and wanted to run tests against it.

When you are using forms like these, all you will get will be probable primes.
You don't need to reinvent the wheel for a sieve for this form either, because it is easily simplified:
6^n - 6^n-1 - 6^n-2..... -6^0 = 6^n - (6^n-1)/5 = (4*6^n+1)/5
for which srsieve package will be vastly faster than any sieve you can write.
You need to sieve with "-pmin 7" (and then remove n=1(mod 5) because these are divisible by 5^k with k>1; except the only prime divisible by 5 at n=1, which is 5); and additionally remove all n=0 (mod 4) because of the Aurifeuillian factorization:
(4*6^4m+1)/5 = (2 * 6^2m + 2*6^m + 1)/5 * (2 * 6^2m - 2 * 6^m + 1).

And then you will find some PRPs with pfgw. Here is how they will start:

(4*6^1+1)/5 is prime
(4*6^2+1)/5 is prime
(4*6^3+1)/5 is prime
(4*6^5+1)/5 is prime
Switching to Exponentiating using GMP
(4*6^15+1)/5 is 3-PRP! (0.0000s+0.0001s)
(4*6^25+1)/5 is 3-PRP! (0.0000s+0.0014s)
(4*6^29+1)/5 is 3-PRP! (0.0000s+0.0011s)
(4*6^73+1)/5 is 3-PRP! (0.0000s+0.0000s)
(4*6^90+1)/5 is 3-PRP! (0.0000s+0.0000s)
(4*6^139+1)/5 is 3-PRP! (0.0001s+0.0000s)
(4*6^194+1)/5 is 3-PRP! (0.0001s+0.0000s)
Switching to Exponentiating using Woltman FFT's
(4*6^242+1)/5 is 3-PRP! (0.0016s+0.0000s)
(4*6^939+1)/5 is 3-PRP! (0.0161s+0.0000s)
(4*6^3518+1)/5 is 3-PRP! (0.2101s+0.0000s)
(4*6^3963+1)/5 is 3-PRP! (0.2434s+0.0001s)
(4*6^4694+1)/5 is 3-PRP! (0.3930s+0.0000s)
(4*6^5570+1)/5 is 3-PRP! (0.5256s+0.0001s)
(4*6^5615+1)/5 is 3-PRP! (0.5309s+0.0000s)
(4*6^6702+1)/5 is 3-PRP! (0.8243s+0.0000s)
(4*6^13962+1)/5 is 3-PRP! (3.8427s+0.0001s)
(4*6^14269+1)/5 is 3-PRP! (3.9239s+0.0001s)
(4*6^16339+1)/5 is 3-PRP! (4.6013s+0.0001s)
(4*6^16882+1)/5 is 3-PRP! (5.0966s+0.0001s)
...

Nice series, pretty dense (at least at its start).

All of these you can prove with Primo, but the larger ones will become unfeasible to prove, so the best place they will go to will be the PRP Top. The new PRP Top cutoff to enter is 20,000 decimal digits, so you'd need n>25702...

(4*6^22582+1)/5 is 3-PRP! (7.0355s+0.0001s)
(4*6^31415+1)/5 is 3-PRP! (13.1867s+0.0002s)
(4*6^105554+1)/5 is 3-PRP! (183.1141s+0.0005s) (82137 digits)
(4*6^120749+1)/5 is 3-PRP! (218.7588s+0.0006s) (93961 digits)
...

These are "near-repunits 444...445 in base 6". I've created an OEIS sequence A248613 for it. Feel free to find additional terms.

[jasong]

Originally Posted by retina

Yes, perhaps. Often people post to several boards at the same time. They get belittled on one board, ignored on a second board, and nanny coddled on a third. It doesn't take a genius to guess which board they direct their attention to.

Where are these nanny-coddling boards? I've never encountered one.

It's the internet, the masks come off, and people, including myself, are revealed as their true selves.

[retina]

Originally Posted by jasong

Where are these nanny-coddling boards? I've never encountered one.

Maybe some people act in such a way that they never get nanny coddled?

Originally Posted by jasong

It's the internet, the masks come off, and people, including myself, are revealed as their true selves.

So the protection of anonymity is at work? Except of course people are not really anonymous. With all the analysis and tracking going on it is almost impossible to remain anonymous without some very disciplined and restrictive behaviours.