... [SNIP] ... Thank you very much for your help, Paul. If I understand right, BPSW is about as fast as a triple Fermat-sprp test using bases 2, 3, and 5,...
... I believe the prp tests which don't prove primality (or are not known to do so) can only be trusted as far as exhaustive testing has gone. This is far ...
Nice to see that the Italian professor got his act together, above a megabit: http://primes.utm.edu/primes/page.php?id=78101 Congrats to Giovanni! David ...
Sorry to intrude here with a trivial question. In another context a question of alternates to decrementing by one in the product to form a "factorial" type...
Congrats to Wataru Sakai for http://primes.utm.edu/primes/page.php?id=78139 There are now 9 primes above 1 megabit credited to OpenPFGW codes with the form...
Let's say I have a (autogenerated) SCRIPT file: <<< SCRIPT // generated by listorial.pl from poworial.lst DIM base=1 DIMS pr SET base, base*2 SET base, base*3 ...
... Let's then say I modify it to: DIM base, 1 Typical - I'd been looking at that script for about an hour, and then within seconds of posting I see my...
Hi, ... Hmmm, almost sounds like some of the things I've ran into. Bug in program, unable to find it. Walk over to project manager, let him sit next to you,...
Mike Oakes and I worked happily together in private to prove many things in the thread that includes http://groups.yahoo.com/group/primeform/message/7523 and...
... Mini-puzzle: (24841^10 + 49681*2^8)^2 + 49682^10 = (24841^10 - 49681*2^8)^2 + 49682^11 is an 88-digit prime of the form p = a^2 + 49682^n = b^2 + 49682^m...
... wrote: > ... To be honest -- and I say this after addressing only the k==1 case -- the puzzle is trivial. Perhaps it would work better as a 100,000+-digit...
... Ah, perhaps you meant k = 1 mod 80 and found a method for k > 81 ? (There is no solution for k=1 or for k=81.) I guess that my failure to handle k = 161 ...
... Ah yes, it /is/ trivial, with max(m,n) = 2, for example. Well spotted Ralph! In that case we could, for example, set k = 80*x + 1 and simply consider the...
... The mini-puzzle was solved, off list, by Chris Caldwell, who found all and only the 3 primes that I had found. Each has more than 70 digits. Congrats to...
In primeform@yahoogroups.com, "David Broadhurst" ... Exactly. My {m,n} = {1,2}, as you guessed. ... Yes, it is, but I don't like using CRT when there are ...
Thanks for the details Ralph! You neatly avoided CRT with a sextic: if((4*x^3+2*x)^2+(4*x^2+1)==(4*x^3)^2+(4*x^2+1)^2,print("bravo!")) bravo! ... Not without a...
... I believe that I have a time-consuming double-loop strategy that would eventually work for n = 3 and for n = 5. Hint: n = 9 seems to be next candidate. ...
... In fact, most n > 5 seem to be possible, with these percentages (I hope!): [ 6, 36.3] [ 7, 30.7] [ 8, 40.0] [ 9, 47.0] [10, 31.5] [11, 38.0] [12, 43.4] ...
... I was extraordinarily lucky to find a gigantic PRP after only about 0.3% of the Poisson mean for the double loop, at 2.8k + 10k digits. Primo is now...
Sorry to have to reply to myself, but I still need an answer on this one. I forgot to mention in my first post that I would like to do this on multiple Linux...
... I now realize that it will not work for k = 33 mod 80. Here is a success with k = 1 mod 80. Let k = 4*x^2 + 1 and p = (x*k^m + x)^2 + k^m. Then p = (x*k^m...
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