Re: [PrimeNumbers] Re: Infallible isprime(p) routine for 0<2^32
- --- On Mon, 3/11/13, djbroadhurst <d.broadhurst@...> wrote:
> 2) at n = 52, Phil's nostrum of floor(28340/4^3) = 442Perhaps I was misremembering
> is an order of magnitude greater than the observed 41
> with the 3 extra spsp tests.
It has been proven ([Monier80] and [Rabin80]) that the strong probable primality test is wrong no more than 1/4th of the time
Are you using the common square root of -1 trick? I could believe that would give an extra power of 2. These are of course not randomly selected bases, but I can't imagine such a skew because of that. Perhaps that 1/4 is a worst-possible scenario, and just by being a 2-SPSP is not the worst possible scenario, so our rejection ratio improves. One would need to delve into the two papers and plug in the wetware, I guess.
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- On 3/13/2013 5:47 AM, Phil Carmody wrote:
> No I couldn't. That's so overly verbose and redundant it makes me twitch, I canApologies, my Pari/GP script-fu is definitely not on par with DJB's. I just
> barely bring myself to repeat it!
> "lift(Mod(p*s, lift(znorder(Mod(2,s))))) == 1" is just
> "p*s % znorder(Mod(2,s)) == 1"
> Having 3 exit conditions to the loop is overkill too.
wanted to provide a script so that people could plug and chug an r,p pair to see
what psp's would be generated. Also, I didn't know that the % (mod) operator
still worked in Pari. I thought everything had to be done with Mod(). Thanks
for that insight.
> The latter worries me a bit, as it might imply wasted effort. I'm trying toAnd apologies here too, I mis-remembered a statement from his Category S page
> picture how these duplicates arise. Given a n, the maximal prime factor p|s is
> uniquely defined, and r as order_2(p) is uniquely defined. Therefore n can only
> appear with pair (r,p)?
and mis-spoke by applying it to the Category E psp's.
On 3/14/2013 11:02 AM, WarrenS wrote:
> 2. Consulting the Cunningham project pages,
> every Mersenne-form number 2^r - 1 now is fully factored if
> r<929. Apparently the first two open cases are r=929 and 947
> yielding 214 and 217 digit numbers to factor.
An update here: M929 has been factored, and a group of people have already
started factoring M947. You can find the factor for M929 here:
And, you can see who is factoring which Cunningham number here:
And more importantly, you can find all known*1 factors for all important*2
numbers in the online factor database here:
Once there, you can type in 2^929-1, and it will show you all the factors of
that number and that it is Fully Factored (FF). Currently, you can type in
2^947-1 and see that it is a CF, composite number, with factors known, but not
yet fully factored.
*1 = All known factors that have been stored into the factordb.
*2 = All numbers that people are interested in and store in the factordb.
Also, the factordb stores prime numbers too. Below 300 digits it will just
prove the number prime, and above that it will accept Primo certificates and
verify them locally.