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Re: [PrimeNumbers] Re: Infallible isprime(p) routine for 0<2^32

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  • David Cleaver
    You can see a good post about using a single spsp test to prove any number
    Message 1 of 37 , Mar 11, 2013
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      You can see a good post about using a single spsp test to prove any number <
      2^32 prime/composite over on the mersenneforum:
      In that post, he has a hash table of size 1024 that works on all numbers < 2^32.

      Warren, I'm curious, would you time this code against yours to see which one
      comes out faster?

      Someone else was able to get n < 2^64 down to 4 spsp tests, (3 fixed, and one
      into a hash). I didn't see anyone reduce it to 3 spsp tests for n < 2^64.
      Hopefully something like that will show up some day soon.

      -David C.

      On 3/10/2013 11:36 PM, WarrenS wrote:
      > If it is your goal to choose X so there are few spsp{2, X},
      > then X=735 does pretty well.
      > There are 57 spsp{2, 735} below 2^32
      > and 1149508 spsp{2, 735} below 2^64.
      > The latter is substantially better than Broadhurst's
      > X=858945; there are 1311216 spsp{2,858945} below 2^64.
      > I make no claim 735 is best, it is merely comparatively good.
      > It would probably be possible to devise an infallible isprime test for
      > <=64-bit integers by performing spsp(2) and spsp(735) tests, and then hashing
      > the 1149508 failures into 8192 bins (each bin representing about 140; need to
      > devise the hash function to get good equidistribution)
      > and then the content of that table entry tells you which base B to use for a
      > final spsp(B)
      > test. (B chosen to kill all fakes in its hash-table bin.)
      > This primality test, in net, would involve three spsp tests and one hash-table
      > lookup in a table with 8192 entries; you could probably compute the table in a
      > few days.
      > I am in fact working on an infallible 64-bit isprime test, but not using the
      > method I just outlined,instead based on combining an spsp(2) test with a Lehmer
      > test.
      > I'll report on that later, if I succeed.
    • David Cleaver
      ... Apologies, my Pari/GP script-fu is definitely not on par with DJB s. I just wanted to provide a script so that people could plug and chug an r,p pair to
      Message 37 of 37 , Mar 14, 2013
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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 can
        > 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.

        Apologies, my Pari/GP script-fu is definitely not on par with DJB's. I just
        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 to
        > 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 apologies here too, I mis-remembered a statement from his Category S page
        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,
        > http://homes.cerias.purdue.edu/~ssw/cun/index.html
        > 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.

        -David C.
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