rem  >  pgms.programsPD.maths.BigNum.Bas6btMapl



Date: Fri, 10 Mar 2000 00:10:18 +1300
From: Don McDonald <dsmcdona@actrix.gen.nz>
Reply-To: Don McDonald <dsmc@actrix.gen.nz>
Newsgroups: sci.math, comp.sys.acorn.apps
Subject: Basic6 myprog beats Maple on tricky bignum!

BEST NEEDS basic64

BASIC64 myprog beats Maple on Factor bignum!
******************
newgroups: sci.math, comp.sys.acorn.apps
date:  09.03.00  23:46  pm. nzdt

myfile > Bas6 v Mapl.

 progm--    fin.soc.acorncpusr.donWn994/2.pi.facthcff. byte/zeta4
don mcdonald,    18.07.99  12:30 PM. 16:00. 22:25

Examples : 1E15-1,  1E18/999 999,  2^37-1,  2^31-1
2^33-9 prime,  6763*10627*29947 Maple IsPrime
1E10-1, 1E11-1,  1E12-1, 1E14-1, 2^32+1 Euler, 13^10+1
1E13-1, 1E16-1,  100 895 598 169 Mersenne
10 662 526 601,  15 527 402 881.


factor 5 byte integer, max 2^40  gets most factors .TEST bugs
enter no. / expression  . Qq/ 0 <CR> = quit  ?195264367847
195264367847 = 195264367847
^G
proc Hhcf(x = 195264367847
  x  > 2 ^31.  slow e.g. minutes.^G
TRY FOR j = 4999  TO SQRT = 441887.279571385705  STEP 2.
factors ?   x1 =  j*x1 =   417559 *  467633 ***************
   centiseconds = 23430^G     4 minutes

 You have 10 seconds to press a key.
417559 prime.      centisec = 24461
467633 prime.      centisec = 24492
417559 * 467633

factor 5 byte integer, max 2^40  gets most factors .TEST bugs
enter no. / expression  . Qq/ 0 <CR> = quit  ?1589083027309
1589083027309 = 1589083027309
^G
proc Hhcf(x = 1589083027309
  x  > 2 ^31.  slow e.g. minutes.^G
TRY FOR j = 4999  TO SQRT = 1260588.36552976328  STEP 2.
factors ?   x1 =  j*x1 =   776057 *  2047637  ******
   centiseconds = 42389^G     **  7 minutes.

 You have 10 seconds to press a key.
776057 prime.      centisec = 43430
2047637 prime.     centisec = 43496
776057 * 2047637

factor 5 byte integer, max 2^40  gets most factors .TEST bugs
enter no. / expression  . Qq/ 0 <CR> = quit  ?
progm facthcff.   e n d.    CLOSEs * RAM SPOOL
(#####*******
               [ previous ] Result 11 of 15 [ list ] [ next ]
                  Re: Math web site (mainly Number Theory)
                           Posted by: Dave Rusin
                      Date: 2000/02/29 Group: sci.math
     _________________________________________________________________
                                      
                                      
                 In article <89f1hr$3qm$1@morgoth.sfu.ca>,
                   Erick Bryce Wong <erick@sfu.ca> wrote:
              >Jeremy Boden <jeremy@jboden.demon.co.uk> wrote:
                  >>Erick Bryce Wong <erick@sfu.ca> writes
    >>>BTW, there are cases where Maple takes forever to favor a "tiny"
                                   number
    >>>(try 195264367847 or 1589083027309 in R4 or R5, dunno about R6),
                                   and I
     >>>suspect it really is an infinite loop in the implementation of
                                  ifactor.
                                     >>
                         >>That's rather peculiar!
      >>I've only got Maple R3 and it takes about 1 second on each of
                     >>195264367847 and 1589083027309.
                                     >
   >Yup, R3 didn't have this problem...But if I recall correctly, R3 did
                                  have an
     >equally bad infinite loop in isprime(), which was fixed in R4. I
                                 know, it's
   >hard to imagine an infinite loop in a probablistic primality test...I
                                   don't
    >remember what the specific number was, but it was only around 9-12
                                  digits.
                                      
    I believe it was R3 which would choke on numbers containing 1093 or
     3511 as prime factors; perhaps you can imagine the primality tests
        being (mis)used which would flag those numbers as special...
                                      
       I don't understand why the numbers shown above defeat R4, R5.
                                    dave

Can I factor these on Acorn A4000 myprog
 UK 1993 computer with 2 MB RAM?  RISC OS 3.11
Yes, Apparently.     // don.


/ don. (loto, ANZbank ATM invisible, nz post prev. occupant unclaimed)
-- 



Date: Fri, 25 Feb 2000 15:47:59 GMT
From: ray steiner <steiner@math.bgsu.edu>
Newsgroups: sci.math
Subject: Wieferich primes

 This post is concerned with Wieferich primes. As some of
you may recall a Wieferich prime is a prime p satisfying
2^(p-1)= 1(mod p^2). It was proved many decades ago that
if this congruence doesn't hold, then the first case
of Fermat's last theorem holds for that p.
Also, the only such primes known are 1093 and 3511
and there are no further primes p < 4*10^12 satisfying
this congruence. A recent article predicted that the
next such prime should satisfy p < 10^26. How was
this prediction made?
    More generally, let p and q be odd primes.
Call p and q a double Wieferich pair if p^(q-1)= 1(mod q^2)
and q^(p-1)=1 (mod p^2).
These pairs are important in the study of Catalan's problem.
Indeed, it has very recently been proved that if p and q
do not form a double Wieferich pair then x^p - y^q = 1 cannot hold
for that p and q.
What double Wieferich pairs are known? Only the following 6:
(p,q)= (2, 1093), (3, 1006003), (5, 1645333507), (83,4871),
(911,318917) and (2903, 18787). Further there are no others
with p < 10^7 and q < 2.77p*log p*(log q +2.34)^2.
Question: Note that if p and q are odd primes at least
one of p, q in all the above pairs is congruent to 3(mod 4).
Is this true in general? That is if p, q are odd primes
and p, q form a double Wieferich pair, is at least
one of p, q congruent to 3(mod 4)? An affirmative
answer would help to narrow Catalan's conjecture
quite a bit.
Regards,
Ray Steiner
--
steiner@math.bgsu.edu

--
steiner@math.bgsu.edu


Date: Fri, 25 Feb 2000 16:08:58 GMT
From: Bob Silverman <bobs@rsa.com>
Newsgroups: sci.math
Subject: Re: Wieferich primes

In article <89687d$qvk$1@nnrp1.deja.com>,
  ray steiner <steiner@math.bgsu.edu> wrote:
>  This post is concerned with Wieferich primes. As some of
> you may recall a Wieferich prime is a prime p satisfying
> 2^(p-1)= 1(mod p^2). It was proved many decades ago that
> if this congruence doesn't hold, then the first case
> of Fermat's last theorem holds for that p.
> Also, the only such primes known are 1093 and 3511
> and there are no further primes p < 4*10^12 satisfying
> this congruence. A recent article predicted that the
> next such prime should satisfy p < 10^26. How was
> this prediction made?

Assume that 2^(p-1) behaves as a 'random' integer with respect to
is divisibility properties.  We know it is divisible by p.  It
should then be divisible by p^2  with probability 1/p.

Thus, the number of primes less than K  which satisfy the criteria
should be O( sum for p < K  of 1/p)  ~ loglog K.  We want this to
be at least 3.  exp(exp(3)) is about 5.3 x 10^8,  so it is
"unlucky" that we have not found another one already.

I'm not sure where "10^26" came from.

--
Bob Silverman
"You can lead a horse's ass to knowledge, but you can't make him think"



Date: Sat, 26 Feb 2000 20:03:55 GMT
From: Nuutti Kuosa <Nuutti.Kuosa@hut.fi>
Newsgroups: sci.math
Subject: Search of the next prime of the form n!+1


I am organizing a search for the next prime of the form n!+1.

Pages of the search project are at : 
http://www.hut.fi/~nkuosa/primeform/

The purpose of this project is to try to find the next prime of the
form
n!+1.
Due to practical reasons it is very likely that the search 
will end when we reach 157,800 digit's lenght.
Then we will probably start the search for the next prime of the form
n!-1.

We are using Chris Nash's Primeform that works in Windows and WinNT.

Here are some estimates on the time required to test one number :
(for probable primality)

P-200    :  72 hours
PII-400  :  15 hours

If you are interested in participating in the project, please visit
the
search pages.


Yours,


Nuutti Kuosa 


[Date: Thu, 09 Mar 2000 14:08:39 +0800
From: "Sophus Lefouque (M)" <fukiyami@yamashita.dontspamme>
Reply-To: vng@letterbox.com
Newsgroups: sci.math
Subject: Re: Area of a Sphere
    [The following text is in the "iso-8859-1" character set]
    [Your display is set for the "US-ASCII" character set]
    [Some characters may be displayed incorrectly]
Leyton Collins wrote:
> 
> Hi all,
> 
> I realize that this may be considered a way too easy question for some
> here but I can't find the answer and I'm not a math whiz.
> 
> Can anybody tell me what the area of a sphere is calculting from the
> known radius?V ViewAttch  48%> Can anybody tell me what the area of a sphere is calculting from the
> known radius?
> 
>  - i.e. if you wanted to know the total area of the earth's surface but
> only had the radius
> 
> Thanks in advance,
> 
> Leyton
=============================================================================
while on this topic, i came across the formula for the area of a spherical
triangle and i decided to try to prove it.
=============================================================================
Area of Spherical Triangle
     Area = ER          (radians)
70%     Area = ER          (radians)
          = (E/)/180R (degrees)
Proof:-
     on axis OA, increase dA produces increase in volume proportional to dA
     to first order in dA.  similarly for dB and dC on their respective 
     axes.  Hence
               V      V      V
          dV ~= ^^ dA + ^^ dB + ^^ dC = kAdA + kBdB + kCdC
               V      B      C
     Integrating gives
           V = kAA + kBB + kCC + H    (for some constant H)
     Considering special cases for A, B, C we arrive at
                 0 = kA/2 + kB/2          + H    ---------- [s1]
                 0 = kA/2          + kC/2 + H    ---------- [s2]
                 0 =          kB/2 + kC/2 + H    ---------- [s3]
        (1/8)4R = kA/2 + kB/2 + kC/2 + H    ---------- [s4]
93%                 0 =          kB/2 + kC/2 + H    ---------- [s3]
        (1/8)4R = kA/2 + kB/2 + kC/2 + H    ---------- [s4]
     [s1]+[s2]+[s3]-2[s4] readily gives H = -R
     [s2]-[s3] allows us to deduce that kA = kB, and similarly
     kB = kC follows from [s1]-[s2].  So kA = kB = kC = k (say)
     [s4] becomes    R = 3k/2 -R, therefore k=R
     So    V = AR+BR+CR-R = (A+B+C-)R = ER        #####
     [This proof sounds a bit like arm-waving, anybody got a better way?]
     
=============================================================================
-- 
Very Nice Guy (Magix Version), Singapore
     ----------------------------------------------------------
       We are often envious of the success of others, but
       not envious of the **** they went through to achieve


