Here is a new record prime 17-tuplet: 356357564387652563232359861 + d, d = 0, 6, 8, 12, 18, 20, 26, 32, 36, 38, 42, 48, 50, 56, 60, 62, 66 (27 digits, November...
21959
Jens Kruse Andersen
jkand71
Nov 1, 2010 8:06 pm
... Congratulations! http://users.cybercity.dk/~dsl522332/math/simultprime.htm is updated. -- Jens Kruse Andersen...
21960
djbroadhurst
Nov 1, 2010 9:31 pm
... Proposition: For every integer pair (p,k), we have V(p, (23*k+11)*3103, 4638985) = p mod 4638985 Proof [using the Sun Tzu Suan Ching]: ...
21961
mikeoakes2
Nov 2, 2010 1:01 pm
... The best pari-GP script I can come up with is /quadratic/ in n_max, so will take about 23360000 secs or 9 months at 3.6 GHz to reach n_max=10^7 and...
21962
djbroadhurst
Nov 2, 2010 2:42 pm
... The code looks good, but could use some more modular theory in V(p,q,n) mod m. For example we may reduce n to n%(m^2-1) for prime m. ... First off, the...
21963
mikeoakes2
Nov 2, 2010 4:14 pm
... So that's what you did... If I change one line of my script from if(fac>1 to if(fac>3 then the run times collapse drastically to /* n_max time 10^4 31...
21964
djbroadhurst
Nov 2, 2010 4:49 pm
... Indeed :-) Here my Chinese speed-up is huge. Please note that I counted *all* the (q,n) pairs with square-free non-Carmichael n, coprime to 6, and more ...
21965
djbroadhurst
Nov 2, 2010 6:25 pm
In primenumbers@yahoogroups.com, ... That was my opinion. The more interesting question, to my mind, is whether there could be a non-Carmichael solution with...
21966
mikeoakes2
Nov 3, 2010 1:33 pm
... I have a cunning script which found these first 7 non-Carmichaels in 9313 msecs. Mike...
21967
bhelmes_1
Nov 3, 2010 8:47 pm
A beautifull day, there are 15 new 40000 digit primes in the collection :-) http://beablue.selfip.net/devalco/Collection/40000/ (needed time 30 days on 6...
21968
Kermit Rose
kermit1941
Nov 4, 2010 5:36 am
http://www.mathpages.com/home/kmath347/kmath347.htm claims that the first exception to the conjecture that n is a prime if and only if f(r**n) = 0 (mod n) ...
21969
djbroadhurst
Nov 4, 2010 9:44 pm
... Excellent: post-hoc speed-ups are always welcome. Yet first credit should always go to the first discovery. Happily this also goes to you: ...
21970
djbroadhurst
Nov 4, 2010 10:28 pm
... Yes. It's very simple, Bernhard, and was long since exposed ... If you sieve any target N to sufficient depth p and and find that N still has no identified...
... Thanks for that, David. For c), I benefited from your Chinese hint: hereby gratefully acknowledged. I think the best thing about a) was that it was truly...
21973
djbroadhurst
Nov 4, 2010 11:54 pm
... Yes. ... Yes. David...
21974
djbroadhurst
Nov 5, 2010 12:18 am
... Indeed I did: http://tech.groups.yahoo.com/group/primenumbers/message/21899 http://tech.groups.yahoo.com/group/primenumbers/message/21898 ...
21975
Maximilian Hasler
maximilian_h...
Nov 5, 2010 3:10 am
I do agree that ... but I do not agree that ... For example, one root of f(x) is r1 = 0.64879067514879204822278147415791168556... and f( r^2 ) =...
21976
mikeoakes2
Nov 5, 2010 9:46 am
... None for n <= 10^7. Took nearly 2 days. Mike...
21977
djbroadhurst
Nov 5, 2010 2:25 pm
Puzzle 2: Find an odd square-free composite integer n < 443372888629441 such that there exist more than 10^12 integers q with n > q > 0 and V(p,q,n) = p mod n,...
21978
mikeoakes2
Nov 5, 2010 2:56 pm
... Typing 443372888629441 into the OEIS Search box gives: http://www.research.att.com/~njas/sequences/index.html?q=443372888629441&language=english&go=Search ...
21979
Kermit Rose
kermit1941
Nov 5, 2010 3:43 pm
... I see. Thank you for pointing out this distinction. It is only in the ring polynomials mod ( x**5 - x**3 - 2x**2 + 1, 2) that (x**2)**5 - (x**2)**3 -...
21980
djbroadhurst
Nov 5, 2010 4:18 pm
... Yes. Moreover there are no pseudoprimes less than 10^5: f(x)=x^5-x^3-2*x^2+1; for(p=2,10^5,if(isprime(p)!=(0==f(Mod(Mod(1,p)*x,f(x))^p)),print(p))); [The...
21981
Kermit Rose
kermit1941
Nov 5, 2010 6:54 pm
... Yes. I am ok with using the language of ring polynomial mod ( particular polynomial, integer). I will not quibble with you about whether or not it is...
21982
djbroadhurst
Nov 5, 2010 8:21 pm
... Indeed. A solution may be found in less than 10 seconds by such a method. The puzzle was set up so that you cannot find the answer simply by looking at a...
21983
paulunderwooduk
Nov 6, 2010 9:46 pm
Hi, referencing: http://www.gpgpgpu.com/gecco2009/6.pdf from: http://www.mersenneforum.org/showthread.php?p=235821#post235821 it is nice to see that a 350-SPRP...
21984
Chris Caldwell
primemogul
Nov 6, 2010 10:15 pm
From: primenumbers@yahoogroups.com [mailto:primenumbers@yahoogroups.com] On Behalf Of paulunderwooduk ... test id good for all n<170,584,961. Excellent--I hope...
21985
Andrey Kulsha
andrey_601
Nov 7, 2010 12:40 am
... A table lookup is still faster IMHO. Best regards, Andrey [Non-text portions of this message have been removed]...
21986
djbroadhurst
Nov 7, 2010 2:34 am
... It can be done a shade faster than that: {ncarm(n)=local(F=factor(n),f=F[,1],m=#f,p,t,d);if(m>3&& sum(j=1,m,F[j,2])==m,t=1;for(j=1,m,p=f[j];if((n-1)%(p-1),...
21987
djbroadhurst
Nov 7, 2010 11:52 am
... Off list, Mike asked for a proof that this works, i.e. that there exists at least one integer Q with n > Q > 0 and V(P,Q,n) = P mod n, for every integer P,...
