Here the timing of a 7993 digits prime certification on PRIMO Version 2.2.0 beta 4 AMD Athlon ~ 2GHz Start Jully 23 2004 … 11 Hours 20 minutes initialisation 3...
... I guess that 10^300 bits is merely a dogga-doggabit, with 10^3000 bits needed for dogga-dogga-doggabit? Hmm, perhaps some ECM might help to progress beyond...
... Please excuse my terminological inexactitude. Many years ago, I was on close terms with a friend from Wogga-wogga and she must have ingrained that name on...
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the primeform group. File : /RMA/RMA.exe ...
primeform@yahoogroups...
Apr 3, 2005 3:52 am
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I have found an even number N such that N|4^N-4 and N > 2^(2^(19824)). With frank abuse of nomenclature, one might call N a "dogga-dogga-doggabit 4-PSP", since...
PS: With A = 2^p +/- 2^((p+1)/2) + 1 we have (A+1)/2 = (8^n -/+ 1)/(2^n -/+ 1) for n=(p-1)/2 [not p-1 as in the previous message]. This makes the advantages of...
David In spite of the rather do[d]gy heading of your post:-) I'm happy that "my" little 10K-digit prime has been of service in the construction of your...
Congrats to Pierre and Marcel for ninth place in http://primes.utm.edu/top20/page.php?id=27 and first place in http://www.ellipsa.net/primo/top20.html David...
Suppose that we find a prime, p > 5, and a positive integer, k, such that k|2^(2*p-1)+1 and p|k+10. We then define A = 4^p + 1, q = 2*(A+1)/k + 1, r =...
It is quite rare to find N|2^N-2 with even N. forstep(n=4,10^7,2,if(Mod(2,n)^n-2==0,print([n,factor(n)[,1]~]))) [161038, [2, 73, 1103]] [215326, [2, 23, 31,...
The Pinch problem seems to be a lot harder than the Underwood problem. Even 2-PSP constructors' challenge [Pinch]: Find a large even N such that N|2^N-2|4^N-4....
Pinch puzzle: Find a large even N such that N|2^N-2 Lemma: Let N = 4^s-2. Then N|2^N-2 if 2*s-1|4^s-3. Proof: If 2*s-1|N-1 then 2^(2*s-1)-1|2^(N-1)-1. Multiply...
... Given that it is a precision error, have you tried raising the precision with `\p 100', or a higher value if 100 is not enough? Décio [Non-text portions...
... Makes sense. Have you tried computing log(a^8000)/log(10)? That's 977 digits, dangerously close to your limit. ... The problem is that you actually need...
\p 32000 a=real(polroots(x^3-x-1)[1]);for(n=521^2,10^6,if(Mod(round(a^n),n)==Mod(0,n),print(n))) did not work -- too much precision I guess. Intrestingly the...
... See my post -- actually, too little precision. Try \p 33200 or a little more. Too *much* precision? That feels like a BOFH excuse to me. Décio [Non-text...
Thanks for the advice Decio. But I think I have a method I will have to test: let A = [0,0,1;1,0,1;0,1,0] then compute A^N mod N finally compute to enough...
Paul and Decio: You seem to have sorted out the truncation issues, but could someone please explain to me (and maybe other helpless bystanders:-) the _math_...
... Going by the subject of the emails in this thread, I have a hunch that he initially believed that his values would generate (all?) primes. Of course, ...
Historical note about 521^2: http://mathworld.wolfram.com/PerrinPseudoprime.html ... I believe that I was among the first people to inform Ian Stewart of his...
... I was thinking about N=x^3-x-1 not being solved for Carmichael and more about there being no N found such that N|x^N-x where x is an integer; then x over...
Paul was ... There is a vital "yet" missing after "not". As I have argued, I hope persuasively, there is no reason to believe that any polynomial that...
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