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2005-09-18, 10:13   #45
T.Rex

Feb 2004
France

32·103 Posts

Quote:
 Originally Posted by T.Rex (A) Is it possible to provide a Pรฉpin's test for all ranks r of F_n,r numbers ?
Looking back to Saouter's proof that there exists a Pรฉpin's test for F_n,2 numbers, my opinion is that this way seems not appropriate for F_n,3 numbers (p=5).

Saouter used the Pocklington's theorem, with $F_{n,2}-1=F*R$ where $F=2^{3^n}$.
Necessity: If there exists a k satisfying Pocklington's conditions, then the prime divisors p_i of F_n,3 have the form $p_i\equiv 1 \ \pmod{2^{3^n}}$, and multiplying p_1 and p_2 prime divisors of F_n,2 leads to a contradiction, and F_n,2 is prime.

If one uses the same way for F_n,3 numbers $F_{n,3} =(2^{5^{n+1}}-1)/(2^{5^n}-1)$ , candidate factors of F_n,3 should have the form$p_i\equiv 1 \ \pmod{2^{5^n}}$. And, using the same idea than Saouter, it only seems able to prove that if a Pรฉpin test does exist for F_n,3 numbers, it can only help to prove that F_n,3 has less than 4 factors, which is not a contradiction and does not help to prove F_n,3 is prime.

Prime divisors p_i of F_n,3 seem to have the form $p_i \equiv 1 \ \pmod{2^{n+2}*3*5^{n+1}} \ , \ n>0$ .

F_0,5 = 31 = 1+2*3*5
F_1,5 = 1082401 = 601 * 1801
601 = 1 + 2^3*3*5^2
1801 = 1 + 2^3*3^2*5^2
F_2,5 = (1+2^4*3*5^3*41*...)*(1+2^4*3^2*5^3*41*...)

Can we mix Pocklington's theorem with the real form of factors of F_n,3 numbers and build a contradiction ?

Any idea ?

Tony

Last fiddled with by T.Rex on 2005-09-18 at 10:27

 2005-09-18, 14:10 #46 alpertron     Aug 2002 Buenos Aires, Argentina 1,423 Posts I've just uploaded the new version of the page with the missing factor found by Saouter.
2005-09-18, 14:16   #47
T.Rex

Feb 2004
France

32×103 Posts

Quote:
 Originally Posted by alpertron I've just uploaded the new version of the page ...
Good idea.
May I suggest you to add which ones (n=0,1,2) are prime ?
Regards,
Tony

2005-09-18, 15:07   #48
Carlos

Jan 2005

D16 Posts

Quote:
 Originally Posted by xilman Nice! You should be given an award for your skill in stating the bleeding obvious Paul
I apologize for asking the question as it seemed to have offended you. Things that are obvious to one person are not necessarily obvious to another.

At least I did not have to resort to cursing to express myself. Please, in the future do not respond to my questions if you must curse to express yourself.

Carlos

2005-09-18, 20:20   #49
xilman
Bamboozled!

"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across

101011100010112 Posts

Quote:
 Originally Posted by Carlos I apologize for asking the question as it seemed to have offended you. Things that are obvious to one person are not necessarily obvious to another. At least I did not have to resort to cursing to express myself. Please, in the future do not respond to my questions if you must curse to express yourself. Carlos
Hey, relax.

My response was to Alex Kruppa's posting, not yours. Review the posting history and you will find that I quoted his words.

The phrase "stating the bleeding obvious" is mildly ironic and contains no perjorative connotations in English (i.e. the language spoken in England). I've no idea whether it exists in American or other variants of English. Alex clearly understood it.

Paul

 2005-10-08, 11:48 #50 alpertron     Aug 2002 Buenos Aires, Argentina 1,423 Posts I've written a program to compute factors of $4^3^n\pm 2^3^n + 1$. In only two days I found many new factors that you can see in my Factors of Modified Fermat Numbers page. I'm optimizing the program and once it is ready for "public consumption" I will post it at the same page.
 2005-10-08, 16:45 #51 T.Rex     Feb 2004 France 32·103 Posts "Feneralized Fermat Numbers" I've had an email discussion with Dr Cosgrave. He is very busy with his students, so he cannot spend too much time searching his papers. Maybe later. Nevertheless, he said that he did not find a new prime "Generalized Fermat Number" (GFN): $F_{n,r(p)}=\frac{2^{p^{n+1}}-1}{2^{p^n}-1}$ . He studied many values of n and r. On his site, there is a paper (Fermat 6) that explains the story. His colleague Wilfrid Keller could provide more information later. Alpertron, I'm interested in your program for finding factors of $F_{n,2}$ numbers. Can it be extended to any GFN ? Tony
2005-10-08, 17:23   #52
alpertron

Aug 2002
Buenos Aires, Argentina

1,423 Posts

Quote:
 Originally Posted by T.Rex Alpertron, I'm interested in your program for finding factors of $F_{n,2}$ numbers. Can it be extended to any GFN ? Tony
I think I can do it. It would take me some days because I'm busy at this time.

 2005-10-10, 00:12 #53 alpertron     Aug 2002 Buenos Aires, Argentina 1,423 Posts I've just uplodaded it to my Web server. You can download it by going to the bottom of: http://www.alpertron.com.ar/MODFERM.HTM Please let me know if there are errors.
2005-10-10, 09:16   #54
ET_
Banned

"Luigi"
Aug 2002
Team Italia

2·41·59 Posts

Quote:
 Originally Posted by alpertron I've just uplodaded it to my Web server. You can download it by going to the bottom of: http://www.alpertron.com.ar/MODFERM.HTM Please let me know if there are errors.
A nice implementation of multi-platform efficient code!

Luigi

2005-10-10, 17:04   #55
T.Rex

Feb 2004
France

32×103 Posts
In ASM ! Bravo !

Quote:
 Originally Posted by alpertron I've just uplodaded it to my Web server. You can download it by going to the bottom of: http://www.alpertron.com.ar/MODFERM.HTM Please let me know if there are errors.
A quick experiment shown that it works well ! Nice work !

I've also done a quick experiment with icc/Linux and it says:
icc genferm.c -o genferm -lm
genferm.c(12): error: invalid combination of type specifiers
typedef long long __int64;
^
compilation aborted for genferm.c (code 2)
Seems icc does not like this. But, is icc useful there ?

Tony

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