mersenneforum.org > Math Stupid Question Re: Fermat Factor search
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2012-10-13, 15:11 #1 c10ck3r     Aug 2010 Kansas 547 Posts Stupid Question Re: Fermat Factor search Okay, so all factors of a Fermat number are of the form k*2^n+1, right? So, theoretically, P-1 would find factors as long as the k is b-smooth? Is the limitation of this the ~1e15 k value, the size of 2^2^n, or something else that I'm missing? Thanks, Johann.
 2012-10-18, 03:52 #2 MattcAnderson     "Matthew Anderson" Dec 2010 Oregon, USA 4A816 Posts Hi, Fermat numbers have the form 2^(2^m) + 1. Factors of Fermat numbers have the form k*2^n + 1, where n >= m+2. My reference is http://www.fermatsearch.org/ Using a trial division software downloaded from the above website, I have been searching n=36 and k>1.4e16 Luigi Morelli has a new software that runs about 50 times faster than this, and can find factors of Fermat and Double Mersenne numbers, but it requires a Graphics Processing Unit, and can be downloaded from - http://www.doublemersennes.org/ Regards, Matt
2012-10-18, 04:23   #3
Dubslow
Basketry That Evening!

"Bunslow the Bold"
Jun 2011
40<A<43 -89<O<-88

722110 Posts

Quote:
 Originally Posted by MattcAnderson Hi, Fermat numbers have the form 2^(2^m) + 1. Factors of Fermat numbers have the form k*2^n + 1, where n >= m+2. My reference is http://www.fermatsearch.org/ Using a trial division software downloaded from the above website, I have been searching n=36 and k>1.4e16 Luigi Morelli has a new software that runs about 50 times faster than this, and can find factors of Fermat and Double Mersenne numbers, but it requires a Graphics Processing Unit, and can be downloaded from - http://www.doublemersennes.org/ Regards, Matt
Thank you Captain Obvious. (To quote Batalov.)

2012-10-18, 05:26   #4
Batalov

"Serge"
Mar 2008
Phi(4,2^7658614+1)/2

3·7·479 Posts

Captain Obvious adds that if "Luigi has a new software", then in that sense everyone has it. The author who had it (as a baby) was one Woltman.

But anyway, that had nothing to do with the OP's question.
Quote:
 Originally Posted by c10ck3r Okay, so all factors of a Fermat number are of the form k*2^n+1, right? So, theoretically, P-1 would find factors as long as the k is b-smooth? Is the limitation of this the ~1e15 k value, the size of 2^2^n, or something else that I'm missing? Thanks, Johann.
To that, the answer is "yes".

In fact, P-1 had been run with high limits already, so one would not expect to succeed where others* failed. Sometimes, though, - as the Russian proverb goes, even a stick shoots.

_______________
*to quote from "Almost Famous": " I'm about to boldly go where... ummm... many men have gone before"

 Thread Tools

 Similar Threads Thread Thread Starter Forum Replies Last Post LaurV Information & Answers 14 2015-06-18 23:37 Uncwilly Lounge 19 2013-03-07 04:44 Biggles Prime Sierpinski Project 3 2006-02-07 22:50 rpropper GMP-ECM 15 2005-11-14 14:43 fropones Math 2 2003-05-28 00:44

All times are UTC. The time now is 01:38.

Tue Feb 7 01:38:25 UTC 2023 up 172 days, 23:06, 1 user, load averages: 1.46, 1.48, 1.53

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔