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 2010-02-09, 19:26 #1 henryzz Just call me Henry     "David" Sep 2007 Liverpool (GMT/BST) 136178 Posts Losing Downguide In a similar vein to this thread. I will now show what 2^2 mutates into. With one factor: 50% If p = 1 mod 4 then we get the downdriver. 50% If p = 3 mod 4 then the power of 2 is 2 or more:25% If p = 7 mod 8 then the power of 2 is 2. 25% If p = 3 mod 8 then the power of 2 is 3 or more:12.5% If p = 11 mod 16 then the power of 2 is 3. 12.5% If p = 3 mod 16 then the power of 2 is 4 or more: With two factors: 50% If one of p and q = 1 mod 4 and the other 3 mod 4 then the power of 2 is 2. 25% If p and q = 3 mod 4 then the power of 2 is 2. 25% If p and q = 1 mod 4 then the power of 2 is 3 or more:12.5% If p and q = 1 mod 8 or p and q = 5 mod 8 then the power of 2 is 3. 12.5% If one of p and q = 1 mod 8 and the other 5 mod 8 then the power of 2 is 4 or more:3.125% If one of p and q = 1 mod 16 and the other 13 mod 16 then the power of 2 is 4. 3.125% If one of p and q = 9 mod 16 and the other 5 mod 16 then the power of 2 is 4. 3.125% If one of p and q = 1 mod 16 and the other 5 mod 16 then the power of 2 is 5 or more: 3.125% If one of p and q = 9 mod 16 and the other 13 mod 16 then the power of 2 is 5 or more: Hopefully I haven't made the stupid mistakes I made in the other thread first time round. Last fiddled with by henryzz on 2010-02-09 at 19:28
 2010-02-10, 19:35 #2 Andi47     Oct 2004 Austria 9B216 Posts Is it possible to catch the 2^2*7 driver directly from the downguide? Last fiddled with by Andi47 on 2010-02-10 at 19:35 Reason: typo
 2010-02-10, 19:57 #3 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22·7·359 Posts Yes, as soon as you hit a value of 2p, where p ≡ 25 mod 56.
2010-02-10, 20:02   #4
TimSorbet
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by Batalov Yes, as soon as you hit a value of 2p, where p ≡ 25 mod 56.
That is correct for morphing from the downdriver to 2^2*7, not from the downguide (2^2).
e.g. http://factordb.com/search.php?se=1&aq=2*137 vs http://factordb.com/search.php?se=1&aq=2^2*137

Last fiddled with by TimSorbet on 2010-02-10 at 20:03

2010-02-10, 20:35   #5
10metreh

Nov 2008

44228 Posts

Quote:
 Originally Posted by Andi47 Is it possible to catch the 2^2*7 driver directly from the downguide?
No, because the sigma of a number with the downguide is always divisible by 7, and the number is not, so the next term (i.e. sigma - the number) is not divisible by 7.

 2010-02-10, 22:42 #6 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 22×7×359 Posts Ah, downguide! Indeed, then. (I misread it.)

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