20100209, 19:26  #1 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2×2,861 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. With two factors:12.5% If p = 11 mod 16 then the power of 2 is 3. 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.Hopefully I haven't made the stupid mistakes I made in the other thread first time round. Last fiddled with by henryzz on 20100209 at 19:28 
20100210, 19:35  #2 
Oct 2004
Austria
2·17·73 Posts 
Is it possible to catch the 2^2*7 driver directly from the downguide?
Last fiddled with by Andi47 on 20100210 at 19:35 Reason: typo 
20100210, 19:57  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10001110100110_{2} Posts 
Yes, as soon as you hit a value of 2p, where p ≡ 25 mod 56.

20100210, 20:02  #4 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17·251 Posts 
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 MiniGeek on 20100210 at 20:03 
20100210, 20:35  #5 
Nov 2008
2×3^{3}×43 Posts 

20100210, 22:42  #6 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·3^{3}·13^{2} Posts 
Ah, downguide! Indeed, then. (I misread it.)

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