Losing Downguide
 

In a similar vein to this [URL="http://mersenneforum.org/showthread.php?t=12682"]thread[/URL]. 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:[INDENT]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:[/INDENT][INDENT][INDENT]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:
[/INDENT][/INDENT]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:[INDENT]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:[INDENT]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:
[/INDENT][/INDENT]Hopefully I haven't made the stupid mistakes I made in the other thread first time round.

Is it possible to catch the 2^2*7 driver directly from the downguide?

Yes, as soon as you hit a value of [I]2p[/I], where [I]p [/I][FONT=Courier New][SIZE=3]≡ [/SIZE][/FONT]25 mod 56.
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[quote=Batalov;205234]Yes, as soon as you hit a value of [I]2p[/I], where [I]p [/I][FONT=Courier New][SIZE=3]≡ [/SIZE][/FONT]25 mod 56.


[/quote]
That is correct for morphing from the downdriver to 2^2*7, not from the downguide (2^2).
e.g. [URL]http://factordb.com/search.php?se=1&aq=2*137[/URL] vs [URL="http://factordb.com/search.php?se=1&aq=2%5E2*137"]http://factordb.com/search.php?se=1&aq=2^2*137[/URL]

[QUOTE=Andi47;205230]Is it possible to catch the 2^2*7 driver directly from the downguide?[/QUOTE]

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.

Ah, down[I]guide[/I]! Indeed, then. (I misread it.)