Losing Downguide

[henryzz]

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.

[Andi47]

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

[Batalov]

Yes, as soon as you hit a value of 2p, where p ≡ 25 mod 56.

[TimSorbet]

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

[10metreh]

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.

[Batalov]

Ah, downguide! Indeed, then. (I misread it.)