Losing Downguide

henryzz · 2010-02-09, 19:26

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 · 2010-02-10, 19:35

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

Batalov · 2010-02-10, 19:57

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

Mini-Geek · 2010-02-10, 20:02

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

10metreh · 2010-02-10, 20:35

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.

Batalov · 2010-02-10, 22:42

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

2010-02-09, 19:26	#1
henryzz Just call me Henry "David" Sep 2007 Cambridge (GMT/BST) 2³·3·5·7² 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 100110110010₂ Posts	Is it possible to catch the 2^27 driver directly from the downguide? Last fiddled with by Andi47 on 2010-02-10 at 19:35 Reason: typo*

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2010-02-10, 19:57	#3
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2²·23·103 Posts	Yes, as soon as you hit a value of 2p, where p ≡ 25 mod 56.

2010-02-10, 22:42	#6
Batalov "Serge" Mar 2008 Phi(4,2^7658614+1)/2 9476₁₀ Posts	Ah, downguide! Indeed, then. (I misread it.)