Well, we should know some time on Monday...

[CODE]linear algebra completed 1010450 of 13454920 dimensions (7.5%, ETA 71h40m) [/CODE]

[QUOTE=henryzz;443851]Assuming two factors 50% I think. The composite is 1 mod 4. This can be made up of two 1 mod 4 factors or two 3 mod 4 factors. If it is the 1 mod 4 factors we will drop to 2^2 otherwise we will stick with 2^3.[/QUOTE]

If we do drop to 2^2, there is a 1 in 6 chance that we'll pickup a 7 as well :-(

[QUOTE=axn;443914]If we do drop to 2^2, there is a 1 in 6 chance that we'll pickup a 7 as well :-([/QUOTE]

And that is why I would prefer to stay at 2^3 until we get 2^3*p with p 1 mod 4. You also can't get 3s or 5s with 2^3.

[QUOTE=axn;443914]If we do drop to 2^2, there is a 1 in 6 chance that we'll pickup a 7 as well :-([/QUOTE]

We dropped to 2^2 and dodged the 7 :smile:

[QUOTE=schickel;443848]The fantastic news is that with it holding steady at 2^3, there has been some downward pressure. What are the odds that the power of 2 will

change it the current line [10664] factors into two primes?[/QUOTE]
Aliquot Sequence 4788 Iteration Number 10664: c182 = 2[sup]3[/sup] × c181
c181 ≡ 1 (mod 4)

[b][u]Two prime factors[/u][/b]
50% chance: 1 (mod 4) × 1 (mod 4): 2[sup]3[/sup] → 2[sup]2[/sup] (Pray no collapse in to 7!).
50% chance: 3 (mod 4) × 3 (mod 4): 2[sup]3[/sup] → 2[sup]3[/sup] (Pray no collapse in to drivers!).

[QUOTE=schickel;443862]Actually my question was worded a little wrong: I actually meant what are the odds the power of 2 would go down if it splits into two.

If it splits into three factors, or it factors into two and the math is wrong, there is a chance that the power of 2 can change to something higher than 3.

To get the downdriver you need a line to factor as 2^n * p, with p being of the proper modular form to drop the power of 2 to 1. (Or 2 * 3^n * p, [n even] if you have

the 2 * 3 driver; very hard to do!)

In this case, there is no way to get the downdriver on the next line, but maybe we can get a reduction to 2^2 which will help a lot with the size of the composites on

each line.[/QUOTE]

Aliquot Sequence 4788 Iteration Number 10664: c182 = 2[sup]3[/sup] × c181
c181 ≡ 1 (mod 8)

[b][u]Three prime factors[/u][/b]
12.5% chance: 1 (mod 8) × 1 (mod 8) × 1 (mod 8): 2[sup]3[/sup] → 2[sup]>3[/sup] (Pray no collapse in to drivers!).
37.5% chance: 1 (mod 8) × 5 (mod 8) × 5 (mod 8): 2[sup]3[/sup] → 2[sup]>3[/sup] (Pray no collapse in to drivers!).
12.5% chance: 1 (mod 8) × 3 (mod 8) × 3 (mod 8): 2[sup]3[/sup] → 2[sup]3[/sup] (Pray no collapse in to drivers!).
12.5% chance: 1 (mod 8) × 7 (mod 8) × 7 (mod 8): 2[sup]3[/sup] → 2[sup]3[/sup] (Pray no collapse in to drivers!).
25% chance: 3 (mod 8) × 5 (mod 8) × 7 (mod 8): 2[sup]3[/sup] → 2[sup]3[/sup] (Pray no collapse in to drivers!).

[QUOTE=Raman;442221]
By the way, if a line factors as 2[sup]3m-1±1[/sup].7[sup]k[/sup].c, or, where by m ≥ 1, k ≥ 1, and then if c is being a composite number,
and then if only let alone assuming c has got two prime factors,
with in the next iteration, 7 will disappear with a given fixed probability of
5/6 if c ≡ 1 (mod 7)
2/3 if c ≡ 2, 3, 4, 5, 6 (mod 7)
and then if they can re-appear if the same condition occurs back again.

By the way, if a line factors as 2[sup]3m-1±1[/sup].7[sup]k[/sup].p, or, where by m ≥ 1, k ≥ 1, and then if p is being a prime number,
and then if only let alone assuming p has got one prime factor,
with in the next iteration, 7 will disappear with a given fixed probability of
0 if p ≡ 6 (mod 7)
1 if p ≡ 1, 2, 3, 4, 5 (mod 7)
and then if they can re-appear if the same condition occurs back again.
[/QUOTE]

Gained a 3 despite 6 prime factors.

[QUOTE=Dubslow;444370]Gained a 3 despite 6 prime factors.[/QUOTE]

Ouch!!

Aliquot Sequence 4788 Iteration Number 10704: 2[sup]2[/sup] × 3 × c184
c184 ≡ 1 (mod 3)

Assuming with two prime factors
50% chances of losing 3 - c184 = 1 (mod 3) × 1 (mod 3)
50% chances of retaining 3 - c184 = 2 (mod 3) × 2 (mod 3)

Assuming with three prime factors
25% chances of losing 3 - c184 = 1 (mod 3) × 1 (mod 3) × 1 (mod 3)
75% chances of retaining 3 - c184 = 1 (mod 3) × 2 (mod 3) × 2 (mod 3)

You cannot be sure about it any time before itself!

Aliquot Sequence 314718 Iteration Number 17164: 2[sup]2[/sup] × 3 × c184
Aliquot Sequence 16100 Iteration Number 17164: 2[sup]2[/sup] × 3 × c184

Iteration Number 17130: Aliquot Sequence 17130 terminates in 8128.
Iteration Number 17490: Aliquot Sequence 17490 terminates in 1264460 / 1547860 / 1727636 / 1305184.

Or how about if or assume that c184 = p184 and then carrying out!

[QUOTE=Raman;444447]Aliquot Sequence 4788 Iteration Number 10704: 2[sup]2[/sup] × 3 × c184
c184 ≡ 1 (mod 3)

Assuming with two prime factors
50% chances of losing 3 - c184 = 1 (mod 3) × 1 (mod 3)
50% chances of retaining 3 - c184 = 2 (mod 3) × 2 (mod 3)

Assuming with three prime factors
25% chances of losing 3 - c184 = 1 (mod 3) × 1 (mod 3) × 1 (mod 3)
75% chances of retaining 3 - c184 = 1 (mod 3) × 2 (mod 3) × 2 (mod 3)

You cannot be sure about it any time before itself!
[/QUOTE]

Should have stopped there.

[QUOTE=Dubslow;444370]Gained a 3 despite 6 prime factors.[/QUOTE]

8 prime factors including 2[sup]2[/sup] not 6 prime factors excluding 2[sup]2[/sup]!
Unsure! :unsure:

Aliquot Sequence 4788 Iteration Number 10681:
Aliquot Sequence 314718 Iteration Number 17141:
Aliquot Sequence 16100 Iteration Number 17141:

[CODE]
2[sup]2[/sup] × 13 × 457 × 1531 × 32982319 × 494035702115239309 × 16895032315382285560931099849261915227930228987063210304384329149657914039458986940231271615739931549325496657964294621271898101762316595310422208017
[/CODE]

[QUOTE=Raman;443217]
[COLOR="blue"]
Computation of Aliquot Sequence 4788 / 16100 / 314718 losing and gaining 3 periodically, as long as the power of 2 is a positive even number!

On the other hand,
Downdriver 2 does not ever gain a 3 unless power of 2 is mutated; Driver 2.3 does not ever lose a 3 unless power of 3 is mutated.
Stable guide 2[sup]2[/sup] does not ever gain a 7 unless power of 2 is mutated; Driver 2[sup]2[/sup].7 does not ever lose a 7 unless power of 7 is mutated.
Stable guide 2[sup]4[/sup] does not ever gain a 31 unless power of 2 is mutated; Driver 2[sup]4[/sup].31 does not ever lose a 31 unless power of 31 is mutated.
Stable guide 2[sup]6[/sup] does not ever gain a 127 unless power of 2 is mutated; Driver 2[sup]6[/sup].127 does not ever lose a 127 unless power of 127 is mutated.
Stable guide 2[sup]12[/sup] does not ever gain a 8191 unless power of 2 is mutated; Driver 2[sup]12[/sup].8191 does not ever lose a 8191 unless power of 8191 is mutated.

Ratio between computation of subsequent Aliquot / Totient sequence next iterations - Consider orbital resonance! Why?
[/COLOR]
[/QUOTE]

Spent my own whole time with this type of thing today!
Why - what! Why - what! Last past time period

[QUOTE=Raman;444464]Last past time period[/QUOTE]
I hate when that happens.