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-   -   Reserved for MF - Sequence 4788 (https://www.mersenneforum.org/showthread.php?t=11615)

 LaurV 2016-09-14 10:58

Well, with the same notation as before, say t=2*F, then its s is 3(1-F*)-2F, therefore this can never be multiple of 3. To "break", it needs to change the power of 2, and that is only possible if F=Qp is a product of an odd perfect square Q and a prime p with p=1 mod 4. It also works if p or q are void. For that to happen below 100 digits the chances (i.e. bad odds) are:

at ~50 digits: below 1%
at ~40 digits: below 1.6%

We are "safe" for now :razz: (yeah, well, I know, "don't jinx it!" hehe).

 unconnected 2016-09-14 11:08

< 100 digits again!

 Dubslow 2016-09-14 11:28

Booooooooooooooooooo. i8359 = 2 * 7^2 * P91

 science_man_88 2016-09-14 11:34

[QUOTE=Dubslow;442507]Booooooooooooooooooo. i8359 = 2 * 7^2 * P91[/QUOTE]

yeah but luckily since sigma(2^3) = 15 we would need a factor of 3 or 5 to drive it up at last check of what's been said. okay now it's is 2^2 and so would need a 7 I think. and now at 90 digits and a single 2.

 Dubslow 2016-09-14 11:51

[QUOTE=science_man_88;442509]and now at 90 digits and a single 2.[/QUOTE]

Huzzah!
[code]
Checked 8417 93 (show) 2687306884...12<93> = 2^2 · 3^4 · 8294157051...13<90>
[/code]
[$]\tau(P90) = 1[/$] and we have recovered the downdriver!

Down below 90! 86 and dropping, now 81. We're at the point where the DB is factoring most of the lines by itself.

[QUOTE=science_man_88;442509]yeah but luckily since sigma(2^3) = 15 we would need a factor of 3 or 5 to drive it up at last check of what's been said. okay now it's is 2^2 and so would need a 7 I think. [/QUOTE]

Yeah these are/were correct.

Edit: 1205Z and I got a refresh down at 37 digits, subsequent refresh attempts are still pending. Exciting! Termination or more disappointment?

 science_man_88 2016-09-14 11:53

[QUOTE=Dubslow;442510]Huzzah!
[code]
Checked 8417 93 (show) 2687306884...12<93> = 2^2 · 3^4 · 8294157051...13<90>
[/code]
[$]\tau(P90) = 1[/$] and we have recovered the downdriver![/QUOTE]
we are about i8433 right now actually. i8458 is this going to be like yesterday where I couldn't keep up with the edit needed to keep it recent lol. edit: yeah probably ... now at or above i8507. and around 74 digits. might update this edit near my end of edit time. driver:2^4*31 now back at over 90 digits and climbing.

 mshelikoff 2016-09-14 12:08

314718 just passed i15,000.

Even though it's a bit (but not extremely) unlikely that the sequence will terminate soon and it's very unlikely that the sequence will catch itself on the way down to create a large cycle, I don't think the probability of those events are in "winning the lottery" territory, and it certainly wouldn't be like the coin flips in Tom Stoppard's "Rosencrantz and Guildenstern Are Dead" at [URL]https://youtu.be/KchhSIVwMdY?t=1m30s[/URL] .

 Dubslow 2016-09-14 12:09

Nope, 1208Z and the downdriver went below 20, cycled up to 50 then below 20 again and now back into the 80s with the D4 = 2^4 * 31. Close to 1500 lines added in the last 30 minutes. Edit: 1210Z and we have passed 10,000 total lines! Congrats Ryan, better than nothing :razz:

 unconnected 2016-09-14 12:11

10000 lines and counting! 2^4*31 isn't easy to survive.

 science_man_88 2016-09-14 12:59

102 digits now i10133 1700 iterations from when I posted in response to dubslow. and 66 minutes roughly. average of roughly 2.33 seconds per iteration roughly. ( to within a minute timing total).

 henryzz 2016-09-14 13:51

What is the highest 2^4*31 has been lost?

 RichD 2016-09-14 14:21

For this sequence, I knocked one off [URL=http://www.factordb.com/sequences.php?se=1&aq=4788&action=range&fr=5160&to=5162]here[/URL].

 science_man_88 2016-09-14 14:23

[QUOTE=henryzz;442520]What is the highest 2^4*31 has been lost?[/QUOTE]

searching factordb for it through the aliquot sequence selection brings up:
[QUOTE]302464450620132002992<21>
36072943892238577277247022778243938144883456<44>
1415767501064859248<19>
3141832894...92<109>
21933224531330162416<20>
1453967358...96<108>
1638782940854073904<19>
4390711203...68<99>
1209528545...12<104>
18412050479184368191100122180112<32>
399880994693565806957517104<27>
3156964483...36<108>
29881985883081475837467776336<29>
5628833728627973643098576<25>
360585660037777060955632<24>
7169123053...64<109>
310797102124039979216<21>
1023816193737241913858384<25>
1417854948980933459921072<25>
1417854948980933459922064<25>[/QUOTE]

as having a factor of it so looks to be about 109 digits for the highest listed there (maybe a limit to the amount of numbers listed though) so maybe see if the sequences involving the higher ones to see if it escaped ? edit: At least one of the sequences looks to have gotten up to 131 digits from a 109 digit start of sequence value before it dropped it all.

 Drdmitry 2016-09-14 14:28

There is a [URL="http://factordb.com/sequences.php?se=1&aq=4788&action=range&fr=10198&to=10199"]chance[/URL]. :)

 henryzz 2016-09-14 14:33

Now on 2^3*3 which is much weaker. The 3 is squared for now as well.

 science_man_88 2016-09-14 14:36

[QUOTE=henryzz;442528]Now on 2^3*3 which is much weaker. The 3 is squared for now as well.[/QUOTE]

the square is no longer:

[QUOTE]Unchecked 10204 114 (show) 6410204042...28<114> = 2^3 · 3 · 300733 · 8881361044...09<107>[/QUOTE]

 henryzz 2016-09-14 17:00

Lost it. Now 2^2*3. Just a case of losing the 3 before it will go down slowly.

 science_man_88 2016-09-14 17:08

[QUOTE=henryzz;442543]Lost it. Now 2^2*3. Just a case of losing the 3 before it will go down slowly.[/QUOTE]

it also involves not getting a 7.

 ryanp 2016-09-14 17:12

So I guess I jinxed it earlier. :smile:

But I can now say I've stuck with this sequence for ~50% of its life... 10K terms and still going!

 henryzz 2016-09-14 17:35

[QUOTE=science_man_88;442546]it also involves not getting a 7.[/QUOTE]

It is impossible to gain a 7 without losing the 2^2. The same arguments can be used to prove this as for losing a 7.

 Raman 2016-09-14 18:12

How about posting all of my opinions over here, rather than in their appropriate threads?

@ Ryan Propper: Have you automated submissions to FactorDB even when you are asleep or outside? Great job!
Factoring a c120 in 2006 would take upto 5 days. In 2016, it is only taking 5 minutes. Or 10 minutes may be? Distributed computing is being going on automatically?
Perhaps you could help out with aliquot sequences of 552, 564, 660, 1512, 1992, 5250, 9120, 11040 besides of 4788 or of 314718.

The open end aliquot sequences below 10000 are
[b]276[/b], 306, 396, [b]552[/b], [b]564[/b], [b]660[/b], 696, 780, 828, 888, [b]966[/b], 996.
Those given in bold are main open end aliquot sequences.
Those not given in bold are side open end aliquot sequences to the above mentioned main open end aliquot sequences.

[QUOTE=10metreh;165862]I hope I live to see it reach 10000 lines![/QUOTE]

Do you mean aliquot sequence 314718 or aliquot sequence 4788? Anyway both aliquot sequences are past 10000 iteration mark right now.
10metreh is not being active these days. Right now aliquot sequence 314718 is at iteration 16744 and counting. Right now aliquot sequence 4788 is at iteration 10284 and counting.

Is 2 the only way of going down by means of a downdriver. Are there no other ways? What about going down by stable guides?
There are many different drivers (perfect numbers - all only known to be even). But there is only one downdriver!
Are there any drivers for odd numbers? Is that statement equivalent to asking for existence of odd perfect numbers?
Any way that an aliquot sequence parity change will occur only for squares or twice squares. They are the only way to terminate an aliquot sequence. Isn't it?

[QUOTE=unconnected;435000]...and now we have squared 3 :no:[/QUOTE]

Say 9, instead of squared 3!
Sometimes that lines like 2[sup]3[/sup].3[sup]2[/sup].p where p is a prime number that is being congruent to 1 (mod 4) can decompose in the next iteration into
2.3 driver, the driver which is the most difficult to escape from - [strike]2[sup]6[/sup].127 driver[/strike]. Some one could pray that this does not happen right now at all!
But, we cannot do anything with it! All is natural! Aliquot sequences is a deterministic algorithm, not a randomized algorithm at all!
The downdriver 2 cannot pick up a 3, the stable guide 2[sup]2[/sup] cannot pick up a 7, the stable guide 2[sup]4[/sup] cannot pick up a 31, the stable guide 2[sup]6[/sup] cannot pick up a 127.
Further more that in the aliquot sequence 4788 iteration 8417, the line factored into 2[sup]3[/sup].3[sup]4[/sup].p where p is a prime number that is being congruent to 1 (mod 12).
Were that p were congruent to 5 (mod 12), the next iteration would have decayed into the 2.3 driver!
Or if aliquot sequence 4788 iteration 10262 or 10284, the line factored exactly into 2[sup]3[/sup].3[sup]2[/sup].p where p is a prime number that is being congruent to 5 (mod 12).

There are exactly 6460 iterations difference of offset between aliquot sequence 4788 and aliquot sequence 314718.

It took 4 years (2012 - 2016) to reach from previous summit of this aliquot sequence to the current summit, at present.
Previous trench of this aliquot sequence was at 16 digits, the current trench, at present was at 12 digits, the next trench will be at 8 digits, 4 digits and 0 digits, termination!
By the way, by that arrange the following in increasing or decreasing order of height / altitude / elevation.
Trench - Ridge - Shallow seas - Sea level - Plains - Summit - Peak.
By the way, by that what is the exact difference between all these things? And then note - notice!

[QUOTE=unconnected;442251]Minimum at 16 digits and now it grows with 2^2*7 driver.[/QUOTE]
2[sup]2[/sup].7 driver is the easiest driver to escape, isn't it? Or is it the 2[sup]4[/sup].31 driver?
Any way that you would need 2[sup]2[/sup].7[sup]2k[/sup], k ≥ 1 to escape from the 2[sup]2[/sup].7 driver, and then
that you would need 2[sup]4[/sup].31[sup]2k[/sup], k ≥ 1 to escape from the 2[sup]4[/sup].31 driver.

A good exercise for analysis:
Arrange the following drivers in the increasing or decreasing order in the probability of their getting escaped from:
2.3 driver, 2[sup]2[/sup].7 driver, 2[sup]4[/sup].31 driver, 2[sup]6[/sup].127 driver.
How about the 2[sup]12[/sup].8191 driver, etc.?

[QUOTE=mshelikoff;442252]It looks like it picked up a 2^2·7 at the 16 digit 1987170707261684 , found the downdriver later, and then picked up another 2^2·7 at 20 digits and this one isn't leaving.[/QUOTE]

The main downdriver for the aliquot sequence 4788 or 314718 from 200 digits exactly ended at 87 digits, eventhough exactly that it did go down upto 16 digits (right now 12 digits!)
The greatest descent for aliquot sequence 1578 or 56440 or 79640 from 110 digits in the Wolfgang Creyaufmüller's page is being given as exactly at 5 digits, although exactly that the main downdriver did end at 58 digits.
By the way by that the greatest ascent for aliquot sequence 1578 or 56440 or 79640 during that peak was exactly at 111 digits, but not exactly 110 digits. However that the greatest ascent for aliquot sequence 4788 or 314718 during that ridge was at exactly 200 digits - exactly.
How do you consider or handle this?

[QUOTE=axn;442302]Currently @ 7960 with 2^3*3 driver (ouch!). Well, at least, it isn't the 2^3*3*5 driver.[/QUOTE]

What is the difference between 2[sup]3[/sup].3 driver and 2[sup]3[/sup].3.5 driver? Or is even 2[sup]3[/sup].5 being a driver?
On the other hand, [strike]by that do[/strike] 2[sup]3[/sup], etc. is not being a driver, but only, let alone, a stable guide.

@ LaurV, what does your D2, D3, etc. mean?
Do they mean the drivers 2.3 driver, 2[sup]2[/sup].7 driver, 2[sup]4[/sup].31 driver, 2[sup]6[/sup].127 driver, 2[sup]12[/sup].8191 driver, etc.?

@ Dubslow, what does your 1205Z, 1208Z, 1210Z, etc. mean?
And do they mean the Greenwich Mean Time (GMT) and the Universal Coordinated Time (UTC) respectively?
By the way, by that do note or notice that they are not being equivalent at all, but only, let alone, during in the autumn / fall / winter times - not - not - but only, let alone, during in the summer / spring times!
Universal Coordinated Time is being UTC - not - not - Universal Coordinated Time is being UCT - not - not!
Note - notice. Note - notice.
Autumn - fall. Spring - summer - autumn / fall - winter.

11 posts per page!

Waiting for lengthy post for the past few days!

Are all the computations and entries in the FactorDB are being correct? Or are there any chances for errors?
Mainly a Carmichael number or a pseudoprime number is being marked as a PRP number or SPRP number, if it were actually a composite number, albeit chances for such false positives are being very low?
By the way by that what does albeit mean - although - eventhough?

[QUOTE=Raman;442221]
How many resources do you have? You are able to crack a p196 very quickly!
[/QUOTE]

Should have been - or
Must have been - or

How many resources do you have? You are able to crack a c196 very quickly!

By the way by that post delete option has been gone away - that ever which ever a way a way.

Law of big numbers:
→ Every big number does try out to keep up its own same driver or downdriver or stable guide without escaping from.
→ It would be interesting if people could predict how much an aliquot sequence can rise or fall.
→ It would be good if an aliquot sequence could link all the open end aliquot sequences below a certain limit.
→ Aliquot sequences tend to find out a level in range to rise to or to fall to.
→ It would be interesting if people could estimate the probability of a certain aliquot sequence that is being getting terminated.
→ It would be interesting if people could consider the probability of a certain aliquot sequence that is being picking up a driver or a downdriver or a stable guide.

Not willing to edit my post any more!

NFS@Home was not active during the year 2014? Factored only one number, namely 3,766+ it seems so! Thus, may be that it was spending out some time idle or taking a break or factoring Fibonacci numbers or Lucas numbers?
Is it possible to get mail notifications from FactorDB if any progress is being made upon any aliquot sequences that I subscribe to? Oops, not to have my inbox getting inundated! However I would enjoy having my inbox getting flooded after having done to other people the same thing!

My user name: Raman, my nickname: Mr. Tuch, my blurb under my user name: Guess what? [strike]caseztuchz[/strike]! [spoiler]OCD Guy[/spoiler]!
Spoiler tag is being the most appropriately used with in the puzzles forum only certainly!

Texas State HPC has reserved that 2,2158M c193 for over six years of time without having it as being getting completed off.
May be that some one else or may be that NFS@Home or @ Ryan Propper could be able to - consider it up - up over!
Good work by @ Paul Zimmermann to finish off that Texas State HPC reserved 7,347+ c188 by using ECM curves running immediately, executing out, finishing off with in that composite number, which would have otherwise got no progress at all - during having been a prime number cofactor!

It would be interesting if people could think about that Euler's Totient / Euler's Phi sequence, but not only, let not alone Sum of Factors / Restricted Divisor Function / Aliquot Sequence, namely - not - not!

Should have been - or
Must have been - or

It is being a good idea to replace all the white colour text font by using attachment tags / spoiler tags / strike through tags.

[QUOTE=Raman;442221]
[url=http://www.poodwaddle.com/worldclock.swf]Like India population is following China population. When ever do you think that India's population will overtake China's population?[/url]
[/QUOTE]

[spoiler]Any way, by that do[/spoiler]
[spoiler]By the way, by that do[/spoiler]
[spoiler]On the other hand, by that do[/spoiler]
Do you think that this incident will eventually happen up?

Right now aliquot sequence 314718 is at iteration 16744 and counting. Right now aliquot sequence 4788 is at iteration 10284 and counting.

FactorDB was down for me from Monday 12 September 2016 8:30 am IST (Indian Standard Time - GMT +0530) to Monday 12 September 2016 12:00 noon IST (Indian Standard Time - GMT +0530).

 science_man_88 2016-09-14 19:38

[QUOTE=Raman;442554]
Is 2 the only way of going down by means of a downdriver. Are there no other ways? What about going down by stable guides?
There are many different drivers (perfect numbers - all only known to be even). But there is only one downdriver!
Are there any drivers for odd numbers? Is that statement equivalent to asking for existence of odd perfect numbers?
Any way that an aliquot sequence parity change will occur only for squares or twice squares. They are the only way to terminate an aliquot sequence. Isn't it?

What is the difference between 2[sup]3[/sup].3 driver and 2[sup]3[/sup].3.5 driver? Or is even 2[sup]3[/sup].5 being a driver?

It would be interesting if people could think about that Euler's Totient / Euler's Phi sequence, but not only, let not alone Sum of Factors / Restricted Divisor Function / Aliquot Sequence, namely - not - not!

[/QUOTE]

you can explain odd squares changing simply be starting with odd+odd=even then noticing that odd*odd=odd then following the fact that for the prime factorization of a number N = p^i * p^j *p^k will result in (i+1)*(j+1)*(k+1) divisors and noting that if each exponent is even all the values will be odd in the product leading to an odd number of divisors that taking N away ( one of the divisors as calculated) you get an even number of odd proper divisors leading to an even sum. the reason both 2^3*3 and 2^3*5 work as drivers is that to drive a sequence the numbers d ( 3 or 5 in this case) must divide sigma(2^n) for some value n ( n=3 in this case) as sigma(2^3) =1+2+4+8 = 15 =3*5 d can be either 3 or 5 or 15.

 firejuggler 2016-09-14 23:08

i10305 : 2^2*3*31

 ryanp 2016-09-15 01:24

And the 3 is gone again, at i10317...

 Dubslow 2016-09-15 05:22

i10343 was a close call with 2^3 * P, but alas P was 15 mod 32 (i.e. [$]\tau(P) = 4[/$], not 1).

 science_man_88 2016-09-15 16:04

2^3*3^2* values at i10412

 LaurV 2016-09-16 03:21

[QUOTE=science_man_88;442646]2^3*3^2* values at i10412[/QUOTE]
That doesn't matter much, it still needs to get all the other factors to be 1 (mod 4), or get down to a single prime. The 3s come and go, but never disappear completely, unless the power of 2 changes, see my former posts in this thread.

As I know you are a "pari guy", try to compare in pari/gp:

[CODE]
gp > d=10; cnt=0; for(i=1,10^4,n=2^3*3^(random(5)+1)*(6*random(10^d)+(2*random(2)-1)); s=(sigma(n)-n); if([COLOR=Red]s%2^3!=0 || s%2^5==0[/COLOR],cnt++; print(n"\t"factorint(n)"\t\t"factorint(s))); printf("...%d / %d (%7.4f%%)... %c",cnt,i,100.0*cnt/i,13))
[/CODE]with

[CODE]
gp > d=10; cnt=0; for(i=1,10^4,n=2^3*3^(random(5)+1)*(6*random(10^d)+(2*random(2)-1)); s=(sigma(n)-n); if([COLOR=Red]s%3!=0[/COLOR],cnt++; print(n"\t"factorint(n)"\t\t"factorint(s))); printf("...%d / %d (%7.4f%%)... %c",cnt,i,100.0*cnt/i,13))
[/CODE](edit: differences[COLOR=Red] marked in red[/COLOR]; make your command prompt box 180 characters wide, or so; right click on its header, properties).

Play with it. Change d (make it 20, 30, 40). Change the powers of 2 and 3 to see how other drivers behave. These things can all be proved with modular calculus too. I did that. But I know you like pari. :razz:

edit: now after running the pari part myself, I see that some "rare" cases are possible too, where the power of 3 being even might help. I didn't consider those in my calculus :blush:
But those cases are rare, and it still needs to get the F's square-free part to a single prime, but in this case it can be any prime, not necessarily 1 mod 4.

[CODE]
2594944565640 [2, 3; 3, 2; 5, 1; [COLOR=Red]7, 2[/COLOR]; 147105701, 1] [2, 2; 3, 2; 5, 1; 487, 1; 1741, 1; 47279, 1]
3069904117464 [2, 3; 3, 2; 426375571[COLOR=Red]87[/COLOR], 1] [2, 2; 3, 1; 112213, 1; 3894691, 1]
3759740514936 [2, 3; 3, 2; 7, 2; 10656860[COLOR=Red]87[/COLOR], 1] [2, 5; 3, 2; 53, 1; 11497, 1; 46073, 1]
14550012257112 [2, 3; 3,[COLOR=Red] 4[/COLOR]; 224537226[COLOR=Red]19[/COLOR], 1] [2, 2; 3, 1; 19, 1; 53, 1; 23557, 1; 92051, 1]
[/CODE]

 LaurV 2016-09-17 06:31

[STRIKE]wow... i10498 is 2^3*3^4*[COLOR=Red]13[/COLOR]*...(p7)[COLOR=Red]53[/COLOR]*...(c155)[COLOR=Red]29[/COLOR], all are 1 (mod 4), so if the "...29" composite splits in two (or more) primes which are all 1 (mod 4), we may lose the 2^3 and go down again![/STRIKE]
Scratch that! I am stupid. Wrong counting. I forgot that if you split a composite in two, you get TWO primes, haha.

so, 2^3*3^4*(3 primes, all 1 (mod 4)) - loses the 2^3
but 2^3*3^4*(4 primes, all 1 (mod 4)) - can not lose the 2^3.

Grrr...

Next time...

 Batalov 2016-09-17 06:35

[QUOTE=LaurV;442819]wow... i10498 is 2^3*3^4*[COLOR=Red]13[/COLOR]*...(p7)[COLOR=Red]53[/COLOR]*...(c155)[COLOR=Red]29[/COLOR], all are 1 (mod 4), so if the "...29" composite splits in two (or more) primes which are all 1 (mod 4), we may lose the 2^3 and go down again![/QUOTE]
ORLY?
[SPOILER]four factors rake in (at least) 2^4, so 2^3 will stay[/SPOILER]

 Dubslow 2016-09-17 06:37

[QUOTE=LaurV;442819]wow... i10498 is 2^3*3^4*[COLOR=Red]13[/COLOR]*...(p7)[COLOR=Red]53[/COLOR]*...(c155)[COLOR=Red]29[/COLOR], all are 1 (mod 4), so if the "...29" composite splits in two (or more) primes which are all 1 (mod 4), we may lose the 2^3 and go down again![/QUOTE]

Hate to burst your bubble, but that aint true. With the power of two being 3, there can be at most 3 powers of two contributed by the sigma of the other non-square primes. Since there are at least 4 other primes, that's at least 4 powers of two, so no mutation.

And even if was 2 or 3 powers of 2, it would mutate to 2^2 (first case) or something higher than 2^3 (second case). And even if it was 2^3 * 3^(2k) * p for some k and prime p == 1 mod 4, it would evolve directly to 2*3 -- which would be even worse.

We have to hope for a mutation to something other than D1 before we can lose the 3 and hope for the downdriver. In the meantime, the size keeps increasing as usual...

 LaurV 2016-09-17 06:51

You guys won't spare me at all.. hehe.
Fortunately I realized before reading your replies, and I edited the post.
I don't know why I was silly, I had my coffee today, so I should not be silly... Or... is that from the coffee? Maybe I need to change the brand... :razz:

edit: @Dubslow: 2^3*3^4*(3 primes, all 1 mod 4) can mutate in everything else with a [U]higher power of 2[/U] only, which is [U]good[/U] for us. It can not mutate into a lower power of 2.

 Dubslow 2016-09-17 09:17

[QUOTE=LaurV;442823]
edit: @Dubslow: 2^3*3^4*(3 primes, all 1 mod 4) can mutate in everything else with a [U]higher power of 2[/U] only, which is [U]good[/U] for us. It can not mutate into a lower power of 2.[/QUOTE]

Yes, that's the second case of my second paragraph. The first case is two or fewer primes. Two primes is fine, one prime leads to D1 -- not good (assuming all are 1 mod 4 of course).

 ChristianB 2016-09-17 12:06

I updated [url]https://www.rechenkraft.net/aliquot/[/url] especially [url]https://www.rechenkraft.net/aliquot/intro-analysis.html[/url] with the current version from Dubslow which should now show the math formulas over https too.

Edit: I stumbled upon the request by accident. It's usually better to send me a PM because I don't follow all threads all the time.

 ryanp 2016-09-18 15:51

c166 blocker at i10526 is GNFS'ing, ETA ~8 hours.

 Raman 2016-09-18 16:56

[QUOTE=LaurV;442823]
edit: @Dubslow: 2^3*3^4*(3 primes, all 1 mod 4) can mutate in everything else with a [U]higher power of 2[/U] only, which is [U]good[/U] for us. It can not mutate into a lower power of 2.[/QUOTE]

Dude, why is it good for you if the power of 2 with in Aliquot Sequence of 4788 mutates into a higher power?

To be considering up on over out up:
It is being the case that:

Aliquot Sequence computation is being a deterministic algorithm, not a randomized algorithm at all.

You cannot get it into taking it into the way what ever that you want it up out.
You will have to obey where ever that it goes up out.

[code]
SNFS cannot be used against Aliquot Sequence computation, but only - let alone GNFS.
How efficient, effective do you think that Coppersmith SNFS - Factorization Factory variant it is being when ever compared in to with in common man's SNFS computation?
Efficient = effective? Skipping it that up out. Snipping it that up out. What ever does Cf. do mean [COLOR="PaleTurquoise"]by meaning it that up out? - does for any way - do in to with in it that up out! up on over in to.[/COLOR]
* sign is being Glossary of Data - Data Under Preparation - Less Important Data - Off Topic Data - what ever references [COLOR="PaleTurquoise"]in to with in it that up out! - does for any way - do in to with in it that up out! up on over in to.[/COLOR]

Mersenne Forum could have post version history - saved storage memory tags.
Not willing to edit my post any more!
Xyzzy's name not appearing at thread's bottom if when ever he does view a given fixed post for any way [COLOR="PaleTurquoise"]- do in to with in it that up out. up on over in to. - does for any way - do in to with in it that up out! up on over in to.[/COLOR]
My own previous posts attachment limits just approaching 128 KB limits for text editor case files confidentially - just below threshold limits for text editor case files confidentially [COLOR="PaleTurquoise"]- does for any way - do in to with in it that up out! up on over in to. - does for any way - do in to with in it that up out! up on over in to.[/COLOR]

Missing out with in some thing?

Good idea in to keeping it that up out lastly! up on over in to.
It is being a good idea to replace all the white colour text font by using attachment tags / spoiler tags / strike through tags.
It is being a good idea to replace all the white colour text font by using attachment tags / spoiler tags / strike through tags / quote tags / code tags / quote tags / code tags.
[/code]

 Batalov 2016-09-18 17:13

[QUOTE=Raman;442907][QUOTE=LaurV;442823]edit: @Dubslow: 2^3*3^4*(3 primes, all 1 mod 4) can mutate in everything else with a [U]higher power of 2[/U] only, which is [U]good[/U] for us. It can not mutate into a lower power of 2.[/QUOTE]Dude, why is it good for you if the power of 2 with in Aliquot Sequence of 4788 mutates into a higher power?
[/QUOTE]
LaurV clearly described the case where sequence [U]escapes[/U] 2^3*3 driver.
"2^3*3^4*(3 primes, all 1 mod 4)" --> 2^n*... where n>=4.
"why is it good for you?"
Because the sequence escapes 2^3*3 driver. Isn't it obvious?

He didn't say "any mutation in everything else with a [U]higher power of 2[/U] is [U]good[/U] for us".

Also, stop posting one reasonable (even if wrong) sentence followed by half a page of pure garbage. You've been warned. Garbage will be removed or the whole post will be [URL="http://mersenneforum.org/showthread.php?t=20588"]moved into trash bin[/URL]. People's eyes bleed when reading your posts.

 Raman 2016-09-19 15:16

[QUOTE=Batalov;442908]LaurV clearly described the case where sequence [U]escapes[/U] 2^3*3 driver.
"2^3*3^4*(3 primes, all 1 mod 4)" --> 2^n*... where n>=4.
"why is it good for you?"
Because the sequence escapes 2^3*3 driver. Isn't it obvious?

He didn't say "any mutation in everything else with a [U]higher power of 2[/U] is [U]good[/U] for us".

Also, stop posting one reasonable (even if wrong) sentence followed by half a page of pure garbage. You've been warned. Garbage will be removed or the whole post will be [URL="http://mersenneforum.org/showthread.php?t=20588"]moved into trash bin[/URL]. People's eyes bleed when reading your posts.[/QUOTE]

What I meant was that it is not in people's hands if an Aliquot sequence computation can lose a driver or gain a downdriver. What is good for people if an Aliquot sequence computation terminates or shoots up. One should always go to the number wherever the Aliquot sequence computation leads to.

You cannot parallelize computation of Aliquot sequences, other than GNFS sieving. Ordinary SNFS or Coppersmith SNFS - Factorization Factory variant cannot be used for Aliquot sequences, but only. Let alone GNFS. You cannot know the next number without completing the current iteration. It is not possible to do computation of Aliquot sequence by treating a semi-prime number as a prime number.

2014: Computation of Aliquot Sequence 4788 / 314718 had been paused?

Closest background colour for mersenne forum posts to hide text is being white. Inside quote and code tags, it is being pale turquoise. Junk attachments are being renamed in to meaningful phrase references - Glossary of Data, Data Under Preparation, Less Important Data, Off Topic Data. No garbage available at all. I know what is being garbage. Garbage should make out meaningless words. All my words with in my latest mersenne forum posts does make out meanings.

Zero views for my latest two - pair of attachments! Why does it say 'Last fiddled with by' rather than instead of 'Last edited by' as it were being previously? What does the former and latter mean out exactly? With in the mersenne forum home page, is number of people viewing being some random number? Random = custom?

My post counts have been suddenly increased by over a 200 during the 2½ year period that I had been inactive. Why? Due to the opening of chess subforums, they have now been included into post counts as like that which they were not before?
Mersenne forum post counts right now shows out as prime factors or like some arbitrary base representation! What ever else new representation is being possible that ever which ever has been available a way a way? Sake purpose process case. Random = custom - beginning out?
Very much tempting enough to starting out for the running up executing out trying attempting admitting attending with in computation of an Aliquot Sequences starting out, beginning out from these numbers very much, right now!

 ryanp 2016-09-19 16:05

We'd better get a downdriver soon, or this one is gonna spiral out of GNFS range...

 Stargate38 2016-09-19 21:37

It currently has 2^4*3^x, so it has a chance of losing the 3.

 henryzz 2016-09-19 22:01

[QUOTE=Stargate38;443021]It currently has 2^4*3^x, so it has a chance of losing the 3.[/QUOTE]

Once that 3 is lost we have hope again. It has stuck around lately though.

 Dubslow 2016-09-20 08:10

Currently being 2^4 * 3^3 * p1 * p2 * C, where p1 and p2 are both 2 mod 9 and C is 1 mod 3, I surmise that (to first order, assuming the primes in C are random and that primes are random mod 3) there is a 1/2 chance of the 3 dropping a power to 3^2 on the next line, or a 1/2 chance of maintaining 3^3, depending on how the C splits (also assuming it is square free with two prime factors). (To put it in notation from my page, [$]\tau_3(p1) = 1[/$], [$]\tau_3(p2) = 1[/$], and [$]\tau_3(C)[/$] is either 0 or 2+ but not 1.)

 ryanp 2016-09-20 19:47

The 3 is gone, and we're on our way down again!

 henryzz 2016-09-20 19:52

[QUOTE=ryanp;443097]The 3 is gone, and we're on our way down again![/QUOTE]

We are currently on 2^3. It is possible to go down with this although slowly. It doesn't take that many small factors to make it rise. We need the power of two to drop. As we have just seen, once there are no 3s or small factors that hang around long term the power of 2 will change more often.

 science_man_88 2016-09-20 20:45

[QUOTE=henryzz;443098]We are currently on 2^3. It is possible to go down with this although slowly. It doesn't take that many small factors to make it rise. We need the power of two to drop. As we have just seen, once there are no 3s or small factors that hang around long term the power of 2 will change more often.[/QUOTE]

the sequence now has 2^2.

 Dubslow 2016-09-20 21:30

Cmon, lets see that 2^2*p (where p is 1 mod 4)!

 schickel 2016-09-21 02:57

Shouldn't the thread title be "Sequence 4788 Episode IV: A New Hope" now?

:razz:

 Raman 2016-09-21 15:10

[QUOTE=Dubslow;443109]Cmon, lets see that 2^2*p (where p is 1 mod 4)![/QUOTE]

A line factoring as 2[sup]2k[/sup].x[sup]2l[/sup].p, p is a prime number candidate congruent to 1 (mod 4), k ≥ 1, l ≥ 1, 3 does not divide x, would be also doing fine by me too!

Why does the computation of Aliquot sequence 4788 / 314718 / 16100 oscillate up and down a lot when ever compared to computation of other Aliquot sequences?

Aliquot sequence 314718 past iteration mark 17000.
Aliquot sequence 4788 / 16100 past iteration mark 10540.
Offset iteration difference 6460.

All of the human history till year 2012 to reach out first 175 digit limit.
Four years to reach the next peak.
Few days with in year 2016 to reach 175 digits next peak.
Congratulations!

 science_man_88 2016-09-21 17:18

it's now caught a 3 as well as 2^2 great it's starting to climb again.... okay yeah I figured out the 3 shouldn't have anything to do with the climb as it doesn't divide 7.

 Dubslow 2016-09-21 23:48

Three does do with the climb, it shoots the abundance (way) above 1. And it's fairly stable too.

 Raman 2016-09-22 10:55

[COLOR="Blue"]Taking a fresh look after I just woke up this morning!
Happy autumnal equinox! Why?[/COLOR]

[QUOTE=henryzz;442223]
[COLOR="DarkOrange"]Please stop using white text I didn't notice on my first read through. Next time you do it on this subforum it will be edited to be clearer.[/COLOR][/QUOTE]

[COLOR="blue"]That is just burden on your side!
I don't think that any of the senior moderators will do that dirty work!
Posts are locked once when they are posted!
Senior moderators will not interfere with posts of other people![/COLOR]

[QUOTE=Dubslow;443201][COLOR="Green"]Three does do with the climb, it shoots the abundance (way) above 1. And it's fairly stable too.[/COLOR][/QUOTE]

[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]

 Dubslow 2016-09-22 11:55

Well... technically it isn't white... maybe even overall less headache inducing than my aliquot pages :smile:

 henryzz 2016-09-22 15:42

[QUOTE=Dubslow;443221]Well... technically it isn't white... maybe even overall less headache inducing than my aliquot pages :smile:[/QUOTE]

I do wonder what is going through his head sometimes. The way he pushes boundaries is like a child.

 Dubslow 2016-09-22 21:56

Lost the 3! Plain 2^2.

 henryzz 2016-09-22 23:59

Hopefully we won't gain it again this time.

 flagrantflowers 2016-09-23 02:08

[QUOTE=henryzz;443234]I do wonder what is going through his head sometimes. The way he pushes boundaries is like a child.[/QUOTE]

Have you considered that this may be medical in nature (mental health). I think that this forum does a poor job at times of trying to engage with people who are obviously experiencing apophenia or other symptoms of psychosis. Numerology seems to be clinically significant in people who are experiencing psychosis. These 'insights' these people glean feel incredibly significant. I don't think that it is malicious, I'm sure that they are probably having a distorted/disordered time of things. Numbers and patterns can feel very engaging and comforting as the are regular and ordered and somewhat bring a sense of normalization.

I understand it is not the place of this forum to offer help to those who are experiencing symptoms of mental illness, but I think that we have an opportunity that we don't actively acknowledge. These posts on the forum are frequently disruptive, but how disrupted is the life of the individual to be engaging 'like a child' the way they do with our community?

I've used this forum for similar reasons when I was personally going through an acute psychotic episode. It's terrifying beyond anything I can describe and the ordered structure of numbers was incredibly comforting -- annoying to everyone here else because it was nonsense, but comforting to me. Some of the worst times I had during this period were when I would post something here I thought was useful and insightful -- it was not, it was pure numerology, pure gibberish -- only to be demeaned and dismissed by the community (their perspective was 100% correct and mine 100% distorted). I used this forum as a way to engage socially when I was unable to do so in daily life, to find a sense of belonging and talk about things I thought were a part of some deeper underlying beauty that resonated with my underlying condition.

I apologize for placing this in this thread, but I have been waiting for an example of this type of interaction. Mental illness and math based apophenia seem to go hand-in-hand, but ostracizing people, for something they feel is incredibly important, in this state can do serious harm. I do like when the forum has given disruptive members their own thread so they can meander, posting as much gibberish as they please.

There is no easy solution, I'm not even saying there is a solution or that anything should, or needs, to change. But I have not seen any conversation on here that has addressed this and the possible good, or harm, our community contributes in these situations.

Apologies again for posting in this inappropriate setting.

 Prime95 2016-09-23 02:31

[QUOTE=flagrantflowers;443267]
There is no easy solution, I'm not even saying there is a solution or that anything should, or needs, to change. But I have not seen any conversation on here that has addressed this and the possible good, or harm, our community contributes in these situations.

Apologies again for posting in this inappropriate setting.[/QUOTE]

No apologies necessary. Raman has well over a decade of this behavior. When he first appeared on the scene his condition was much, much, much worse. Many attempts were made to contact him, his parents, his school, etc. to get him some help. I assume this eventually succeeded as he was better for several years. At the risk of making a long-distance layman's diagnosis, I believe his medications need some adjusting. Hopefully, he will see to that.

 henryzz 2016-09-23 08:24

I had failed to consider that. Apologies Raman.

 Dubslow 2016-09-23 21:50

Gah!

 science_man_88 2016-09-23 22:00

[QUOTE=Dubslow;443330]Gah![/QUOTE]

11 iterations with a 3 again oh joy ...

 ATH 2016-09-24 00:55

Episode V: The Driver strikes back?

 axn 2016-09-24 03:00

[QUOTE=ATH;443346]Episode V: The Driver strikes back?[/QUOTE]

These are not the drivers you're looking for!

[2^2*3 is not a driver. 3 is independent of 2^2, and can come and go as it pleases without any change in the power of 2]

 ATH 2016-09-24 11:03

Yeah I saw on the intro analysis link. It should be a 7 with 2^2 for it to be a driver.

But maybe 3 does not know it's not supposed to be a driver, at least it has been "driving" it up higher for a while now (step 10565-10574 and 10586-10605).

 henryzz 2016-09-24 12:16

[QUOTE=ATH;443364]Yeah I saw on the intro analysis link. It should be a 7 with 2^2 for it to be a driver.

But maybe 3 does not know it's not supposed to be a driver, at least it has been "driving" it up higher for a while now (step 10565-10574 and 10586-10605).[/QUOTE]

It doesn't have to be a driver to go up consistently. 3 is an awkward case with 2^2 since it fixes the power of 2 to 2. Since [$]2^3|\sigma(2^2*3*x)[/$] and [$]2^2||2^2*3*x[/$], [$]2^2||\sigma(2^2*3*x)-2^2*3*x[/$]. x can be composite although gcd(x,6)=1.
The only way out of this is to change the power of 3.
The 3 can be eliminated if there are no factors of [$]x \equiv 2 mod 3[/$]. Each factor has a 50% chance of meeting this condition so the chance of losing the 3 is 1 in 2^a where a is the number of factors >3. As n gets larger on average the number of factors increases so it is less likely that the 3 will be lost.
The other alternative is to raise the power of the 3. This requires [$]\sigma(2^2*3*x)\equiv 2^2*3*x mod 9[/$]. Only one of the factors of x can be 2 mod 3 otherwise [$]9|\sigma(2^2*3*x)[/$]. I believe that there is a 50% chance of increasing the power of 3 if only one of the factors of x is 2 mod 3.
As you can see it isn't easy to loose the 3. We have been lucky so far.
[$]\sigma(2^2*3)=28 >24=2*(2^2*3)[/$] which means it will always rise while we have it.

 Batalov 2016-09-24 21:52

Argh. [URL="http://factordb.com/sequences.php?se=1&aq=4788&action=range&fr=10612&to=10613"]Close but no cigar[/URL].

 science_man_88 2016-09-25 00:19

[QUOTE=Batalov;443397]Argh. [URL="http://factordb.com/sequences.php?se=1&aq=4788&action=range&fr=10612&to=10613"]Close but no cigar[/URL].[/QUOTE]

now has 2^10 on the last iteration shown.

 Raman 2016-09-25 09:38

[QUOTE=science_man_88;443405]now has 2^10 on the last iteration shown.[/QUOTE]

Not only that out.

Did any one else certainly notice away with in this unusual similarity only up?

Aliquot Sequence 660 Iteration Number 971 = 2[sup]5[/sup] × 2069 × c193.

Aliquot Sequence 4788 Iteration Number 10616 = 2[sup]10[/sup] × 2069 × c175.

GCD(Aliquot Sequence 660 Iteration Number 971, Aliquot Sequence 4788 Iteration Number 10616) = What?

[QUOTE=Raman;442554]
@ Ryan Propper: Have you automated submissions to FactorDB even when you are asleep or outside? Great job!
Factoring a c120 in 2006 would take upto 5 days. In 2016, it is only taking 5 minutes. Or 10 minutes may be? Distributed computing is being going on automatically?
Perhaps you could help out with aliquot sequences of 552, 564, 660, 1512, 1992, 5250, 9120, 11040 besides of 4788 or of 314718.

<snip>

away out off up down my own - that ever which ever a way a way ever.
away out off up down my own - that ever which ever a way a way ever.
[/QUOTE]

Aliquot Sequence 4788 Iteration Number 10615: 2[sup]6[/sup] × 113 - hopefully - not - 2[sup]6[/sup] × 127!

Consecutive Prime Numbers 113 And 127!

 science_man_88 2016-09-25 13:19

[QUOTE=Raman;443433]Not only that out.

Did any one else certainly notice away with in this unusual similarity only up?

Aliquot Sequence 660 Iteration Number 971 = 2[sup]5[/sup] × 2069 × c193.

Aliquot Sequence 4788 Iteration Number 10616 = 2[sup]10[/sup] × 2069 × c175.

GCD(Aliquot Sequence 660 Iteration Number 971, Aliquot Sequence 4788 Iteration Number 10616) = What?

Aliquot Sequence 4788 Iteration Number 10615: 2[sup]6[/sup] × 113 - hopefully - not - 2[sup]6[/sup] × 127!

Consecutive Prime Numbers 113 And 127![/QUOTE]

using PARI/GP to try to find the gcd of the composites that haven't been factored
( after reducing it enough I tried gcd of the two last results and got 1. so [TEX]2^5\cdot2069[/TEX] edit: okay I used iteration 971 for 660 but the point is the same.

 ryanp 2016-09-27 17:10

Plodding along slowly. That 2^2 (and now 2^6) is hard to shake...

 schickel 2016-09-27 23:49

Amazing, I was just re-reading this thread that showed up in the "Similar threads" section, and 5 years ago we took ~2 weeks getting ready for a c172. Now we're having this happen:
[QUOTE=ryanp;432854 (on 4/30)]I'll handle the next C180 as well. :)[/QUOTE]

[QUOTE=ryanp;432959 (on 5/2)]OK. Next C180 is done:

[CODE]Mon May 2 13:11:50 2016 prp63 factor: 852828185224582024294795864244283338007551456801212478326125773
Mon May 2 13:11:50 2016 prp117 factor: 704131409692002901688336495667355706435373199424596628222345220771103914385162349468158740451400667823527752200435477[/CODE][/QUOTE]And

[QUOTE=ryanp;432991 (again, on 5/2)]Working on the C170 now.

[CODE]linear algebra completed 733239 of 5927756 dimensions (12.4%, ETA 10h 9m)[/CODE][/QUOTE]

[QUOTE=ryanp;433019 (on the next day, 5/3)]Awww. Could've had this one by ECM if I had tried harder...

[CODE]Tue May 3 08:55:24 2016 prp50 factor: 10343669734301041373937055383512171230895724608491
Tue May 3 08:55:24 2016 prp120 factor: 993739759889626805545135903141953923233113832717632966003694092677384994673520180146397186251215841015082033396156750101[/CODE][/QUOTE]We just need ryanp to get mad at [B]all[/B] the sequences!

Thanks for the yuuge assist, Ryan!!

:bow wave::bow wave:

 LaurV 2016-09-28 01:58

[QUOTE=schickel;443643]We just need ryanp to get mad at [B]all[/B] the sequences!
[/QUOTE]
Well... I already made a diabolic plan: I am waiting for this sequence to get to 202 digits (and hopefully with 2^3*3*5 driver :razz:), and then I will quote Ryan's post where he said that "he is pissed off" and "he will kill it", and I will ask "did you say something?" :grin:

(well, you all jinxed it that it will terminate, and it didn't, so I try the other way around... hehe)

 ryanp 2016-09-28 02:00

[QUOTE=LaurV;443650]Well... I already made a diabolic plan: I am waiting for this sequence to get to 202 digits (and hopefully with 2^3*3*5 driver :razz:), and then I will quote Ryan's post where he said that "he is pissed off" and "he will kill it", and I will ask "did you say something?" :grin:[/QUOTE]

I'll find a way... :smile:

 flagrantflowers 2016-09-28 03:44

[QUOTE=ryanp;443651]I'll find a way... :smile:[/QUOTE]

Just so that things are jinxed in a clearly defined way; what is your limit, in terms of factor size? C200? C210?

 legendarymudkip 2016-09-28 15:01

[QUOTE=flagrantflowers;443660]Just so that things are jinxed in a clearly defined way; what is your limit, in terms of factor size? C200? C210?[/QUOTE]

In September 2013, he factored RSA210. Given that was (almost exactly) 3 years ago, assuming a rate of Moore's Law at 18 months, he should be able to factor a C220 by now. From this it should be at least C220!.

 VBCurtis 2016-09-28 16:45

Such extrapolation works fine for the sieving step, but the hurdle to really big jobs is handling the matrix. Clusters may have gotten faster in 3 years, but cluster time likely hasn't gotten easier to find. RSA numbers are interesting enough to make a case for time on a nice cluster, but "I want to kill this %$^&%ing Aliqueit sequence" might not gain similar access. On the other hand, 64GB memory desktops are now available, and a year on one of those might solve a GNFS-215 matrix without a cluster. So, perhaps you're right about 215+, given enough patience.  RichD 2016-09-28 20:18 So is it safe to say, "You can't over-sieve your way to less memory requirements."  GP2 2016-09-28 21:28 [QUOTE=VBCurtis;443710]On the other hand, 64GB memory desktops are now available, and a year on one of those might solve a GNFS-215 matrix without a cluster. So, perhaps you're right about 215+, given enough patience.[/QUOTE] In the cloud you can get 61 GiB of memory on a 4-core Xeon @ 2.5 GHz at about$1000 for a year's worth of computing time at the cheapest current spot prices. You can get 244 GiB of memory on a 16-core for maybe three times that amount.

The drawback is, spot prices fluctuate, so much of the time the spot price would exceed what you'd want to pay. So a year's worth of computing time might be way longer in wall-clock time...

Edit: Google cloud prices are similar, 4-core 52 GB preemptible @ $73/month or 16-core 208 GB preemptible @$292/month, nonfluctuating prices and maybe a better chance of getting interrupted less often.

Are there any interesting problems that need humungous memory but considerably less than a year to solve?

 VBCurtis 2016-09-29 04:44

[QUOTE=RichD;443728]So is it safe to say, "You can't over-sieve your way to less memory requirements."[/QUOTE]

More sieving -> less memory, yes generally; but we're already assuming enough sieving for a density-140+ matrix. There is a diminishing-returns aspect to extra sieving, and a fuzzily-known bound where more sieving ceases to reduce matrix effort at all (or perhaps it was that a matrix won't even build if one over-over-sieves- that happens on smallish problems somewhat often if a script gets carried away).

 henryzz 2016-09-29 08:59

It is always possible to reduce the large prime bound. This should reduce the size of the matrix although it may significantly increase the sieving time.

 Batalov 2016-09-29 20:07

:direction:

:rolleyes:

 schickel 2016-09-29 21:25

[QUOTE=Batalov;443841]:rolleyes:[/QUOTE]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?

 henryzz 2016-09-29 21:46

[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]

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.

 flagrantflowers 2016-09-29 22:12

[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]

And then you need to hit the same again to reduce 2^2 down to 2? And this only works if the split is two factors?

 schickel 2016-09-29 22:57

[QUOTE=flagrantflowers;443855]And then you need to hit the same again to reduce 2^2 down to 2? And this only works if the split is two factors?[/QUOTE]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.

 flagrantflowers 2016-09-30 00:27

[QUOTE=schickel;443862]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]

Thank you, that is a very useful play-by-play. I can't be asked to learn the maths on, but am entertained by following the progress/thread. I mean, I'm try to not get too into numerology lately.

But if I'm understanding. Assuming it splits into two it can be either two 1 mod 4 factors or two 3 mod 4 factors; I would assume the 'odds' would be 1:1?

I'm try to learn.

 Dubslow 2016-09-30 07:12

[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]

:tu:

In general, flagrantflowers, if you write the line as:

2^b * (q1^k1 * q2^k2 * q3^k3)^2 * (p1 * p2 * p3)

...you can use the following rules to determine any possible mutations:

1) Ignore any primes whose power is even (like the q's above)

2) For the remaining primes, we'll assume they have no power, since getting a prime cubed (or higher) is extraordinarily unlikely (except for small primes like 3 or 5 or 7).

3) Look at the power of 2 that divides p+1 for each prime p.

4) If the sum of those powers is greater than b, no mutation will occur.

5) If the sum of those powers is exactly b, then the power b will increase (an upwards mutation).

6) If the sum of those powers is less than b, then the power will be reduced to this lower sum (a downwards mutation).

Since all primes are odd (except 2 of course), p+1 automatically is even, so each prime contributes at least 1 power of two. To a zeroth order approximation, half of primes+1 are divisible by two and not four, a quarter of all primes+1 are divisible by four but not eight, and an eighth of all primes are divisible by eight but not sixteen, etc.

So right now, we have 2^3 * C, where C has no small factors and is thus likely to split into two (large) factors p1 and p2 (or rather less likely, three factors p1 p2 and p3).

We know that C is 1 mod 4. Therefore, since 1*3 = 3 mod 4, we know that both of its factors are either both 1 or both 3 mod 4 -- but it can't be that one is 1 mod 4 and the other is 3 mod 4.

If they are both 1 mod 4, then we know that p1+1 and p2+1 are 2 mod 4, hence their power of two is precisely 1 (since they are not 0 mod 4, their power of two can't be 2 or more), for a total power of two of 2. Thus, 2 < b=3, so by rule 6) we would mutate from 2^3 * p1 * p2 to 2^2 * ...., and 2^2 is "better" in the sense that it's less abundant than 2^3, i.e. each line goes down more.

If they are both 3 mod 4, then p1+1 and p2+1 would be 0 mod 4, so they each have a power of two of at least 2, so the total power of two would be at least 4, which is more than b=3, so no mutation would occur.

To zeroth order approximation, it's about equally likely that they are both 1 or both 3 mod 4, so around a 50% chance this line mutates to 2^2 or stays at 2^3.

As schickel mentioned, with these rules you can see that the only way to get 2^1, as necessary for the downdriver, is to get a line 2^b * (primes to even power) * p, where p is 1 mod 4, so that the power of two in p+1 is only 1, whence by rule 6) we get 2^1.

For details on the [i]why[/i] of these rules, see my [URL="http://www.rechenkraft.net/aliquot/intro-analysis.html"]page here[/URL] (and let me know if something confuses you :smile:)

 henryzz 2016-09-30 09:57

Something that we have both ignored is that a prime can be 7 mod 8. This doesn't particularly affect this case since we are dealing with 2^3 with C = 1 mod 4. However, in other situations such as with C = 3 mod 4, it is necessary to consider mod 8 or above as is done at the top of Dubslow's last post.
It is also worth noting that 5) needs adjusting for primes other than 2. I think that the correct statement is:
5) If the sum of those powers is exactly b, then the power b will increase (an upwards mutation) with probability 1/(p-1).
This is due to N and sigma(N) having to match mod p^n(assuming we are mutating from p^(n-1) to p^n). We already know that neither are 0 mod p^n but are 0 mod p^(n-1) so the probability is 1/(p-1) rather than 1/p. This is just 1 for p=2 which makes the original statement correct for p=2.

 Dubslow 2016-09-30 10:47

I was only talking about the power of 2, yes.

 ryanp 2016-09-30 14:36

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

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

 axn 2016-09-30 14:51

[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 :-(

 henryzz 2016-09-30 17:43

[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.

 Prime95 2016-10-03 16:36

[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:

 Raman 2016-10-03 19:37

[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]

 Dubslow 2016-10-06 07:02

Gained a 3 despite 6 prime factors.

 henryzz 2016-10-06 11:20

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

Ouch!!

 Raman 2016-10-07 07:48

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!

 firejuggler 2016-10-07 12:29

[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.

 Raman 2016-10-07 13:13

[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

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