Nice split! ;)

I think this is one of the largest gnfs factorizations ever done, isn't it? (beside of the RSA numbers, and few other big thingies who in fact involved a lot of [U]collective[/U] effort), this may be one of the largest gnfs factorizations succeeded by an individual (!?).
The actual C177 looks like a piece of cake, in comparison. Congrats! 
The c177 staunchly resisted ECM, but is in GNFS now. ETA to factor: sometime tomorrow.

c187 (15465249...) LA is in progress now:
[CODE]Tue Aug 16 16:43:13 2016 commencing linear algebra Tue Aug 16 16:43:15 2016 read 12431234 cycles Tue Aug 16 16:43:55 2016 cycles contain 36336667 unique relations Tue Aug 16 17:23:44 2016 read 36336667 relations Tue Aug 16 17:25:29 2016 using 20 quadratic characters above 2147483490 Tue Aug 16 17:29:05 2016 building initial matrix Tue Aug 16 17:59:30 2016 memory use: 5410.3 MB Tue Aug 16 17:59:44 2016 read 12431234 cycles Tue Aug 16 17:59:48 2016 matrix is 12431056 x 12431234 (6478.2 MB) with weight 1969154012 (158.40/col) Tue Aug 16 17:59:48 2016 sparse part has weight 1549035458 (124.61/col) Tue Aug 16 18:04:58 2016 filtering completed in 2 passes Tue Aug 16 18:05:04 2016 matrix is 12430700 x 12430878 (6478.1 MB) with weight 1969133778 (158.41/col) Tue Aug 16 18:05:04 2016 sparse part has weight 1549027613 (124.61/col) Tue Aug 16 18:07:15 2016 matrix starts at (0, 0) Tue Aug 16 18:07:19 2016 matrix is 12430700 x 12430878 (6478.1 MB) with weight 1969133778 (158.41/col) Tue Aug 16 18:07:19 2016 sparse part has weight 1549027613 (124.61/col) Tue Aug 16 18:07:19 2016 saving the first 48 matrix rows for later Tue Aug 16 18:07:23 2016 matrix includes 64 packed rows Tue Aug 16 18:07:25 2016 matrix is 12430652 x 12430878 (6351.3 MB) with weight 1714974825 (137.96/col) Tue Aug 16 18:07:25 2016 sparse part has weight 1540650919 (123.94/col) Tue Aug 16 18:07:25 2016 using block size 8192 and superblock size 4423680 for processor cache size 46080 kB Tue Aug 16 18:09:06 2016 commencing Lanczos iteration (48 threads) Tue Aug 16 18:09:06 2016 memory use: 5371.8 MB Tue Aug 16 18:09:56 2016 linear algebra at 0.0%, ETA 109h 1m Tue Aug 16 18:10:12 2016 checkpointing every 120000 dimensions[/CODE] 
Great work. Maybe we'll see this sequence soon in the "terminations and merges" thread :) Should we start a poll as to how long this downdriver lasts?

[QUOTE=Prime95;440178]Great work. Maybe we'll see this sequence soon in the "terminations and merges" thread :) Should we start a poll as to how long this downdriver lasts?[/QUOTE]
How many ways can an even sequence end? Perfect number, amicable pair/X, ?zero? 
[QUOTE=flagrantflowers;440179]How many ways can an even sequence end?
Perfect number, amicable pair/X, ?zero?[/QUOTE] more like sociable cycle ,1, or it might not. by sociable cycle I mean N numbers that are the proper divisor sum of each other with N=1.. infinity. 
[QUOTE=science_man_88;440181]more like sociable cycle ,1, or it might not. by sociable cycle I mean N numbers that are the proper divisor sum of each other with N=1.. infinity.[/QUOTE]
So odd sequences cannot end in one and even sequences cannot end in zero? 
All aliquot sequences end in some sort of generalized aliquot cycle. A cycle of length zero is just a prime number:
some n > p > 1, and the aliquot function of 1 is not defined (since 1 has no proper divisors). A cycle of length one is a perfect number: some n > P > P > P > ... A cycle of length two is an amicable pair: some n > P > Q > P > Q > P > ... Sociable numbers are those who form a cycle of length 3 or more: some n > P > Q > R > P > Q > R > ... or some n > P > Q > R > S > P > Q > R > S Any aliquot sequence that doesn't end in some sort of cycle (length 0 or more) must, by necessity, have infinitely many terms (which thus by necessity are arbitrarily large). We don't know if any such sequences exist, and answering that question is the point of this forum. 
[edit: crosspost, I was replying to the guy on the former page, idea is that no sequence ends in zero]
You have to sum the proper divisors and 1, to get to the next term. For example if your term is 10, you sum 1+2+5=8, your next term is 8. (10 is not a proper divisor of 10). No sequence ends in zero, as you at least have to add 1 to the sum. When you get to a prime (or start with a prime), then 1 is the only thing you add, as there are no other proper divisors, and the sequence ends in 1. All sequences end in 1 or in a sociable chain [edit: what is called above "generalized aliquot cycle"]. That is because I believe the famous conjecture is true. If you believe that the famous conjecture is false, then some sequences go to infinity. 
c187 finished a bit earlier than expected:
[CODE]Fri Aug 19 23:59:24 2016 prp83 factor: 81686847473504758006224548165679412739660168673296296131927624175642936458151919111 Fri Aug 19 23:59:24 2016 prp104 factor: 18932362263513586512787823670585462560896906573207638060424099856662477952354307120700802524336003290743[/CODE] 
Congrats on losing the first (hardest) 10 digits from the top, Ryan! No more c19Xies in the near future.
10 done, 190 to go! :rolleyes: 
Just trimmed another digit off... :smile:

don't count the fish!

The c188 blocker at index 5358 has thus far resisted ECM, firing off GNFS...
[CODE]linear algebra completed 15687 of 14408801 dimensions (0.1%, ETA 85h53m) [/CODE] 
c188 done:
[CODE]Thu Aug 25 17:26:51 2016 p64 factor: 3876216633055061419501730313594001806409661150743884472980900647 Thu Aug 25 17:26:51 2016 p124 factor: 8032181759747525191246997851763484249697240356342526142842271225485394373575011306918512486599998419192876604004337942300257 [/CODE] 
Congrats on shedding the next 10 digits!
[SPOILER](the intervals between these congrats will soon be shorter, and shorter, and shorter... For every ten digits, the gnfs jobs become 4 times faster! The only remaining ingredient is luck. So, good luck on the ride down!)[/SPOILER] 
[QUOTE=Batalov;441282]re:spoiler[/QUOTE]
No congrats under 170 digits. After a C196 gnfs, those are piece of cake for him. So, no congrats for him anymore... :razz: edit: only one congrats, when it terminates! yeaaaa! 
I've decreased the update interval for 4788 on [url]http://www.rechenkraft.net/aliquot/AllSeq.html[/url] to twice a day for those who don't follow it directly on factordb.com.

Still on the way down, and almost at 170 digits now! :)

Woohoo  below 170 digits now!
P.S. I hit refresh on the factordb page several times a day watching this number go lower and lower :smile: 
shhhh! you two are going to spoil it! :rant:
(edit: PS: same here) 
Now below 160 digits!
And yes, I am also checking the status multiple times a day. :smile: 
[QUOTE=rajula;442007]Now below 160 digits!
And yes, I am also checking the status multiple times a day. :smile:[/QUOTE] It is becoming quite the addiction:smile: 
There are two things that one never tires of watching:
swimming women and other people working. 
Wrote a small program to help automate my main factoring script and posts to factordb.com, so the updates should come a bit more quickly now. :)

Er... you mean like aliqueit? lol.

And so we are below 150. Soon this will become fast, and hopefully until the end!

:drama:
Is the number of values less than a million available that are known to be attached to the 4788 genealogy tree? I see from [URL]http://www.mersenneforum.org/showpost.php?p=286377&postcount=65[/URL] that 1022 sequences are in that tree but I think those are the sequences that always stay below a million. I think there are more that sneak above a million and then come back down or grow much larger and come back down like 314718. I guess I'm wondering about getting a sense of the odds, if the sequence falls below a million again like 314718 did, that it will intersect itself to make a sociable cycle with thousands of numbers in it. 
[QUOTE=mshelikoff;442180]:drama:
Is the number of values less than a million available that are known to be attached to the 4788 genealogy tree? I see from [URL]http://www.mersenneforum.org/showpost.php?p=286377&postcount=65[/URL] that 1022 sequences are in that tree but I think those are the sequences that always stay below a million. I think there are more that sneak above a million and then come back down or grow much larger and come back down like 314718. I guess I'm wondering about getting a sense of the odds, if the sequence falls below a million again like 314718 did, that it will intersect itself to make a sociable cycle with thousands of numbers in it.[/QUOTE] well using: [url]http://factordb.com/aliquot.php?type=1&aq=4788&big=1[/url] to figure out the values in the sequence as calculated form 4788 I got: Checked 0 4 (show) 4788 = 2^2 · 3^2 · 7 · 19 Checked 1 4 (show) 9772 = 2^2 · 7 · 349 Checked 2 4 (show) 9828 = 2^2 · 3^3 · 7 · 13 Checked 3 5 (show) 21532 = 2^2 · 7 · 769 Checked 4 5 (show) 21588 = 2^2 · 3 · 7 · 257 Checked 5 5 (show) 36204 = 2^2 · 3 · 7 · 431 Checked 6 5 (show) 60564 = 2^2 · 3 · 7^2 · 103 Checked 7 6 (show) 105420 = 2^2 · 3 · 5 · 7 · 251 Checked 8 6 (show) 233268 = 2^2 · 3 · 7 · 2777 Checked 9 6 (show) 389004 = 2^2 · 3 · 7 · 11 · 421 Checked 10 6 (show) 745332 = 2^2 · 3 · 7 · 19 · 467 however for intersections with other sequences would take the merge thread and a few more at least probably. Checked 0 4 (show) 8092 = 2^2 · 7 · 17^2 Checked 1 4 (show) 9100 = 2^2 · 5^2 · 7 · 13 Checked 2 5 (show) 15204 = 2^2 · 3 · 7 · 181 Checked 3 5 (show) 25564 = 2^2 · 7 · 11 · 83 Checked 4 5 (show) 30884 = 2^2 · 7 · 1103 Checked 5 5 (show) 30940 = 2^2 · 5 · 7 · 13 · 17 Checked 6 5 (show) 53732 = 2^2 · 7 · 19 · 101 Checked 7 5 (show) 60508 = 2^2 · 7 · 2161 Checked 1827 6 (show) 543242 = 2 · 7 · 38803 Checked 1828 6 (show) 388054 = 2 · 194027 Checked 1829 6 (show) 194030 = 2 · 5 · 19403 Checked 1830 6 (show) 155242 = 2 · 77621 Checked 1831 5 (show) 77624 = 2^3 · 31 · 313 Checked 1832 5 (show) 73096 = 2^3 · 9137 Checked 1833 5 (show) 63974 = 2 · 29 · 1103 Checked 1834 5 (show) 35386 = 2 · 13 · 1361 Checked 1835 5 (show) 21818 = 2 · 10909 Checked 1836 5 (show) 10912 = 2^5 · 11 · 31 Checked 1837 5 (show) 13280 = 2^5 · 5 · 83 Checked 1838 5 (show) 18472 = 2^3 · 2309 Checked 1839 5 (show) 16178 = 2 · 8089 came up after searching a bit of the merge and termination thread. 
The link in my last post included a link to the 4788 staysunderamillion genealogy at [url]http://www.rieselprime.de/Others/4788_1000000.png[/url] .

[QUOTE=mshelikoff;442185]The link in my last post included a link to the 4788 staysunderamillion genealogy at [url]http://www.rieselprime.de/Others/4788_1000000.png[/url] .[/QUOTE]
okay well what else do you want ? the last sequence I quoted was 1001880 but it eventually goes under 1000000 and hits 8092 which merges with 4788 
Nevermind.

[QUOTE=rajula;442171]And so we are below 150. Soon this will become fast, and hopefully until the end![/QUOTE]
at 140 digits now 
Aliquot Sequence 4788 is now at 135 digits... which means that if it will go far low then its progress can be made faster.
Only one day taken  Aliquot Sequence 4788  150 digits  140 digits. Aliquot Sequence 933436 is the longest computed sequence at 12559 iterations. Aliquot Sequence 314718 is following it at 12092 iterations. Seems that it will overtake aliquot sequence 933436 sooner. [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] Aliquot Sequence 11040 had been under downdriver for a while, but it is now stable, without a driver, but having a guide, under the influence of it. Who is pursuing aliquot sequence 11040? Of the open end aliquot sequences below 10000, the aliquot sequences 552, 564, 660, 1512, 1992, 4788, 5250, 9120 are stable, not under (without) the influence of a driver at all. Aliquot Sequence 4788 was once had been stable, without (not under) the influence of a driver at all, but it has now acquired the downdriver, and now it has fallen down up from 200 digits to 135 digits, and then counting if... Aliquot Sequence 4788 acquired downdriver all at 175 digits and then 200 digits. Are there being no new aliquot sequence terminations since Thursday, 23 April 2015? Or Wolfgang Creyaufmüller went "Missing In Action", for a while, just like Wilfred Keller, for a while? What is the probability that the aliquot sequence 4788 at this moment would terminate? It would be very interesting if it would catch into a very big cycle  16100 loop back again. [color="white"]ever ever ever ever ever ever ever ever ever ever ever ever ever ever ever ever ever[/color] [QUOTE=ryanp;433367] The c196 on index 5283 (1742276725...) looks to be a tough nut to crack. I've already run 10K curves at B1=26e7 and another 10K at B1=76e8, but no luck so far. Everyone else is encouraged to run curves too... I will be starting GNFS soon. [/QUOTE] [color="white"]a an the this that that which these those is was are were who why whom whose how where what which when ever[/color] How many resources do you have? You are able to crack a p196 very quickly! . is frequently often misused for multiplication, although it has got a special meaning with in  the decimal point. Are you working out with in university resources? What university? What post / position are you with in? Well, let me know very well about this thing around  variably. [color="white"]think stuff style item step idea[/color] [QUOTE=henryzz;433373] 1 mod 4 so 50% chance of raising the power of 2 [/QUOTE] [color="white"]that ever which ever a way a way[/color] If a line factors as 2[sup]2[/sup].c196 where c196 is congruent to 1 (mod 4), and then 50% of raising the power of 2 only if the 196 digit composite cofactor has got only exactly two prime factors. How do you say with certainty before hand that the 196 digit composite cofactor has got only exactly two prime factors? If given the amount of ECM curves done, and then it should have been evident. By the way, if a line factors as 2[sup]m[/sup].x, with in the next iteration, power of 2 will become 1 if x is a prime congruent to 1 (mod 4) Let x = p[sub]1[/sub][sup]a[sub]1[/sub][/sup].p[sub]2[/sub][sup]a[sub]2[/sub][/sup]...p[sub]n[/sub][sup]a[sub]n[/sub][/sup] Let t be the highest power of 2 dividing (1+p[sub]1[/sub]+p[sub]1[/sub][sup]2[/sup]+...+p[sub]1[/sub][sup]a[sub]1[/sub][/sup]).(1+p[sub]2[/sub]+p[sub]2[/sub][sup]2[/sup]+...+p[sub]2[/sub][sup]a[sub]2[/sub][/sup])...(1+p[sub]n[/sub]+p[sub]n[/sub][sup]2[/sup]+...+p[sub]n[/sub][sup]a[sub]n[/sub][/sup]) = ((p[sub]1[/sub][sup]a[sub]1[/sub]+1[/sup]1)/(p[sub]1[/sub]1)).((p[sub]2[/sub][sup]a[sub]2[/sub]+1[/sup]1)/(p[sub]2[/sub]1))...((p[sub]n[/sub][sup]a[sub]n[/sub]+1[/sup]1)/(p[sub]n[/sub]1)) With in the next iteration, power of 2 will decrease to t if t < m. With in the next iteration, power of 2 will increase if t = m. With in the next iteration, power of 2 will remain same at m if t > m. [color="white"]@ ramshanker  How old are you at all ??[/color] [QUOTE=unconnected;435000]...and now we have squared 3 :no: [/QUOTE] [color="white"]that which they are being as like into that which it is being as like into[/color] 3 is not a component of driver, only let alone a stable guide as long as the power of 2 stays / remains all at an even number. No need to worry about it at all. 3 is preserved due to the prime factors other than 2, which are congruent to 2 (mod 3), that which they are being raised to an odd power, as long as the power of 2 is staying / remaining all at an even number. 3 will disappear if there are no prime factors congruent to 2 (mod 3), that which they are being raised to an odd power, and then if they can reappear if the same condition occurs back again. [color="white"]and then if only let alone assuming c has got two prime factors and then if only let alone assuming c has got two prime factors[/color] The only way to escape the 2.3 driver is to have a line factoring / factorizing as like into 2.3[sup]2m[/sup].p where by m ≥ 1, p is being a prime number congruent to 1 (mod 4). [color="white"]and then if only let alone assuming p has got one prime factor and then if only let alone assuming p has got one prime factor[/color] On the other hand, lines factoring / factorizing as like into 2[sup]m[/sup].3[sup]k[/sup].x where by m ≥ 2, k ≥ 1 3 can be escaped, power of 2 can be mutated if t ≤ m, (and then t as defined as statement that which it is being mentioned above), If the power of 3 is being an even number, and then it already gives away with a value of t at least is being ≥ 2. By the way, any way, the aliquot sequence computation algorithm is being a deterministic algorithm  not  not  is being a randomized algorithm. [color="white"]and then if only let alone assuming c has got two prime factors and then if only let alone assuming c has got two prime factors[/color] By the way, if a line factors as 2[sup]3m1±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 reappear if the same condition occurs back again. [color="white"]and then if only let alone assuming p has got one prime factor and then if only let alone assuming p has got one prime factor[/color] By the way, if a line factors as 2[sup]3m1±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 reappear if the same condition occurs back again. [url=http://www.poodwaddle.com/worldclock/pop7/]By the way, the future population clock / watch, what ever do you think will happen  fit the best? High Projection? Medium Projection? Low Projection? Why?[/url] By the way, why the last page of the mersenne forum thread / post is  not  not  being shown some times? Down up from 200 digits to 135 digits, and then counting if... 
We dodged a bullet at iteration 5657. 2*p. 50% chance of losing down driver.
Raman we didn't know that it only had 2 factors for certain. However we were pretty certain based upon ecm. 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. 
Why not edit it now? I vote for a temporary ban if he continues to use white text. Not acceptable.

[QUOTE=VBCurtis;442233]Why not edit it now? I vote for a temporary ban if he continues to use white text. Not acceptable.[/QUOTE]
Didn't Batalov just warn Raman for such behavior? (or was it for something else?) In any case, the white text does irritate. On a positive note: now below 120 digits and going down fast. :smile: 
115 digits! :smile:
105 ~30min later 
[QUOTE=unconnected;442236]115 digits! :smile:[/QUOTE]
113 actually right now ( just refreshed my screen). edit:and now about 108 ... 105, 
I'm still writing about this topic after I wrote "nevermind," so I apologize for writing "nevermind" because obviously I'm still thinking about it, and I'm fairly new to the forum and might be making a mistake by asking this question now. But...
What are the odds of this sequence meeting itself and creating a very large cycle if the sequence dips below a given number of digits (I'll use 7) and then terminates? Maybe it's an unlikely precondition before even considering the chances of the event if the precondition takes place. But I'm curious about it and wondering if I'm thinking about it the right way. I'm not talking about when a merger is found but how the final sequence at factordb will appear. If the count of nonuntouchable numbers above 10^7 with an immediate descendant below 10^7 were used as a denominator and the count of the subset of those numbers that belong to the 4788 genealogy were used as the numerator, would that fraction give a reasonable estimate of the odds? I'm using the word "odds" as if this were a horse race instead of the given fixed probabilities in Raman's post, but that seems to be part of the appeal of following these things. 
[QUOTE=VBCurtis;442233]Why not edit it now? I vote for a temporary ban if he continues to use white text. Not acceptable.[/QUOTE]
Mainly because I was busy. Handing my dissertation in tomorrow. I may got back and edit that one when I have time. 
[QUOTE=mshelikoff;442239]I'm still writing about this topic after I wrote "nevermind," so I apologize for writing "nevermind" because obviously I'm still thinking about it, and I'm fairly new to the forum and might be making a mistake by asking this question now. But...
What are the odds of this sequence meeting itself and creating a very large cycle if the sequence dips below a given number of digits (I'll use 7) and then terminates? Maybe it's an unlikely precondition before even considering the chances of the event if the precondition takes place. But I'm curious about it and wondering if I'm thinking about it the right way. I'm not talking about when a merger is found but how the final sequence at factordb will appear. If the count of nonuntouchable numbers above 10^7 with an immediate descendant below 10^7 were used as a denominator and the count of the subset of those numbers that belong to the 4788 genealogy were used as the numerator, would that fraction give a reasonable estimate of the odds? I'm using the word "odds" as if this were a horse race instead of the given fixed probabilities in Raman's post, but that seems to be part of the appeal of following these things.[/QUOTE] I think that this is an incredibly hard question to answer. I think that each number has a different number of numbers that can precede it. This implies to me that it is not a simple formula as 4788 has a certain subset of the numbers below 10^7. I think that working this out would be exceedingly difficult. Working out all the numbers that can precede a number in a sequence is not easy. 
In step 6295 it reached:
108779027694592579666495916767464 (33 digits) and then it went up again. Now it is back to 88 digits in step 6666. 
And now we're going down again, as of terms 6700 onward...

got down as low s 18 digits, now it went up to 70 digits

[QUOTE=firejuggler;442249]i 72007300 it's hovering around 2025 digits[/QUOTE]
and at 7581 it's back up to over 60. edit: and moving so fast I can't keep this up to date 7639 and 71 digits now edit2: changed already. 
Minimum at 16 digits and now it grows with 2^2*7 driver.

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.

This is almost heartbreaking.

Just as heartbreaking as [URL="http://factordb.com/sequences.php?se=1&aq=4788&action=range&fr=3925&to=3999"]the previous bottom[/URL].
But then again look at AllSeq page  there are constantly 3040 downdrivers but only 1 per few months terminations/merges... Not all is lost just yet! 
Minimum was step 7189:
1987170707261684 (16 digits) 
[QUOTE=henryzz;442244]I think that this is an incredibly hard question to answer. I think that each number has a different number of numbers that can precede it. This implies to me that it is not a simple formula as 4788 has a certain subset of the numbers below 10^7. I think that working this out would be exceedingly difficult.
Working out all the numbers that can precede a number in a sequence is not easy.[/QUOTE] If the number of sequence transitions from above to below 10^7 (instead of the numbers below 10^7) were considered in the 4788 genealogy compared to all such transitions then wouldn't that roughly account for the different number of numbers that can precede a number in a sequence? I don't know if that would be a way to get an orderofmagnitude estimate. Further edit: Maybe 10^7 is too large because the numbers that precede would need to be worked out. A smaller 10^n might be able to test the idea of whether it estimates the likelihood of a much greater number falling into a certain genealogy. 
[QUOTE=mshelikoff;442262]If the number of sequence transitions from above to below 10^7 (instead of the numbers below 10^7) were considered in the 4788 genealogy compared to all such transitions then wouldn't that roughly account for the different number of numbers that can precede a number in a sequence? I don't know if that would be a way to get an orderofmagnitude estimate.[/QUOTE]
the problem is it depends on what you mean by transitions below did you know that because 4787 is prime that 4787^2 leads to 4788 ? that number is over 22 million. and that's not on the linked genealogy to my knowledge of what you said so any that branch back from that number will also lead to 4788. the number of numbers that can precede a number in a sequence is the number of partitions that are a proper divisors list for another number. E.g. 6 = 1+5 = 1+2+3 are the only possible arrangements using 1 as the start ( a necessity to be a divisors list) this leads to 25 and 6 being the numbers that can get to 6. 6 has been repeated so let's check 25 and we get: 95 119 143 but that would have started with something like 66 possible partitions that have strictly increasing members starting at 1 that could ( before inspection further) possibly have lead to 25. 
And there, 2^2 * 7^2 in i7825 with proper two primes (1 mod 4), and 2^2 * 7 is lost!

Uh oh. And now factordb appears to be down. Did I break it? :sad:

can we get an update on the status on 4788 since factordb is down?

[QUOTE=ryanp;442280]Uh oh. And now factordb appears to be down. Did I break it? :sad:[/QUOTE]
It is probable that your work attracted too much interest and people were refreshing the page very frequently. Maybe just continue offline and keep us updated :smile: 
grrrr... 2^2*7, I told you that you are jinxing it! A fisherman never counts his fish! :razz:
OTOH, D2 is a "good driver", with only a mild increase (compared with D3 or others). P.S., FDB seems down from this part of the world too. 
Status
[CODE]n Digits Number 7926 121 [URL="http://factordb.com/index.php?showid=1100000000866145455"](show)[/URL] [URL="http://factordb.com/index.php?id=1100000000866145455"]3936113967...80[/URL]<121> = 2^2 · 5 · 3163 · 11273 · [URL="http://factordb.com/index.php?id=1100000000866145457"]5519490243...71[/URL]<112>[/CODE] 
[QUOTE=Batalov;442287]Status
[CODE]n Digits Number 7926 121 [URL="http://factordb.com/index.php?showid=1100000000866145455"](show)[/URL] [URL="http://factordb.com/index.php?id=1100000000866145455"]3936113967...80[/URL]<121> = 2^2 · 5 · 3163 · 11273 · [URL="http://factordb.com/index.php?id=1100000000866145457"]5519490243...71[/URL]<112>[/CODE][/QUOTE]That's 16 more lines than I have. Maybe we can get lucky again! 
The easiest way to picture it is [URL="http://factordb.com/aliquot.php?type=1&aq=4788"]the graph[/URL].
Obviously the xaxis is iterations rather than time, so much less time is spent at low levels of y. 
4788 has become the 8th longest open Aliquot sequence below 1e6, and it has all the chances to become 6th in the near future. However if we take merges into account then the sequence 314718 with currently 14420 terms has become the longest Aliquot sequence below 1e6.
Great job! 
Currently @ 7960 with 2^3*3 driver (ouch!). Well, at least, it isn't the 2^3*3*5 driver.

I'm still whacking away at it, now that factordb is back up (and being careful not to put too much load on the site myself).
Any predictions/estimates for how long the 2^3*3 driver will persist? 
[QUOTE=ryanp;442314]I'm still whacking away at it, now that factordb is back up (and being careful not to put too much load on the site myself).
Any predictions/estimates for how long the 2^3*3 driver will persist?[/QUOTE] well until we get a factorization that doesn't lead to a multiple of 24 would be the logical place to start. I don't know the conditions for such to happen but I bet someone does. 
[QUOTE=ryanp;442314]
Any predictions/estimates for how long the 2^3*3 driver will persist?[/QUOTE] The 3 can not disappear. That is because all the terms of the sigma which are not multiple of 3 are 1, 2, 4, 8, and F* times 1, 2, 4, 8, where F* is all possible combinations of factors in F, which are not 2 or 3. (i.e. the factorization 2^3*3^x*F, where x>0 and F is a product of primes not 2, 3). You can lose a 2 or gain a 2 (or more) in special cases, which depend of the size of the numbers. For 10 digits in F, you have about 18% to lose the exponent 3 of 2^3. For 20 digits numbers, your chances go close to 10% (below it). For 30 digits numbers in F, the chances to get a sigma 2^x*3^y*F' with x!=3 and y>=1 are about 7%. If you go higher, your chances to lose the 2^3 get slimmer. If you limit y to 1, your chances to lose 2^3 are about 3.5%, at 20 digits. If y=2, your chances to lose 2^3 are about 22% at 20 digits. If y=3 or larger, you get again, 10% at 20 digits. They also get slimmer for larger numbers. 
[QUOTE=science_man_88;442316]well until we get a factorization that doesn't lead to a multiple of 24 would be the logical place to start. I don't know the conditions for such to happen but I bet someone does.[/QUOTE]Not exactly true, this driver usually breaks by getting a multiple of 48, which is also a multiple of 24, see my comment above. You can not lose the 3 here.
edit: I know the numbers because I am in kinda' the same shit with my [URL="http://factordb.com/sequences.php?se=1&eff=2&aq=225900&action=last20&fr=0&to=100"]225900[/URL], to which I am currently working actively, see its evolution in the last weeks. 
[QUOTE=ryanp;442314]
Any predictions/estimates for how long the 2^3*3 driver will persist?[/QUOTE] I don't have a prediction or estimate. But I feel that a YouTube video from The Sound of Music (1965) of Peggy Wood as Mother Abbess, dubbed by Margery McKay, singing "Climb Ev'ry Mountain" is appropriate. [url]https://www.youtube.com/watch?v=EoCPuhhE6dw[/url] That could be taken the wrong way for a huge number of reasons, but it's meant in the spirit of encouragement. 
[QUOTE=LaurV;442328]Not exactly true, this driver usually breaks by getting a multiple of 48, which is also a multiple of 24, see my comment above. You can not lose the 3 here.
edit: I know the numbers because I am in kinda' the same shit with my [URL="http://factordb.com/sequences.php?se=1&eff=2&aq=225900&action=last20&fr=0&to=100"]225900[/URL], to which I am currently working actively, see its evolution in the last weeks.[/QUOTE] ah but you use usually that's the thing that means it doesn't always break it because it hasn't escaped being a multiple of 24 yet. 
[QUOTE=LaurV;442325]The 3 can not disappear. That is because all the terms of the sigma which are not multiple of 3 are 1, 2, 4, 8, and F* times 1, 2, 4, 8, where F* is all possible combinations of factors in F, which are not 2 or 3. (i.e. the factorization 2^3*3^x*F, where x>0 and F is a product of primes not 2, 3).[/QUOTE]
I don't know anything about the math behind Aliquot Sequences, but it seems like it has lost a 3 several times before near the peaks in the graph? Are these situations special because of the 2^6 and 2^2 infront of the 3? Is it only the 3 after 2^3 that cannot be lost? [CODE]Step 4064: 2^6 · 3^3 · 11^2 · 2280345883<10> · 17383086933815011<17> · 155927685066589177241898435529<30> Step 4065: 2^6 · 67 · 17207650673<11> · 539744611637742767<18> · 72407948320983644687137288253<29> Step 5306: 2^2 · 3 · 23 · 43 · 2113 · 1124214194939693956866751<25> · 1021217636...83<169> Step 5307: 2^2 · 3^3 · 139 · 877 · 1614463 · 966231493 · 1532527068...71<71> · 1365298954...49<108> Step 5308: 2^2 · 17 · 53 · 5519 · 20879 · 23912690842817779000064290538036192241983<41> · 3832885252...49<63> · 1822375450...77<86> Step 6691: 2^2 · 3 · 1297 · 43920156157693<14> · 1500555007933194276969235039038229<34> · 64626509745706158780051904878300730111033<41> Step 6692: 2^2 · 2212646575...89<92> Step 7245: 2^2 · 3^2 · 7 · 43 · 1181 · 68053 · 2850476466709771<16> Step 7246: 2^2 · 3 · 7 · 37 · 1489 · 226564231 · 4637314011607<13> Step 7247: 2^2 · 7 · 11 · 1619 · 32810003 · 517277073001771<15>[/CODE] 
Yes. This is due to sigma(2^3)=15=3*5
It is impossible for us to gain or lose a 5. Our only hope is to change the power of 2. 
Ok, but if the power of 2 changes then we can loose the 3 later.
I misunderstood it as it was never going away again, which did not match the previous steps. 
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[QUOTE=mshelikoff;442335]I don't have a prediction or estimate.
But I feel that a YouTube video from The Sound of Music (1965) of Peggy Wood as Mother Abbess, dubbed by Margery McKay, singing "Climb Ev'ry Mountain" is appropriate. [url]https://www.youtube.com/watch?v=EoCPuhhE6dw[/url] That could be taken the wrong way for a huge number of reasons, but it's meant in the spirit of encouragement.[/QUOTE] One can also "climb ev'ry mountain" in this problem  [url]https://projecteuler.net/problem=569[/url] ...again, in the spirit of encouragement. :rolleyes: 
[QUOTE=Batalov;442359]One can also "climb ev'ry mountain" in this problem  [URL]https://projecteuler.net/problem=569[/URL]
...again, in the spirit of encouragement. :rolleyes:[/QUOTE] What's the scale and how tall is Tenzing? I add just under 2 meters to those peaks if I stand on top...:smile: 
Lost the 3 in step 8051 and now it starts going down in steps 80518054:

[QUOTE=ATH;442346]Ok, but if the power of 2 changes then we can loose the 3 later.
I misunderstood it as it was never going away again, which did not match the previous steps.[/QUOTE] Sure, otherwise we won't need work any aliquot sequence, if we would have a proof that 3 never disappears, we would have a sequence which goes up to infinity. Fortunately (or unfortunately) this is not the case. If, for example, the power of 2 will switch from 2^3 to 2^2, and the sequence do not get a 7, then we will have larger chances to lose the 3: Assume we have t=2^2*3^y*F, with s=sigma(t), then the terms in s which do not contain 3 (i.e. the divisors of t/(3^y) are 1, 2, 2^2, F*, 2F*, 4F*, where F* is all combinations of primes in F. Therefore, s\3 (s without the terms with 3) is (1+2+4)(1+F*). As all terms in F* do not contain a 3, (but their sum can contain one) we must have F*=1 (mod 3) or F*=0 (mod 3) to lose the 3. Example, (1) F=p (a single prime): t=2^2*3^y*p, t\3=2^2*p, s=1+2+4+p+2p+4p+(terms with 3), s\3=7(1+p). As p can not be a multiple of 3, it is either 1 or 2 (mod 3), in this case the sequence will lose a 3 if p is 1 (mod 3), which happens in 50% of the cases, regardless of the size of t. (2) F=p*q (product of 2 primes): t=2^2*3^y*p*q, t\3=2^2*p*q, s\3=1+2+4+p+2p+4p+q+2q+4q+pq+2pq+4pq, so s\3=7(1+p+q+pq)=7(1+F*). As p, q can not be a multiple of 3 (they are primes) F* needs to be 0 or 1 (mod 3), for the sequence to lose a 3, which happens in 25% of the cases, regardless of the size of t. That because we can have either p=q=1 (mod 3) or p=q=2 (mod 3) or p=1, q=2 (mod 3), (or viceversa, which we may ignore due to symmetry). Therefore we have, respectively (mod 3): [CODE] [FONT=Courier New]p q  pq F*=p+q+pq 1+F* 1 1  1 3=0 1 1 2  2 5=2 0 2 1  2 5=2 0 2 2  1 5=2 0[/FONT] [/CODE]which means that the parenthesis is divisible by 3 always, except when all primes in F are 1 (mod 3). You can try as an exercise with 3 or 4 primes. Then put all together to see for the general case. This case (i.e t=2^[COLOR=Red][B]2[/B][/COLOR]*3^y*F) is not dependent of the size of t, and not much of the y (power of 3). But when 2^3, and general odd power of 2, the size of t counts (due to distribution of primes, they get rarer upper, so F=pq, F=pqr, etc, is more often than F=p). 
[QUOTE=ATH;442388]Lost the 3 in step 8051 and now it starts going down in steps 80518054:[/QUOTE]
2^4 without 31 is "steady", it does not go down. To go down needs the next prime after 2 to be higher than 31 always, and there are a lot of random odd numbers which are multiples of either 3, 5, 7, 11, 13, 17, 19, 23, or 29. With 2^4, the threes can freely come and go. So the sequence oscillates up and down in small limits, but average is "steady" until it changes the power of 2, or gets a 31. 
[QUOTE=ATH;442342]I don't know anything about the math behind Aliquot Sequences, but it seems like it has lost a 3 several times before near the peaks in the graph?
[/QUOTE] You would make an excellent guinea pig for my allaboutthetheoryofAliquotsequences page! (Ernst, you watching this thread? :razz:) [url]https://www.rechenkraft.net/aliquot/introanalysis.html[/url] Admittedly there's a substantial rewrite available that hasn't been pushed to the link above, but it should still be something at least. (ChristianB, could you pull the most recent changes from my GitHub? Edit: You can view the most recent version [URL="https://htmlpreview.github.io/?https://github.com/dubslow/MersenneForumAliquot/blob/master/website/html/introanalysis.html"]here[/URL]) It is very long winded though, and works slowly in order to aid development of intuition properly motivate each new idea. Edit: Btw, for the record, what was the lowest number reached (so far) since 200 digits? EditEdit: Looks like the following is the smallest, coming on the second primary downdriver run: [code]Checked 7188 16 (show) 3974341414523362<16> = 2 · 1987170707261681<16> Checked 7189 16 (show) 1987170707261684<16> = 2^2 · 7 · 2897 · 24497888299<11>[/code] 
[QUOTE=Dubslow;442397]You would make an excellent guinea pig for my allaboutthetheoryofAliquotsequences page! (Ernst, you watching this thread? :razz:)
[URL]https://www.rechenkraft.net/aliquot/introanalysis.html[/URL] Admittedly there's a substantial rewrite available that hasn't been pushed to the link above, but it should still be something at least. (ChristianB, could you pull the most recent changes from my GitHub? Edit: You can view the most recent version [URL="https://htmlpreview.github.io/?https://github.com/dubslow/MersenneForumAliquot/blob/master/website/html/introanalysis.html"]here[/URL]) It is very long winded though, and works slowly in order to aid development of intuition properly motivate each new idea. Edit: Btw, for the record, what was the lowest number reached (so far) since 200 digits? EditEdit: Looks like the following is the smallest, coming on the second primary downdriver run: [code]Checked 7188 16 (show) 3974341414523362<16> = 2 · 1987170707261681<16> Checked 7189 16 (show) 1987170707261684<16> = 2^2 · 7 · 2897 · 24497888299<11>[/code][/QUOTE] When I go to either referenced page (with IE) I find things like: [CODE] we also have to include $2 \cdot (2^n  1)$, $2^2 \cdot (2^n  1)$, ..., $2^{n1} \cdot (2^n  1)$. (If $2^n  1$ were not prime[/CODE]which is somewhat difficult to understand. What is needed to read the page as intended? 
You may have a "mathjax dollar" problem, the forum itself here had one in the past :)
Your IE may not recognize the $ sign as matjax delimiter, or it is not enabled on the destination site. The owner of the site (Dubslow) has to enable matjax to render the single dollars as math delimiters. Can you see equations which are not "inside text", i.e. they are single on a text line? Then rightclick on one of them and see your mathjax settings. Edit: same here, with (newest) Firefox. The problem is not from you, but from Mathjax. The forum solved the issue by substituting the inline dollars with "\ (" and respectively "\ )" for inline math (with no quotes and no spaces between, grrr...) 
Looks like a problem with the HTML preview site. When I look at rechenkraft (or a local git clone), the inline math displays correctly.
Antonio, you sure it happens at rechenkraft too, not just the HTML preview thingy? 
I don't see ANY math at rechenkraft (only see dollars everywhere!)
I see standalone equations on the other site, but not the inline, which are displayed unrendered (with dollars). If I copy the file from [url]https://github.com/dubslow/MersenneForumAliquot/blob/master/website/html/introanalysis.html[/url], select all, ctrl+c, ctrl+v, save locally to aaa.html, open it with the browser (any browser) then [B]everything looks fine[/B]. The problem is that prevview/github has not enabled the single dollar in matjax rendering. There is a lot of arguing about those mathjax dollars on the web. 
[QUOTE=Dubslow;442403]Looks like a problem with the HTML preview site. When I look at rechenkraft (or a local git clone), the inline math displays correctly.
Antonio, you sure it happens at rechenkraft too, not just the HTML preview thingy?[/QUOTE] I looked again , and it magically sorted it's self on both sites when I told IE to display insecure content on the rechenkraft site (didn't complain about this on my previous visit, as far as I recall). 
I think we might all be suffering from caching issues. Browsers trying to be smart and just making life more difficult.
At any rate, I had changed the MJ link to http months ago because RKN didn't used to be https. Git has it now back to https, so when ChristianB pulls the changes they should be in agreement again (and hopefully less issues as a result). 
[QUOTE=Dubslow;442408]I think we might all be suffering from caching issues. Browsers trying to be smart and just making life more difficult.
At any rate, I had changed the MJ link to http months ago because RKN didn't used to be https. Git has it now back to https, so when ChristianB pulls the changes they should be in agreement again (and hopefully less issues as a result).[/QUOTE] yeah changing https in my browser to http made it work. 
We have downdriver again @i8078! :max:

[QUOTE=unconnected;442430]We have downdriver again @i8078! :max:[/QUOTE]
:popcorn: 
[QUOTE]... Ford every stream
... Follow every rainbow...[/QUOTE] :xmastree: 
I will run this thing in the ground, so help me. It has officially pissed me off now... :smile:

And so it begins again. :popcorn:

[QUOTE=ryanp;442444]I will run this thing in the ground, so help me. It has officially pissed me off now... :smile:[/QUOTE]
You are by far not the first one... hehe 
[QUOTE=LaurV;442465]You are by far not the first one... hehe[/QUOTE]
But I assure you, I will be the last. :smile: 
[QUOTE=ryanp;442466]But I assure you, I will be the last. :smile:[/QUOTE]
Shhh... don't jinx it again! :razz: 
[QUOTE=Dubslow;442408]I think we might all be suffering from caching issues. Browsers trying to be smart and just making life more difficult.
At any rate, I had changed the MJ link to http months ago because RKN didn't used to be https. Git has it now back to https, so when ChristianB pulls the changes they should be in agreement again (and hopefully less issues as a result).[/QUOTE] The problem with https links to rechenkraft is this (according to Chrome debugger): [QUOTE]Mixed Content: The page at 'https://www.rechenkraft.net/aliquot/introanalysis.html' was loaded over HTTPS, but requested an insecure script 'http://cdn.mathjax.org/mathjax/latest/MathJax.js?config=TeXAMSMML_HTMLorMML'. This request has been blocked; the content must be served over HTTPS.[/QUOTE] So you just need to make the link to MathJax.js use https. 
[QUOTE=Sergei Chernykh;442500]
So you just need to make the link to MathJax.js use https.[/QUOTE] The second part of my quote you included discuses that :smile: In my GitHub it's been https for a while, just waiting on rechenkraft to pull the changes. 
Let's see what happens this time. Again below 120 digits with the downdriver. :rolleyes:

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