20160910, 23:21  #2553 
"Marc Shelikoff"
Dec 2014
Cambridge, MA
F_{16} Posts 
The link in my last post included a link to the 4788 staysunderamillion genealogy at http://www.rieselprime.de/Others/4788_1000000.png .

20160910, 23:51  #2554  
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
Quote:


20160911, 02:24  #2555 
"Marc Shelikoff"
Dec 2014
Cambridge, MA
F_{16} Posts 
Nevermind.

20160911, 10:43  #2556 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 

20160911, 12:17  #2557  
Noodles
"Mr. Tuch"
Dec 2007
Chennai, India
2351_{8} Posts 
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. Like India population is following China population. When ever do you think that India's population will overtake China's population? 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. ever ever ever ever ever ever ever ever ever ever ever ever ever ever ever ever ever Quote:
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. think stuff style item step idea that ever which ever a way a way If a line factors as 2^{2}.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^{m}.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_{1}^{a[sub]1[/sub]}.p_{2}^{a[sub]2[/sub]}...p_{n}^{a[sub]n[/sub]} Let t be the highest power of 2 dividing (1+p_{1}+p_{1}^{2}+...+p_{1}^{a[sub]1[/sub]}).(1+p_{2}+p_{2}^{2}+...+p_{2}^{a[sub]2[/sub]})...(1+p_{n}+p_{n}^{2}+...+p_{n}^{a[sub]n[/sub]}) = ((p_{1}^{a[sub]1[/sub]+1}1)/(p_{1}1)).((p_{2}^{a[sub]2[/sub]+1}1)/(p_{2}1))...((p_{n}^{a[sub]n[/sub]+1}1)/(p_{n}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. @ ramshanker  How old are you at all ?? that which they are being as like into that which it is being as like into 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. 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 The only way to escape the 2.3 driver is to have a line factoring / factorizing as like into 2.3^{2m}.p where by m ≥ 1, p is being a prime number congruent to 1 (mod 4). 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 On the other hand, lines factoring / factorizing as like into 2^{m}.3^{k}.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. 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 By the way, if a line factors as 2^{3m1±1}.7^{k}.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. 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 By the way, if a line factors as 2^{3m1±1}.7^{k}.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. 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? 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... Last fiddled with by Raman on 20160911 at 12:50 

20160911, 13:01  #2558 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
13356_{8} Posts 
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. 
20160911, 16:51  #2559 
"Curtis"
Feb 2005
Riverside, CA
3^{4}×59 Posts 
Why not edit it now? I vote for a temporary ban if he continues to use white text. Not acceptable.

20160911, 17:34  #2560  
"Tapio Rajala"
Feb 2010
Finland
3^{2}×5×7 Posts 
Quote:
On a positive note: now below 120 digits and going down fast. 

20160911, 17:35  #2561 
May 2009
Russia, Moscow
13×197 Posts 
115 digits!
105 ~30min later Last fiddled with by unconnected on 20160911 at 18:19 
20160911, 17:37  #2562 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
113 actually right now ( just refreshed my screen).
edit:and now about 108 ... 105, Last fiddled with by science_man_88 on 20160911 at 18:13 
20160911, 18:14  #2563 
"Marc Shelikoff"
Dec 2014
Cambridge, MA
3×5 Posts 
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. 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Reserved for MF  Sequence 3366  RichD  Aliquot Sequences  470  20210422 02:17 
Reserved for MF  Sequence 3408  RichD  Aliquot Sequences  474  20210307 20:28 
Reserved for MF  Sequence 276  kar_bon  Aliquot Sequences  127  20201217 10:05 
Assignments are reserved but not showing up  prism019  GPU to 72  6  20200921 22:11 
80M to 64 bits ... but not really reserved  petrw1  Lone Mersenne Hunters  82  20100111 01:57 