20090610, 14:51  #45 
May 2009
Dedham Massachusetts USA
1513_{8} Posts 
I have some statistics about the numbers that reach certain values before merging or termination.
To get the limit from 300k to 5 million, at least* 2672 numbers of the 67390 terminated or merged before 2^64, so less than 65k numbers to check. A count of the numbers that reach 10^15 (1 Quadrillion): For numbers <=300000, the count is 3740 or about 1.2% For numbers <= 1 million, the count is 12916 or about 1.3% For numbers <= 5 million, the count is 67390 or about 1.35% A count of the numbers that reach 10^12(1 trillion) but don't reach 10^15 before merging or termination: For numbers <=300000, the count is 340 or about .113% For numbers <= 1 million, the count is 1271 or about .127% For numbers <= 5 million, the count is 7070 or about .141% A count of the numbers that reach 10^9(1 billion) but don't reach 10^12 before merging or termination: For numbers <=300000, the count is 662 or about .22% For numbers <= 1 million, the count is 2493 or about .25% For numbers <= 5 million, the count is 16274 or about .34% *  A sequence stopped if it went over 2^64 or if it had a (possibly prime) cofactor > 2^52 with no primes under 2^26 dividing it. The second condition means I didn't test every number up to 2^64 so some might merge or terminate before 2^64 would be reached. 
20090615, 10:30  #46  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2·4,729 Posts 
a fivedigit number squared
Quote:
3379 . 84959577112658511249647834283135668 = 2^2 * 383 * 25357^2 * 86249748121303912397851 

20090615, 11:42  #47  
"Robert Gerbicz"
Oct 2005
Hungary
7·211 Posts 
Quote:


20090615, 14:26  #48 
May 2009
Dedham Massachusetts USA
3·281 Posts 
Ya I have seen those  I was mostly contributing the 5 million case which I think is new with the 1 million and 300k for comparison (since I didn't go to 30 or 80 digits). I was hoping to make it up to 10 million, but ran out of memory around 6 million.
Since his values total a bit over 10,440 for the first 1 million, then 2476 terminated between 1 quadrillion and 80100 digits. Ya, I am also thinking it is false. 
20090709, 20:58  #49 
Mar 2006
Germany
2·3^{2}·7·23 Posts 
71digit twin
as in post #44 now i got 2 factors (twins) in consecutive indices:
seq 247840: Code:
698. 1947292834179145293256748179082560222447097478093195308505991094604768308 = 2^2 * 7 * 69546172649255189044883863538662865087396338503328403875213967664456011 699. 1947292834179145293256748179082560222447097478093195308505991094604768364 = 2^2 * 7 * 69546172649255189044883863538662865087396338503328403875213967664456013 
20090712, 16:55  #50 
May 2009
Dedham Massachusetts USA
1513_{8} Posts 
2^2  the weak downguide
I was noticing with 2^2 that it works sort of as a 'downguide' if you don't have a 3 or 5. I was wondering why it didn't pick up a 7 along the line until I realized it can't  if you have 2^2  it will never pick up a 7 unless it changes 2 exponent first  this follows from the fact that 2^2 makes the siqma always be divisible by 7 so sigma  n is never divsible by 7  fairly obvious if you think about it (which I hadn't)
So that means 2^2 is only lost when you get a prime not equal to 7 mod 8, or two primes equal to 1 mod 4. It is somewhat difficult to pick up a power of 3 because there are usually 4 odd terms so the chances are one of them will be divisible by 3 meaning sigma  n isn't. 5 is more likely to appear (and disappear) Therefore 2^2 will generally lead to small a reduction in size if 3 or 5 isn't present. With higher values, the chances of two or less primes is reduced and you can keep your 2^2 longer. Picking up a 3 or 5 will send it up again, but value can also be lost again for another downrun. For 2^3 (and larger), I think it less likely to have no primes less than 15 and primes like 3 and 5 send the sequence up faster, so I don't think this will work as a 'down guide' except for a few steps. Last fiddled with by Greebley on 20090712 at 16:56 
20090712, 16:58  #51 
Nov 2008
2×3^{3}×43 Posts 

20090712, 17:08  #52 
May 2009
Dedham Massachusetts USA
3·281 Posts 
Actually it just occurred to me that with 2^3 you can never pick up 3 or 5 for similar reasone (sigma 2^3 = 15) so you only have to worry about 7, 11, or 13. Maybe it can have a chance at a down run with it and my statement above about it only being a few steps isn't quite correct.
I think it also twice as likely that you will lose a 3 than gain it with 2^2. The reason is that you lose it if all the sigma terms aren't divisible by 3, but to gain it you need all the sigma terms not divisible by 3 AND the remainder (mod 3) is correct (so one out of two). I bet this is the other reason 2^2 seems to work as a down driver (not have a 3 or 5). For 5 you have 4 times the chance of losing it compared to gaining it. 
20090719, 14:23  #53 
May 2009
Dedham Massachusetts USA
3·281 Posts 
As you probably know, you have to manually fix any term that is a square over 2000. I have fixed a lot of these  they are usually between 20007000.
Well I just found a potential record breaker: Sequence 495246 at index 256 contains 444793^2. That is very high compared to the rest. I have not yet found a third power over 2000. 
20090721, 16:30  #54 
Mar 2006
Germany
2·3^{2}·7·23 Posts 
highest squares and driver
the highest squares for all open seqs upto k=1M:
495246:i256 444793 570690:i996 193771 352440:i778 150517 224250:i37 112297 85176:i740 94543 364924:i608 79367 814890:i1334 71263 180768:i344 61253 864666:i320 56489 679932:i349 53591 559176:i1086 52433 424020:i282 50833 685146:i727 48497 417336:i168 46511 717696:i95 44417 the only seqs <1M with the current driver 2^9*1023 (= 2^9*3*11*31) are: 363270 and 604560 there's no seq <1M with the driver 2^12*8191. 
20090721, 16:40  #55 
May 2009
Dedham Massachusetts USA
1513_{8} Posts 
Interesting to know.
Did you search for 3rd powers? I am curious if there is one over 2000 somewhere in the db for the < 1M seqs. My guess is 'no' but I find it hard to guess how big the biggest third power (or greater?) would be. 
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