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Old 2009-06-10, 14:51   #45
Greebley
 
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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.
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Old 2009-06-15, 10:30   #46
Batalov
 
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Default a five-digit number squared

Quote:
Originally Posted by schickel View Post
Here's one from 110432:
Code:
 1613 .  883866359463668463521525972362866107318450543014047075048557464839182863055364072700728788483776 = 2^6 * 17 * 439 * 1607^2 * 1733309 * 413414251171984964928943833274782930952904102537554377365457011053267334799073
From 134856:
3379 . 84959577112658511249647834283135668 = 2^2 * 383 * 25357^2 * 86249748121303912397851
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Old 2009-06-15, 11:42   #47
R. Gerbicz
 
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Quote:
Originally Posted by Greebley View Post
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.
On http://www.aliquot.de/aliquote.htm there is also some stat, for example: "The extension of calculation limit from C60 up to C80 reduces the number of OE-sequences about 2 - 2.5%. " or see the work: http://christophe.clavier.free.fr/Al...e/Aliquot.html raising the mean size from 103 digits to 135 digits on 74 sequences resulted none of them is eliminated. That's why it is called these type of problems law of small numbers and very probable that Catalan's conjecture is false.
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Old 2009-06-15, 14:26   #48
Greebley
 
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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 80-100 digits.

Ya, I am also thinking it is false.
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Old 2009-07-09, 20:58   #49
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Default 71-digit 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
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Old 2009-07-12, 16:55   #50
Greebley
 
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Default 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 2009-07-12 at 16:56
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Old 2009-07-12, 16:58   #51
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Quote:
Originally Posted by Greebley View Post
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.
In addition, 11 and 13 will also cause an increase.
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Old 2009-07-12, 17:08   #52
Greebley
 
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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.
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Old 2009-07-19, 14:23   #53
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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 2000-7000.

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.
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Old 2009-07-21, 16:30   #54
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Default 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.
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Old 2009-07-21, 16:40   #55
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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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