Losing downdriver

[CRGreathouse]

Originally Posted by Greebley

I don't think the fact that we can have a square adds much. For example the chance of getting a square of 5 is only 1/25. Since we can't get a square of 2 or 3, the chance of a square part will be bigger than 1/25, but not a lot bigger.

One can add 1/25 + 1/49+ 1/121 to get an aproximation of the chance of escape to be around 1/430 for a 100 digit number.

1/25 + 1/49 + 1/121 = 0.0686...
P(2) - 1/4 - 1/9 = 0.0911...

So the larger primes actually add a reasonably substantial part. I get a 1/230 chance without considering squares, a 1/215 chance considering 5, 7, and 11, and a 1/210 chance considering all primes > 3.

[henryzz]

Originally Posted by CRGreathouse

1/25 + 1/49 + 1/121 = 0.0686...
P(2) - 1/4 - 1/9 = 0.0911...

So the larger primes actually add a reasonably substantial part. I get a 1/230 chance without considering squares, a 1/215 chance considering 5, 7, and 11, and a 1/210 chance considering all primes > 3.

you seem to have got results that are twice as likely as Greebley's
which one of you is correct?

[CRGreathouse]

Originally Posted by henryzz

you seem to have got results that are twice as likely as Greebley's
which one of you is correct?

Greebley, I think. I forgot to divide by 2.

My only point was to show the contribution of the larger primes.

[Greebley]

So 1/420 sounds right. That means the chance of escape is about 1/(4.2*<Number of Digits>) - does that sound right?

[Greebley]

Ironically, over the weekend my run got a down-driver at 105 digits and escaped going from 101 to 100 digits - so now I know the odds.

Now its wandering around driverless which has been happening to me a lot recently. Its making the 700 digit range much slower for me than the 500 where I usually had a driver.

[gd_barnes]

Originally Posted by Greebley

Ironically, over the weekend my run got a down-driver at 105 digits and escaped going from 101 to 100 digits - so now I know the odds.

Now its wandering around driverless which has been happening to me a lot recently. Its making the 700 digit range much slower for me than the 500 where I usually had a driver.

Wow, you have sequences at 500 or 700 digits? That's pretty amazing. Can you share them with us?

Also, wouldn't a 700 digit number always be slower testing than a 500 digit number?

[Greebley]

Sorry the 700k starting numbers vs the 500k starting numbers. So nothing exciting.

[Mini-Geek]

Originally Posted by gd_barnes

Wow, you have sequences at 500 or 700 digits? That's pretty amazing. Can you share them with us?

Were...you being sarcastic? It's impossible using current hardware and publicly known factoring algorithms to get anywhere near that high in a remotely reasonable amount of time.

Originally Posted by gd_barnes

Also, wouldn't a 700 digit number always be slower testing than a 500 digit number?

Nope. Extreme example: Try factoring 334! (which has 700 digits; of course not taking into account already that you know it's 334! and therefore a trivial factorization) vs this c500, which I'll let you know factors as p250*p250:

1695747911822354484381997577023402594846452149212181351156213445442416262308068696079790604630029894105542185365466081449964443131762673104304035139393232272369582138589765587532659519591908235314587877026382804760079271221862312110364323119218439335777073033168326538360116202826513345943023996895274024267423059275853356727456540594693562824161951564600609901278422824134229912663192334776839612879867665622274041104933288916718009110132205998960417046549816257966584380574695456919776114939713881

But yes, on average a 700 digit number is far slower than a 500 digit number, so it's practically guaranteed that an Aliquot sequence at 700 digits would be far harder than an Aliquot sequence at 500 digits. The fact that (if NFS would be necessary) it's currently impossible to do any of 250, 500, or 700 digits in a reasonable amount of time is beside the point. (the largest general factorization so far was 200 digits, and took an enormous amount of work, something like 80 CPU years for the sieving alone)

[henryzz]

I seem to be able to glean from the discussion in this thread that the probability of a number being prime is ln(N).
Am i correct? If not how have i gone wrong?
I came to this conclusion based on the fact that earlier on in this thread in post #6 it is said that the probability of a number being 2*p with p=1mod4 is ln (10^(number of decimal digits))*2. I removed the *2 because i am not now concerned with whether the prime is 1mod4 or 3mod4.
How do we work it out if we have trial factored upto a certain number?
I have found http://www.mersenneforum.org/showthread.php?t=12356
RDS's answer for questions b and c seem to contain the answer i need, but i cant see how to input the size of the number into that formula unless it is ln (10^(number of decimal digits))/(exp(gamma)/[(number of decimal digits of trial factoring) * log(10)])

[10metreh]

Originally Posted by henryzz

I seem to be able to glean from the discussion in this thread that the probability of a number being prime is ln(N).
Am i correct? If not how have i gone wrong?
I came to this conclusion based on the fact that earlier on in this thread in post #6 it is said that the probability of a number being 2*p with p=1mod4 is ln (10^(number of decimal digits))*2. I removed the *2 because i am not now concerned with whether the prime is 1mod4 or 3mod4.
How do we work it out if we have trial factored upto a certain number?
I have found http://www.mersenneforum.org/showthread.php?t=12356
RDS's answer for questions b and c seem to contain the answer i need, but i cant see how to input the size of the number into that formula unless it is ln (10^(number of decimal digits))/(exp(gamma)/[(number of decimal digits of trial factoring) * log(10)])

http://en.wikipedia.org/wiki/Prime_Number_Theorem

[Mini-Geek]

To sum up: you're right, the chance a number is prime is about 1 in ln(N).
http://primes.utm.edu/howmany.shtml#3 has more interesting related info.