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Old 2006-01-20, 12:16   #265
Greenbank
 
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Jul 2005

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Default

Excellent, thought I'd be missing something.

n=56 still not done yet, can't be far off as I've got 266121 so far and the estimate was 280214.

After that I'll be finding all of the previously found Octoproths and saving the records out into separate files. If I get time over the weekend I'll fix the DB and upload them and open up the cgi-bin to view/download them.

So much to do, so little time...
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Old 2006-01-20, 16:03   #266
Greenbank
 
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Default n=56 complete!

OK, n=56 is complete!

Estimate was 280214.

Final count after isprime() PARI checking script: 285415

2^n = 72057594037927936 so I used a kmax of 72060T.

Output:

n=56, kmin=0T, kmax=72060T, version=5.0, here T=10^12
Starting the sieve...
Using the first 11 primes to reduce the size of the sieve array

[285415 lines removed]

The sieving is complete.
Number of Prp tests=1503700301
Time=187990 sec.

So that's 2 days, 4 hours, 13 minutes, 10 seconds.

That's the end of it, not going to start n=57, there are too many of them!
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Old 2006-01-24, 21:51   #267
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

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Default new octo program!

In this program now there is no limitation on k value. I've tried all three optimization flags ( among -O1, -O2, -O3 ) and -O1 was the fastest ( try this one ). Just to see that the program correctly identify large k values, I've tried:

C:\>octo_6_0 72 10000000T 10000010T
You can also find the k n values in results_octo.txt file ( These are 3-probable primes )
n=72, kmin=10000000T, kmax=10000010T, version=6.0, here T=10^12
Starting the sieve...
Using the first 8 primes to reduce the size of the sieve array
10000002989467112895 72
10000000479372125565 72
10000007742413127075 72
10000007506574893605 72
10000009359875458875 72
10000007773909145265 72
10000003164141619125 72
10000001353327652625 72
10000000828720374885 72
10000008277743037575 72
10000002428249499645 72
10000001785208783145 72
10000003434544089015 72
10000007626546970445 72
10000002610159188375 72
The sieving is complete.
Number of Prp tests=591004
Time=539 sec.

Correct! These k values are larger than 10^19 ( >2^60, previous limit ).
I've also modified some parts in the code.

See the attachment for the c code. Or download the exe for windows from: http://www.robertgerbicz.tar.hu/octo_6_0.exe
Attached Files
File Type: txt octo_6_0.txt (16.6 KB, 188 views)
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Old 2006-01-26, 20:21   #268
robert44444uk
 
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Jun 2003
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Default Still mystified

I am still mystified why some values of n give many more octoproths and dodecaproths than others. I am also mystified as to why Robert and others are able to forecast the number of octos. If it was variable k I would understand, (modular arithmetic), but I am a bit stumped here.

Can someone furnish a non code explanation?

Regards

Robert Smith
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Old 2006-01-26, 21:07   #269
R. Gerbicz
 
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"Robert Gerbicz"
Oct 2005
Hungary

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Default number of octoproths for n

Quote:
Originally Posted by robert44444uk
I am still mystified why some values of n give many more octoproths and dodecaproths than others. I am also mystified as to why Robert and others are able to forecast the number of octos. If it was variable k I would understand, (modular arithmetic), but I am a bit stumped here.

Can someone furnish a non code explanation?

Regards

Robert Smith
OK here it an explanation: note that this is only a conjecture, because we can't make much weaker statment that f(n)>0 because it would mean that we know that there are infinitely many twin primes, however there is no proof for that

By prime number theorem a random k integer is prime by about 1/log(k) probability, because there are about k/log(k) prime numbers up to k. If we are searching for octoproths then we are searching 8 prime numbers where k<2^n, so if these numbers are independent then there are about 2^n/(log(2^n)^4*log(2^(2*n))^4)=2^n/((n*log(2))^8*16) octoproths ( because 4 forms are about 2^n in almost all cases the other 4 forms are much larger: about 2^(2*n) )

But these numbers are not independent! We have to remultiple by some factors, count the remainders modulo p..., this factor for each n is the weight of n ( for octoproth )

More information at:
http://www.ltkz.demon.co.uk/ktuplets.htm#mathbg
See the famous Hardy-Littlewood Prime k-tuple Conjecture.

Actually an octoproth isn't a k-tuplet but it is very similar to that.
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