[QUOTE=gd_barnes;166363]A huge amount of work completed Kenneth! You completed it long before our Riesel base 3 mini-drive I completed k<100M to n=100K. Great work! :smile:

Administrative and balancing note: Kenneth had previously sent 26 primes for n=54K-68K. His cores had reached different limits at different times. So there are officially 111 primes for n=54K-100K for Riesel base 3 k=100M-200M and n=54K-100K. All will be shown on the Riesel base 3 primes n>25K page shortly.


Gary[/QUOTE]

Yes it was a huge amount of work. Now for future reference we can say that it will take 4-6 months of CPU time to complete a 100M range with n>25K-n<=100K. Looking forward to see the primes on your prime page. And maybe in the future I can take another 100M range :smile:

Regards

Kenneth

[quote=KEP;166388]Yes it was a huge amount of work. Now for future reference we can say that it will take 4-6 months of CPU time to complete a 100M range with n>25K-n<=100K. Looking forward to see the primes on your prime page. And maybe in the future I can take another 100M range :smile:

Regards

Kenneth[/quote]

Thanks for posting the time that it took. Needless to say, if someone wants to take a k=100M range for n=25K-100K, they should probably put a full quad on it.

I had already added the primes to and removed the k's from the pages yesterday. :-)

The best thing that would help right now is to finish the Riesel base 3 mini-drive I to get k<100M up to n=100K also.


Gary

Regarding Riesel base 3 k=3677878, I'm the first thing in the morning putting the LLR effort to a stop, to switch for sieving. The sieving is at least 4 times more efficient at current pace, than LLR, so I will sieve for a couple of months on the dual core, before switching back to LLR. k=3677878 is currently paused (for at least 2 months) for sieving at n=553K.

Regards

KEP

Very impressive tenancity on that k Kenneth!

I hope you find a true monster prime for it. That would be very cool. :smile:

I hope so too :smile:

Currently there is 8764 k/n pairs remaining for n<1M. I've now planned to sieve to at least p=46T, which should be completed in about 60 days on the dual core.

Regarding the prime, I'm starting to loose the confidence that there is actually hiding a prime for k=3677878 for n<=1M, but if a prime is in fact hiding in the remaining bunch of k/n pairs, the odds is pretty good since there is now a 1:8764 chance to find a prime at next tested n :smile: or about 0.011 % chance of finding a prime, which is actually pretty good odds. But I guess only time can tell if a prime is found and at what n it is hiding :smile:

Kenneth!

I think you can predict the odds of a prime with n<=1M by summing (sieve bits depth) * 2 / (n * log2(3)) for the remaining n<=1M.

If you can give me your current sieve depth when you had the 8764 candidates remaining, I can give you an almost exact odds calculation of a prime using a spreadsheet that I generally only use for base 2. But I've set it up to use for any base so it should work for base 3.

For sake of calculations, I'll assume that you've currently sieved to P=10T. If so, the chance of a base 3 n=553K candidate being prime is ~1 in 11400 and the chance of an n=1M candidate being prime is ~1 in 20600. That will improve as you sieve deeper but your # of candidates will drop so the overall chance of finding a prime is still the same.

Calculus is needed to calculate the exact chance of a prime over such a wide n-range, but I can get around that with a reasonable estimate by calculating the expected # of primes for each n=10K interval using the average # of candidates within each interval. You cannot use the average of the 2 chances shown above (for n=553K and n=1M) because the chance of a candidate being prime does not drop in a linear fashion as the n-value increases. Therefore I have concluded:

Assuming that you have sieved to P=10T and have 8764 candidates remaining until n=1M, you should, on average, find 0.564 primes -and- your chance of finding at least one prime is 43.11%.

Unfortunately at this point, it does appear that the odds are now against you unless I am way off on how far you have currently sieved to obtain the 8764 candidates remaining. But...at some point the odds will have to be against you but you'll find a prime. It may be for this k or some other k at a high n-value.


Gary

I have been sitting a bar the entire day, so to avoid confusion I'll shorten my answer and hope I get it straight :smile:

The actual sieve depth with 8764 candidates remaining was 1.768 Trillion.

My calculations of the odds was simply based on the hope to find at least 1 prime for n<1M, and then since I had 8764 candidates remaining, the odds would according to my humble oppinion be equal to 1 in 8764... not sure if thats the exact right way to do it, well according to Garys answer it appears not to be the right way to do it, however it gave me an idea of the odds at least :smile:

Take care everyone and happy holiday! At least where I come from tomorrow is a day off and an extended weekend holiday :smile:

Regards

Kenneth

[QUOTE=jrk;172696]I think you can predict the odds of a prime with n<=1M by summing (sieve bits depth) * 2 / (n * log2(3)) for the remaining n<=1M.[/QUOTE]
Mistake. I meant to say you would get the estimated # of primes that way, not the odds.

Here is a Pari/GP for estimating both for base 3:

[code]? a=553e3;b=1e6;c=8764;d=3;e=log(1.768e12)/log(2);f=0;g=0;for(h=0,c-1,i=a+(b-a)*h/c;j=e*2/(i*log(d)/log(2));f=f+j;g=g+(1-g)*j;);print(f);print(g);
0.59630858639522664388930502211808276182
0.44917023517202043213062131604016633431[/code]

So there are estimated 0.60 primes and 45% that there is at least one.

'a' is the starting n,
'b' is the ending n,
'c' is the count,
'd' is the base,
'e' is the sieved bit depth.

This assumes the n-value increases linearly from start to end of the range. Have I calculated correctly?

I haven't specifically checked your calculation but it appears correct to me and it's nice to have my calculation independently confirmed by a different method. I had 0.56 of a prime and a 43% chance. The difference, I suspect, is because I had to assume a sieve depth, which I estimated at P=10T. You were able to use his exact depth.

Yes, the remaining candidates would be equal across the n-range since the entire range is sieved to the same depth at the moment. That is the same assumption that I used.


Gary

[QUOTE=gd_barnes;172961]I had 0.56 of a prime and a 43% chance. The difference, I suspect, is because I had to assume a sieve depth, which I estimated at P=10T. You were able to use his exact depth.[/QUOTE]
If I re-do my calculation assuming a sieve of 10T (10e12) I get 0.63 primes and 47%.