20050714, 00:32  #1 
Jul 2005
2·3^{2} Posts 
Combined Sieving?
Hi,
I want to know whether I can do combined sieving for a bunch of k together with newpgen or some other tool? The issue is not to save time but save a lot of bother doing it for each k independently. jaat 
20050716, 03:04  #2 
Mar 2003
New Zealand
13·89 Posts 
I don't know of any sieving program that will do this yet.
But with pfgw you can use an ABC2 file to do trial factoring and prp testing for a number of k values. As an example, put this in myfile.txt: Code:
ABC2 $a*5^$b1 //{number_primes,$a,1} a: in { 1002 2004 } b: from 100 to 1000 This is convenient, but will be much slower than sieving with newpgen unless the range for each k is quite small. 
20051012, 10:34  #3 
Jun 2003
2^{5}·5·31 Posts 
Maybe its time we gave serious though to combined sieving.
Can someone get in touch with Mikael Klasson to get a modified version of proth_sieve that can handle base 5? No idea how tough it'll be to do this, but i guess its worth a shot. 
20051026, 17:55  #4  
Jul 2005
18_{10} Posts 
Quote:
If there is any hope for this project to gather some momentum, this is a must. jaat 

20051130, 19:05  #5 
Jul 2005
2×193 Posts 
Even if you can get this done it's going to be much slower.
You lose all of the Quadratic Residue filtering. All of the smallsteps will have to be calculated with powmod instead of simple bit shifts. The order(2) filtering will be gone (although you might be able to adapt this to base 5. Better off finding a program that implements base 5 sieving now as there would have been some attempts at optimising this sieving. proth_sieve only implements base 2, and therefore it is only optimised for base 2. 
20060418, 02:26  #6 
Mar 2003
New Zealand
13·89 Posts 
Riesel5 candidate k=151026 is the only one satisfying k = 0 (mod 3). This makes it the only remaining candidate that could have primes for both odd and even exponents. The possibilities for any other candidates are:
If k = 1 (mod 3) then k*5^even1 and k*5^odd+1 are composite If k = 2 (mod 3) then k*5^odd1 and k*5^even+1 are composite [3  k*5^n+/1 whenever 3  k*5^(n2)+/1 because 5^2 = 1 (mod 3)]. This fact could be exploited by a siever, for example to halve the memory required for a bitmap of candidate n (assuming k=151026 is not in the sieve). NewPGen does not seem to take advantage of this. 
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