## Re: Fast prime generators

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• Hello Mark, i wrote a prime sieve with Cuda on a Gpu Geforce 450 with 192 cores. All work is done on the Gpu. I get as result only the numbers of primes in
Message 1 of 49 , Mar 27, 2012
Hello Mark,

i wrote a prime sieve with Cuda on a Gpu Geforce 450 with 192 cores.
All work is done on the Gpu.

I get as result only the numbers of primes in arithmetic progressions.
The Programm needs approximately 1 month for counting all primes below 10^15

I am not sure if this is a good result, but i could give you the code,
if you want, it is quite nice and uses not the sieve of Eratosthenes,
but more a sieve of Ulam in some derivated form.

Does it make any sense to calculate the numbers of primes in arithmetic progressions up to 10^15 ?

Nice greetings from the primes
Bernhard
• Hi, I had a methodology(may be not efficient had a intuition it will be useful) to generate all odd composite. Factors of all odd composites are predetermined
Message 49 of 49 , May 7, 2012
Hi,
I had a methodology(may be not efficient had a intuition it will be useful) to generate all odd composite. Factors of all odd composites are predetermined through multiples of 4. Let A(n) = 4n where n = 1,2,3... Then A(n) = P1Q1 = P2Q2 = ..... = PkQk be the 'K' possible set of distinct even factors(means both Pk,Qk  should even) such that Pk > Pk-1 > Pk-2 > ....> P1 then  (Pk  + 1) (Qk  + 1),   (Pk-1 + 1)(Qk-1  + 1), ......   (P1 + 1)(Q 1  + 1)    will eventually generate all odd composite numbers. For e.g. A(1) = 4 = 2x2 here P1 = 2, Q1  = 2. Then (P1 + 1)(Q 1  + 1) = (2 + 1)(2 +1) = 9 First odd composite. A(2) = 8 = 2x4 here P1 = 2, Q1  = 4. Then (P1 + 1)(Q 1  + 1) = (2 + 1)(4 +1) = 15 second odd composite A(3) = 12 = 2x6 => 3x7 = 21  third odd composite A(4) = 16 = 2x8 = 4x4 => 5x5 =25 fourth odd composite and 3x9 = 27 fifth odd composite like that it continues. Based on this i found a methodology to
sieve Primes in given range 'a' to 'b' a,b both odd numbers. Write all the odd composite between 'a' to 'b' Find 'm' even such that 3(m + 1) <= 'a' < 3(u + 3) Now v= u/2 Calculate 4v = P1Q1 = P2Q2 = ..... = PkQk Then omit the following numbers  (Pk  + 1) (Q k  + 1),   (Pk-1 + 1)(Q k-1  + 1), ......    (P1 + 1)(Q 1  + 1)  which falls in the range 'a' to 'b' Now increment 'k' and repeat the procedure till (Pk  + 1) (Q k  + 1) > b. Now the remaining numbers are prime.

With regards