primenumbers at Yahoo! Groups

Re: Prp vs. pfgw.

mgrogue@... — Mon, 09 Nov 2009 15:08:00 GMT

... For base 2, I recommend LLR, since it will automatically do a Proth test for k*2^n+1 numbers. You could also use PFGW, but you would need to force a

Re: Prp vs. pfgw.

Chris Caldwell — Sun, 08 Nov 2009 19:19:24 GMT

... Others will correct me if I am wrong, but I think they use the same arithmetic engine now. Of course prp.exe does not prove primality. LLR might be the

Prp vs. pfgw.

Di Maria Giovanni — Sun, 08 Nov 2009 19:12:41 GMT

Hi I must test numbers in the form k*2^n+1. Is Pfgw faster than prp.exe? Thank Giovanni Di Maria

A Ramanjan Prime Corollary

reddwarf2956 — Sat, 31 Oct 2009 21:07:09 GMT

A Ramanjan Prime Corollary: 2*p_(i-n) > p_i for i > k where k = primepi(p_k) = primepi(R_n). That is, p_k is the n'th Ramanujan Prime, R_n, and the k'th prime.

Re: Is this a convergent series and if so what is its sum?

djbroadhurst — Fri, 30 Oct 2009 19:46:05 GMT

... In fact, polling the odd primes really helps: I found 10,000 good digits in 7 minutes, by consulting the 24 odd primes p < 100. The result is in

Re: Is this a convergent series and if so what is its sum?

djbroadhurst — Fri, 30 Oct 2009 15:41:01 GMT

... Finally, here is how to obtain 70 good digits, using no odd prime at all. Observe that the formulas below involve only the numbers 2 and Pi:

Re: A slight problem with proof of convergence

djbroadhurst — Thu, 29 Oct 2009 18:34:32 GMT

... Your intuition was good, Julien. The key to the proof of finiteness (and also to the very fast evaluation) is that you were using a Dirichlet character:

Re: Is this a convergent series and if so what is its sum?

djbroadhurst — Thu, 29 Oct 2009 17:09:00 GMT

... To obtain, 70 good decimal digits, in 9 milliseconds, it suffices to use only the first two odd primes p = 3,5. This is so fast that, to time it

Re: Is this a convergent series and if so what is its sum?

djbroadhurst — Thu, 29 Oct 2009 10:40:42 GMT

... I used only primes p < 400 {ans(A,N)=local(S=1/2.,L); forprime(p=3,A,S=S+if(p%4==1,1,-1)/p); for(s=1,N,L=if(s==1,Pi/4,if(s%2==0,zeta(s)*(1-1/2^s),

Re: Is this a convergent series and if so what is its sum?

marku606 — Thu, 29 Oct 2009 06:15:23 GMT

... Rats, my computer doesn't know what to do with a dvi file. While in my ignorance, I can't fathom how it can take merely 174 milliseconds to compute 300

Re: Is this a convergent series and if so what is its sum?

djbroadhurst — Thu, 29 Oct 2009 02:27:03 GMT

... Here are the first 300 digits:

Re: sum of twinprimes

marku606 — Wed, 28 Oct 2009 15:46:15 GMT

... Yes it's a known fact. All primes > 3 have the form 6n +/- 1. Thus a twin prime is always of the form (6n-1,6n+1) . The sum is 12n, which is divisible by

sum of twinprimes

Robdine — Wed, 28 Oct 2009 15:24:00 GMT

Is it a known fact that the sum of primes of a twinprime pair is always divisible by 12 ? (divisible by 4 is evident) or is there a counterexample? gr. Rob

Re: Is this a convergent series and if so what is its sum?

andrew_j_walker — Wed, 28 Oct 2009 08:47:00 GMT

... The sum is a variation on sum (9) at http://mathworld.wolfram.com/PrimeSums.html It starts (1/3)-(1/5)+(1/7)+(1/11)-(1/13)-(1/17) and equals 0.3349813253

A slight problem with proof of convergence

julienbenney — Wed, 28 Oct 2009 01:08:15 GMT

Re ... I can easily see that the positive terms would converge. However, there is a problem with showing that the series noted above converges absolutely. Term