I try something and the result is a great surprise for me Did a mathematician ( as i am not ) can explain me why it seems to work ? let N(1)=2 first prime...
I find it hard to believe that you have improved on Riemann's approximation. Take a look at Ribenboim, page 238, table 27, which shows how pi(x) wobbles about...
Unfortuneately a few computer/power outages have slowed my run down but hopefully in the next few days I'll be able to complete it. For fun I extracted a list...
This bug (and others) have been fixed, and a updated development version of PFGW has been placed into the OpenPFGW yahoo groups' file folder. I will not...
Congrats to Jens and Hans. Here the problem is that we cannot completely rely on hardware+software to avoid false negatives. Jens and Hans do the best they...
... Thanks. ... Indeed. It is described here for the curious: http://hjem.get2net.dk/jka/math/primegaps/residuemismatch.txt ... We also met the conditions of...
Hi Michael, I just updated my Cunningham Chain record list in the primenumbers group. You can find it there under Files > Prime Tables. The current records...
Thanks Stephen. More generally: a^(N-1) = a^(a-1) mod N if a|N with N/a prime and gcd(N/a,a)=1. In default PRP mode, with a=3, this shows up as a residue of ...
... Yes, in this case that is of course a very good and allways actual source. In my file you can find some additional information about Cunningham chain...
I found two more Lucas series that yield x^3 + y^3 + z^3 = 1, namely those with [p, q] = [9, -1] and [p, q] = [55, 1]. Here are two small primality proofs that...
... Rather, the original question was: why is the residue small? The answer, spotted by Stephen, is that (2^512-19)/3 is prime. At the back of my mind was the...
When I built the June 3th Dev version of PFGW, I errantly did not copy in the EXE files (duh!). Also went on vacation over the weekend, so I have not gotten...
The formula N^(N*P-1) mod (N*P) = N^(N-1) is an elementary property of congruences, illustrated here for 2 and 3: 2^(p-1) mod p = 1 so 2^(2(p-1)) mod p = 1...
Milton Brown
miltbrown@...
Jun 7, 2004 8:03 pm
4440
Milton claimed to have proven that ... for natural N and prime P, and hence that 8 mod 4 = 2...
[N-1, Brillhart-Lehmer-Selfridge] Calling Brillhart-Lehmer-Selfridge with factored part 33.34% (34*U(81839)+404*lucasU(9,-1,256)^2+44*lucasU(9,-1,512)+9)^3 ...
Thanks David, This seems to be an exception. Notice 2^(p-1) mod p = 1 so 2^(2(p-1)) mod p = 1 and 2*2^(2p-2) mod (2p) = 2 or 2^(2p-1) mod (2p) = 2^1 When p=2,...
Milton Brown
miltbrown@...
Jun 8, 2004 3:31 am
4443
Of course, a mod p can not be 0 (as is 2 mod 2 = 0) Looks like we both forgot. So with this condition the proof is valid D'accord? Milton L. Brown miltbrown...
Milton Brown
miltbrown@...
Jun 8, 2004 9:18 am
4444
... Wrong again! I explicitly avoided the trap that Milton fell into: http://groups.yahoo.com/group/primeform/message/4436 ... ........................! I...
Thanks for the updates Jim. I've now also finished the k*3^n+1 run up to n=10 million and k=8191. The only difference this time is I added a few more fermat...
I finally found a Lucasian family of Diophantine solutions to x^3 + y^3 + z^3 = 1 based on U/V(1618,1,n) with fundamental discriminant D = (1618^2-4)/72^2 =...
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