... ? dbl(226,437,1469) ... ? gcd(437^2-1,n) 1 But, since Q is symmetric around 0 with -Q, I needed to only test gcd(Q-1,n). Now I am testing with dbl() with...
Hi! Code for Aribas: " for i := 3 to 999 do p := i**2 + (i+1)**2; if (not rab_primetest(p)) then m := factor16(p); if (m > 0) then n := p div m; writeln(i, "...
... Ah, it is indeed easy to generate double PSPs of that rather special type: For example: [6, 286, 779] (Underwood) [50, 550, 1769] [77, 484, 2231] [6, 532,...
... Thnaks for those examples. I have tested upto n=5000 and will go further when I have done some program profiling. Hand-wavingly: If gcd(P,n)!=1 then n=p*R,...
... Correction: Note [0,-Q;1,0]^2==[-Q,0;0,-Q] If p==1 (mod 4) then [0,-Q;1,0]^(2*(n-1)/2)==Id*(-Q)^((n-1)/2)==Id*Q^(n-1)/2 which could be +-Id (depending on...
... Sorry, wrong. This is better: the test becomes [0,-+Q;1,0]^(2*(n-1)/2)==Id*(-+Q)^((p*R-1)/2)==(-+Q)^((R-1)/2==Id (mod p) (Correct me if I am wrong again!...
Hi, I'm new here, but like many of the group members in this forum, I'm fascinated by prime numbers and really would like to understand "why the primes seem to...
... are ... Ouch. Well spotted, David. I forgot to mention that the "twin set" (3,5) should be excluded from this conjecture. This is the only pair where...
... Harvey Dubner kindly sent me a copy of his paper http://direct.bl.uk/bld/PlaceOrder.do?UIN=106073168&ETOC=RN which extends the upper limit to 2*10^11: ...
... The sequence 1, 48, 201, 258, 393, 453, 558, 573, 633, 678, 1623, 2103 is calculated as "Numbers n such that triples generated by {2*(n- 1),2*n,2*(n+1)}...
Hi, Group. ...a 2nd-kind Proth number... (2^16667)*(2^16667+28507)+1 is prime! using pfgw -t -q ... but how many digits is it? I think over 10,000. Bill...
... It has 10,035 digits and 3 small factors: C:>pfgw -f0 -od -q"len((2^16667)*(2^16667+28507)+1)" PFGW Version 1.2.0 for Windows [FFT v23.8] No factoring at...
... paul@pp4:~/pfgw$ ./pfgw -q"(2^16667)*(2^16667+28507)+1" PFGW Version 1.2.0 for Pentium and compatibles [FFT v23.8] (2^16667)*(2^16667+28507)+1 is...
Hi David, Pleased to see that CHG is still being used in new and interesting ways. Thank you for giving credit for my minor role in this impressive proof. ...
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