... Counterexamples are easy enough to find: n = 79786523; a = 2982522; s = 0; t = 36290422; n = 97676723; a = 14888402; s = 0; t = 60052383; David...
24617
djbroadhurst
Oct 31, 2012 8:55 pm
... Thanks for acknowledging your error. ... It can be made stronger, for no extra effort. This better condition (6) may be rather hard to fool: (1) to (5), as...
24616
bhelmes_1
Oct 31, 2012 5:19 pm
Dear David, Thanks a lot for your links, and sorry for the wrong description. I have tried to analyze the counterexample and i think the following test is not...
24615
Norman Luhn
n.luhn
Oct 30, 2012 9:35 pm
12 Years ago, I had have also wrote a factoring algorithm. For 10**16+37 , I get after 133 steps 168040027 and after 12637 steps : 59509631 But it works not...
24614
Kermit Rose
kermit1941
Oct 30, 2012 7:52 pm
I converted Bernhard's test 6 for prime numbers into a factoring algorithm. Here is one successful factoring which makes use of it. ... [168040027, 59509631,...
24613
djbroadhurst
Oct 29, 2012 10:13 pm
... It may jump, considerably, just by increasing n by 1: {f(n)=local(s); forprime(p=2,n-1,s+=numdiv(n-p));[n,s,1.*s/n];} ...
24612
djbroadhurst
Oct 29, 2012 6:12 pm
... Since 1923, we have a had a very precise conjecture for the asymptotic density of primes of the form x^2+1. See Shanks' review ...
24611
djbroadhurst
Oct 29, 2012 3:35 pm
... In your desperate efforts to add "wriggles";, you have achieved the singular feat of devising a "primality test" that no \prime\ can satisfy. It's hard to...
24610
kad
yourskadhir
Oct 29, 2012 11:26 am
What are all the logic lie behind the worse running time complexity of factorization of Odd Semi-prime N = pq?...
24609
bhelmes_1
Oct 29, 2012 6:21 am
A beautiful day, i am looking for counterexamples for the following prime test only for p = 1 mod 4 1. gcd (a, p)=1 2. Let jacobi (a, p) = -1 3. let jacobi...
24608
bhelmes_1
Oct 28, 2012 12:30 pm
Dear David, i did a little investigation concerning the distribution of primes concerning the polynom f(x)=x^2+1 ...
24607
Sebastian Martin Ruiz
s_m_ruiz
Oct 28, 2012 7:52 am
Hello all : I need an asimtotic aproximation fot this: Sum[d(n-p), p prime p<n] d(i)=number of divisors of i. (like sum [d(n) n<x] =xlogx) Can...
24606
WarrenS
warren_d_smi...
Oct 26, 2012 9:10 pm
A famous old conjecture is that infinitely many primes of form n^2+1 exist. I have no idea how to settle that, but can one instead settle this easier problem: ...
24605
paulunderwooduk
Oct 24, 2012 8:02 pm
... Thanks for your efforts, David (and gremlins), for dispelling many 3/4/5 selfridge tests. Also your script: ...
24604
djbroadhurst
Oct 24, 2012 4:20 pm
... costs, generically, 3 + 3 = 6 selfridges since you ask for a pair of strong Frobenius tests and CP Theorem 3.5.8 shows how to do Frobenius in 3 selfridges....
24603
djbroadhurst
Oct 24, 2012 12:24 am
... They used 2 rather cheap tricks to produce http://physics.open.ac.uk/~dbroadhu/cert/underw65.txt Below I show how a couple of extra gcds screen out those...
Dear David, thank you for your work and the counterexample. I have noticed that both counterexamples are of the form n = 3 mod 4 I asked myself if there is a...
24600
Robin Garcia
sopadeajo2001
Oct 22, 2012 10:19 pm
I was wondering if we could find more coherent-defined sequences -and useful- whose first term begins with 137 or with a first term beginning with 137 - t or...
24599
djbroadhurst
Oct 22, 2012 9:44 pm
... In all circumstances, the command "if(A,B)" in Pari-GP will execute B if and only if A, on its completion, has not evaluated to zero. Your example followed...
24598
Lélio
lelio_73
Oct 22, 2012 6:22 pm
2012/10/22 Sebastian Martin Ruiz said ... elaborating on it we get: gap^2 < 4*p + 2*gap the table in http://mathworld.wolfram.com/PrimeGaps.html implies that ...
24597
Robin Garcia
sopadeajo2001
Oct 22, 2012 5:45 pm
David said: "What you asked for is what you got." I say: "Everything you need you got it" (Roy Orbison) ;with humour.. Yet, once i assumed i missed the equal...
24596
Sebastian Martin Ruiz
s_m_ruiz
Oct 22, 2012 5:38 pm
Hello all: Can anyone prove this? ((q-p)/2)^2-(q+p)/2<0 for all p and q consecutive prime numbers p<q Sincerely Sebastian Martin Ruiz ...
24595
djbroadhurst
Oct 22, 2012 5:05 pm
... That bad code is equivalent to {for(p=2,10,for(q=p,10,for(r=q,10, q=p*q*r;print([p,q,r]))));} ... .... What you asked for is what you got. David...
24594
sopadeajo2001
Oct 22, 2012 4:18 pm
The precedent message must be read: who is failing ?...
... I ran your linked program here for prime p<100000 and found: [226437979, 4683227] [342055369, 50400111] [1920594079, 307611949] [1894955311, 169658692] ...
24591
djbroadhurst
Oct 22, 2012 1:42 pm
... In the case of the 1+1+1+2 selfridge tests of this thread, the gremlins found it easier to manufacture counterexamples on the fly. For example ...
24590
paulunderwooduk
Oct 22, 2012 12:48 am
... http://www.cecm.sfu.ca/Pseudoprimes/index-2-to-64.html In order to crack my 1+1+2+2 and 2+2 tests, I am contemplating taking Jan Feitsma's list of...
24589
djbroadhurst
Oct 21, 2012 9:23 pm
... The gcd of p-1 and q-1 was chosen for help with the 3 Fermat tests in your 1+1+1+2 method. Then one has to keep slogging to be lucky enough to satisy one...
