When I was a graduate student of physics at the University of Texas: as late as 2000 our department of computational relativity STILL used Fortran to do all...
21929
Chris Caldwell
primemogul
Oct 27, 2010 1:09 pm
... as late as 2000 our department of ... of colliding neutron stars! As an undergraduate I learned Fortran, Compass, and had a course in Algol, because "it is...
21930
leavemsg1
Oct 27, 2010 3:57 pm
not giving up... I took the decision tree out for L here's the T-Sequence in its most raw form, and it's more efficient like David Broadhurst wanted. the t- ...
21931
djbroadhurst
Oct 28, 2010 2:06 am
... At last you have discovered the difference between a weak and a strong Lucas test. By demanding that one of your Lucas tests is strong, you are on the way...
21932
paulunderwooduk
Oct 28, 2010 4:53 am
... With the help of a good programmer, Vincent Diepeven, and some computation collaborators, Carlos Pinho and Jeff Gilchrist, we have pushed up the tested...
21933
djbroadhurst
Oct 28, 2010 11:20 am
... That's an impressive number of tests. How many times was it necessary to invoke your (seemingly) ad hoc restrictions that the target is coprime to 5 and 7...
21934
paulunderwooduk
Oct 29, 2010 12:52 am
... The number tests, taking into consideration the symmetry, is of the proportion (2-1)/2*(3-1)/3*(5-1)/5*(7-1)/7/2 of squared n That is 0.114285714*n^2. For...
21935
paulunderwooduk
Oct 29, 2010 1:25 am
... I think I am off by a factor 4... Paul...
21936
paulunderwooduk
Oct 29, 2010 1:51 am
... Drats. It is only n<3.9*10^7 (39 million). So the number of individual tests is about: 0.228571429*(n/2)^2 ~= 0,87*10^14 Paul...
21937
Alan Eliasen
aeliasen
Oct 29, 2010 3:38 am
... Whenever you're implementing this algorithm, you need to be warned that different programming languages implement different conventions for the "mod"...
21938
Alan Eliasen
aeliasen
Oct 29, 2010 7:00 am
... Oops. That was intended to read BigInteger. -- Alan Eliasen eliasen@... http://futureboy.us/...
21939
djbroadhurst
Oct 29, 2010 1:12 pm
... Faulty "deduction", it seems to me. Consider n = 7*373*383 = 1000013 Then we may satisfy all three of these conditions {cond(a,n) = gcd(6*a,n) == 1 && ...
21940
djbroadhurst
Oct 29, 2010 9:25 pm
... Great story. Thanks, Alan. Jens has a good quote, somewhere, along the lines that failures of primality tests are not matters of life and death, whatever...
21941
paulunderwooduk
Oct 29, 2010 11:47 pm
... I observe: 373 = 2 (mod n) 383 = -2 (mod n) ... Again, all a = +-2 (mod) ... It seems that n=7^m*p^2 or n=7^m (m odd and p>7) (at least) never yield a...
21942
paulunderwooduk
Oct 30, 2010 12:13 am
... If you add a "break" you'll quickly see "p" is not necessarily prime, Paul...
21943
djbroadhurst
Oct 30, 2010 1:04 am
... Obviously not: 7^m*p^2 (m odd and p>7) is equivalent to 7 as far as Kronecker is concerned. ... Obviously not: 7^m*p^2 (m odd and p>7) is equivalent to 7 ...
21944
paulunderwooduk
Oct 30, 2010 1:37 am
... Guilty as charged. My deduction about 7*N was flawed. The only reason for discounting 7 was oodles output from pari where a kronecker pair could not be...
21945
paulunderwooduk
Oct 30, 2010 1:45 am
... I did check numbers divisible by 5 or 7 for n<11364001, but cut them out of being tested for larger n, because of the lack of kronecker pairs for some 5*N...
21946
djbroadhurst
Oct 30, 2010 2:19 am
... Thanks, Paul. ... As official representative of the factor 7, may I cordially invite you to test (n,a) pairs with n > 11364001 and 7|n ? Best regards David...
21947
mikeoakes2
Oct 30, 2010 5:17 pm
Back in May, in http://tech.groups.yahoo.com/group/primenumbers/message/21508 ... (David gave a nice example of a large n where condition (c) meant that Q was...
21948
djbroadhurst
Oct 30, 2010 9:10 pm
... Let's try to check your "non-Carmichael solution". Here, for a given (q,n), I count the number of p's, up to p_max, for which V(p,q,n) != p mod n : ...
21949
djbroadhurst
Oct 30, 2010 10:29 pm
... Oh it's that wretched problem of existential quantification. You wrote: "for Q a multiple of 1247". I understood: "for Q any multiple of 1247". You meant:...
21950
djbroadhurst
Oct 30, 2010 11:18 pm
... Here is the proof, a la Chinese: v(p,q,n,m)=2*polcoeff(lift(Mod((p+x)/Mod(2,m),x^2+4*q-p^2)^n),0); t(p,k,m)=lift(v(p,(11*k+3)*29*43,11*29*37*43,m)-p); ...
21951
kraDen
kradenken
Oct 31, 2010 4:23 am
Hi All, After 7 years the Multifactorial Prime Search http://mfprimes.free-dc.org/ has achieved its set goals Viz. find all primes of the form n!k+/-1 where...
21952
djbroadhurst
Oct 31, 2010 9:08 am
... But why stop at k = 25 ? Here is a case where k has 127 bits: http://primes.utm.edu/primes/page.php?id=67139 and an easy BLS proof: ...
21953
djbroadhurst
Oct 31, 2010 11:38 am
... Definition: The integer pair (q,n) is a "Lucas super-pseudoprime" (LSPS) pair if and only if n is composite, n > q > 0, and V(p,q,n) = p mod n, for every...
21954
djbroadhurst
Oct 31, 2010 12:08 pm
... I remark that after this multifactorial prime was found, OpenPFGW was adapted so as to be able to parse and prove such primes. Thus ... may be obtained,...
21955
mikeoakes2
Oct 31, 2010 12:42 pm
... [I replied to this 4 hours ago, but Yahoo seems to have lost it] My program output (which I didn't think worth posting) was [Q=3741 n=507529] [Q=17458...
21956
djbroadhurst
Oct 31, 2010 12:59 pm
... Because 10^6 is a small number. Try these non-Carmichaels: 1080905, 1739089, 1992641, 2110159, found in only a few minutes. David...
21957
djbroadhurst
Oct 31, 2010 1:23 pm
... A few minutes later these turned up: 4013569, 4638985. If you will only think Chinese, such numbers may be found in GHz-minutes, rather than GHz-years :-) ...
