Jean, Thank you for the new version in almost no time! I tried to test Kynea numbers. It's working but it's slower than pfgw. Two examples are given below. Now...
15666
Paul Underwood
paulunderwooduk
Dec 1, 2004 10:50 am
... The way to speed up searching Carol/Kynea numbers is through modular reduction. When reducing over 2^n+-2^k-1 ( k<70% of n ingeneral ) we can use additions...
Thank you for your tests ! I am not surprised if LLPP4 deterministic test is slower than pfgw PRP one, because the "Computing U0" loop is more time consuming...
15668
pminovic
Dec 1, 2004 1:23 pm
... PRP ... This is true, it takes about 50 minutes to compute U0, I'll append the lresults.txt file tomorrow. ... No! To prove primality of a PRP using "pfgw...
15671
jim_fougeron
Dec 1, 2004 9:42 pm
I have a 2004 addition for CPAP-4 (yes, it was very easy to find) 78006074.883#+k*2004+R is prime for k = [0...3] Using the special crafted R of: ...
15672
Jens Kruse Andersen
jkand71
Dec 2, 2004 12:32 am
... How fortunate that the year is divisible by 3#. Don't expect me to wait until 2010 to beat it :-) Congratulations on beating my ... erm ... improving your...
15673
Ken Davis
kraden
Dec 2, 2004 1:19 am
The following is posted on behalf of "Jean Penne" who sent his reply to "primenumbers-owner" instead of "primenumbers" by mistake. Cheers Ken ... pfgw ... the ...
15674
patience_and_fortitude
patience_and...
Dec 2, 2004 2:42 pm
Are perfect cubes + 1 ever prime? i.e.: Is any number of the form (a^3)+1 ever prime? ... It seems like there should be eventually... ... Better yet, is any...
15675
mikeoakes2@...
mikeoakes2
Dec 2, 2004 2:53 pm
In a message dated 02/12/2004 14:46:19 GMT Standard Time, ... No. If q is any odd number > 2, then a^q + 1 = (a+1) * (a^(q-1) - a^(q-2) + ... +1) and so has...
15676
Ray Chandler
rayjchandler
Dec 2, 2004 4:39 pm
So after 1^q+1=2 there are no further primes for odd q. -Ray Chandler ________________________________ From: mikeoakes2@... [mailto:mikeoakes2@...] ...
15677
Paul Leyland
xilmanuk
Dec 2, 2004 4:48 pm
... We can say even more than that. All primes of the form (a^q + 1) are either 2 (i.e. a=1) or of the form (b^{2^n}+1) where b and n are integers. When...
15678
patience_and_fortitude
patience_and...
Dec 2, 2004 7:29 pm
...and a similar rational holds for (a^q)-1 counterparts. Thanks, Shawn...
15679
Didier van der Straten
didiervander...
Dec 2, 2004 11:48 pm
Dear interested primers Have appreciated feedbacks about my initial draft. I start disclosing "my" Goldbach partition count evaluation method and formula. ...
15680
gulland68
Dec 3, 2004 12:06 am
Someone told me that one of either Goldbach or Riemann had been proven unprovable by current knowledge of mathematics. Can someone inform me as to which it...
15681
Jose Ramón Brox
ambroxius
Dec 3, 2004 12:50 am
My two cents... Original English Text: My love you are no where, and I want you now here Translating and translating back... My love you are no where, and I...
15682
Jim Doyle
ozyjim2004
Dec 3, 2004 12:52 am
Hi I'm not sure if this is a new idea but I have not come across it . Its based on the fact that apart from 2 and 5 all primes have a last digit of 1,3,7 or 9....
15683
Robin Garcia
sopadeajo2001
Dec 3, 2004 2:45 am
Robin Garcia <sopadeajo2001@...> wrote: Robin Garcia <sopadeajo2001@...> wrote: Would anybody try to speed up the algorith implemented in a simple...
15684
Jud McCranie
judmccr
Dec 3, 2004 4:02 am
... As far as I know, neither of them has been proven undecidable. But either one of them could conceivably be undecidable....
15685
Jud McCranie
judmccr
Dec 3, 2004 4:02 am
This is true, isn't it? For every epsilon > 0, let x = 1+epsilon. There is always a prime between x^n and x^(n+1) for all n>m, where m depends on epsilon....
15686
Sren Nielsen
srentospace
Dec 3, 2004 4:06 pm
Under visualization there are four links, I couldn't open. I know that at least one link (my own) is closed for ever. I have tried to tell it to Chriss...
15687
Chris Caldwell
primemogul
Dec 3, 2004 10:08 pm
... Perhaps I can; but what page are you talking about? It always helps to be specific. Perhaps you mean: http://primes.utm.edu/links/visualization/ This is...
15688
Jens Kruse Andersen
jkand71
Dec 3, 2004 10:26 pm
... I also chose to search an easy improvement with a "special" d: 46313478 * 1201#/1302643 + x498 + 2310n, for n = 0..3 Using the special crafted x498 at the...
15689
cino hilliard
hillcino368
Dec 4, 2004 5:13 am
Hi, ... By Ohakm's Razor, the long division (x^n+1)/(x+1) is the best proof that x+1 divides. However We also can also argue as follows using Baconian...
15690
primenumbers@yahoogro...
Dec 4, 2004 9:35 am
Hello, This email message is a notification to let you know that a file has been uploaded to the Files area of the primenumbers group. File :...
15691
Norman Luhn
nluhn
Dec 4, 2004 11:22 am
Hello members ! The start of a record-breaking run of 170 consecutive integers (4339207185939-4339207186108) that are non-semiprimes. No greater runs less than...
15692
Michael Gian
pastmyprime2
Dec 5, 2004 10:19 pm
As I have read, for prime numbers, p, 2^p-1 are called Prime- Exponent Mersenne numbers. These are sub-group of 2^k-1, k positive integers, which are called ...
15693
Décio Luiz Gazzoni...
deciogazzoni
Dec 6, 2004 1:11 am
... In response to this astonishing (for me at least) claim, I had to try and prove that this is true. I recommend anyone else to try to prove it before ...
15694
David Cleaver
wraythex
Dec 6, 2004 10:39 am
Hello everyone, I visited the ECMNET page located at: http://www.loria.fr/~zimmerma/records/ecmnet.html and saw that the latest version of ecm available is...
15695
Ignacio Larrosa Ca...
ilarrosa
Dec 6, 2004 10:55 am
Sunday, December 05, 2004 11:19 PM [GMT+1=CET], ... The sum of the first n odd cubes is S(n) = Sum((2k-1)^3, k, 1, n) = 2n^4 - n^2 = n^2(2n^2 - 1) If n = 2^m,...
15696
Mazzarello Gianni
mzzgv
Dec 6, 2004 11:55 am
It seems correct, but we can say it in an equivalent way: the set of prime numbers = 2, 5 PLUS all odd numbers > 5, not ending by 5 and not composite. ...
