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...
... 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...
... 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...
... 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...
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 ...
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...
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...
So after 1^q+1=2 there are no further primes for odd q. -Ray Chandler ________________________________ From: mikeoakes2@... [mailto:mikeoakes2@...] ...
... 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...
Dear interested primers Have appreciated feedbacks about my initial draft. I start disclosing "my" Goldbach partition count evaluation method and formula. ...
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...
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...
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....
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....
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...
... 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...
... 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...
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...
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 :...
primenumbers@yahoogro...
Dec 4, 2004 9:35 am
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Hello members ! The start of a record-breaking run of 170 consecutive integers (4339207185939-4339207186108) that are non-semiprimes. No greater runs less than...
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 ...
... 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 ...
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...
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,...
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. ...
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