Dear List, Three new conjectures are proposed here : http://www.primepuzzles.net/conjectures/conj_050.htm You can send your comments/proofs/counterexamples to...
Hi all; Maarten van Thiel, northern Netherlands, but born in the south. I am not at all a mathematic, on the contrary, I worked as a chef for years, got...
... Welcome. ... A situation very similar to what you mention has already been met. Back in the days of only few prime hunter at universities with batch-jobs...
A Pari program to find all primes 1 or 7 mod 8. For any k=1,2,3,4,5,....Choose the k value and the range u and m you want. Slow algorithm, though. Hope Phil...
... Be forewarned - I'm now *way* OT. I've known about Riesel, and the Swedish machine 'BESK' (Binär Elektronisk SekvensKalkylator = "Binary Electronic...
... The library in question was DJB's primegen, and the issue was that gcc on the Power architecture seems to like having _unsigned_ chars by default (unlike...
Hi Everybody, I have two sets (A and B) of n positive integers each, and I need to subtract each element of B from each element of A. Obviously, the ...
I'm not sure how this relates to primes, maybe the elements of A and B are all prime? *wishful thinking* Anyway, wouldn't this be an O(2n-1) = O(2n) = O(n)...
Hello Kaveh, ... Consider the sets A = { N, 2N, 3N, ... N^2 }, B = { 0, 1, 2, ... N-1 }. The set obtained by subtracting each element of B from each element of...
... I think he's talking about making the nxm table of all the differences of (an element of A) - (an element of B). What I wonder about is what you want to do...
The elements of my sets are random positive integers, and I need all n^2 values of (an element of A) - (an element of B). In fast multiplication using fast...
... Hi Kaveh, I think you misunderstood Peter's point (quoted above). With fast multiplication, n digits times m digits, the result will only have n+m digits,...
Hi Kaveh, My method would be O(n^3) for what you want to do, but the idea is: instead of computing anything ahead of time, if you ask "is k in the set?" look...
Hi , all , I was looking at this output from GMP-ECM : Found probable prime factor of 1 digits: 3 focusing on the word probable , when this occurred to me . ...
Walter Nissen
wnissen@...
Nov 10, 2006 5:50 pm
18416
A beautifull evening, I try to give you a sufficent proof for primes p:=x^2+x+1 p:=x^2+x+1=x*(x+1)+1 with p = 3 mod 4 => 2 appears only one time as divisor of...
Hello everyone Over a year an a half ago I wondered how much improvement I could make on Fermat factoring, by using mod 8, 3, 5, 7, etc for more quickly...
... I remember doing the same. I was sitting at an HPUX workstation at Nokia, so I guess that was 6-7 years ago. ... It's a constant factor speedup. See the...
Let A and B be two sets of n and m positive integers, respectively. We want to know if {(an element of A) - (an element of B)} covers the entire interval of...
... The FFT does not do nm work it only does max(n,m) work, as it's a convolution, which throws away large quantities of information compared with what's ...
... Even n, that is. Negative odd n values can force all values to be even, such that one of the terms is 2. (As spotted by Ronny off-list - good catch.) Phil ...
Ah, this seems a little different from your last problem statement. It seemed last time you wanted to make sure all numbers in the interval [0,c] were...
Consider factoring of 91 = 10^2 - 3^2 = (10 - 3) * (10 + 3) This took one step by Fermat method whereas the difference of the factors is 13 - 7 = 6. I wondered...
... Posted by: "Phil Carmody" thefatphil@... thefatphil Date: Sun Nov 12, 2006 12:40 am ((PST)) Note that to fail to cover the range, all you need is...
Hi! Maybe Kaveh could use a Monte Carlo-type test - i.e. pick a fixed number of randomly chosen k's to test and look whether the respective difference exists....
Hi, I am new to this group, and I have been doing some work on prime numbers that I feel I should share with others interested in prime numbers. I have...
In the real life of computation, my B set is flexible. I should generate k*m numbers is B until the [0,nm] range is covered. So actully the real problem is to...
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