... Yes the signs are important for these trinomials. The most genral case I have is x^n+-x^k-1 (x,n,k all positive integers x>1,n>2,n>k>0 and discounting...
9007
mohamed saleh
mastermindax
Oct 1, 2002 7:54 am
hiiiiiiiii friends i am just new member in your group and I glad to be that I am mathmatical interester from Egypt spieal in number threoy I joined since...
9008
P A L M A email|palma...
palmaaaaaa
Oct 1, 2002 9:50 am
I never received anything. Please send me the proof, without examples. thank you ... ADVERTISEMENT [Image] ... -- GENUG SCHOYN ...
9009
David Broadhurst
djbroadhurst
Oct 1, 2002 2:10 pm
N=2^64695-2^15-1 is prime. Paul Underwood discovered that N is probably prime, as part of a large trawl of candidates of his favourite form f(a,b)=2^a-2^b-1 ...
9010
David Oscari
conjectures
Oct 1, 2002 2:26 pm
Are there a proof of Conjecture Goldbach? Ahora podés usar Yahoo! Messenger desde tu celular. Aprendé cómo hacerlo en Yahoo! Móvil:...
9012
jbrennen
Oct 1, 2002 2:48 pm
... I was wondering after I wrote this, and thought maybe somebody would have an answer, or maybe I'm just crazy... Assume we have a large odd composite number...
9014
Greg@...
mazur1947
Oct 1, 2002 4:58 pm
In a message dated 9/27/02 2:30:03 PM Eastern Daylight Time, jack@... ... Reply: When you talk about limits of pi(2n)/pi(n) it seems similar to a proof...
9016
David Broadhurst
djbroadhurst
Oct 1, 2002 5:52 pm
... Err, it seems to me that it has all bits but one set to 1: ? b=binary(2^64695-2^15-1); ? print(sum(k=1,64695,b[k])) 64694 It is a "base-2 near repunit", ...
9017
David Broadhurst
djbroadhurst
Oct 1, 2002 5:57 pm
... Not really. Everyone knows that Hans is doing something huge, well above 6k digits. So why spend cycles to become a poor second in 2003 :-? David...
9018
Jon Perry
jon_perryuk
Oct 1, 2002 6:01 pm
... But it is: 2^15[2^64680-1]-1 and perhaps this may yield a name. Jon Perry perry@... http://www.users.globalnet.co.uk/~perry/maths BrainBench...
9019
David Broadhurst
djbroadhurst
Oct 1, 2002 6:04 pm
... That was unfortunate. I had a registered Titanix (thanks!) which did all the batch except the p1012 and then used latest Primo 2.0.0 beta3 for this, the...
9021
David Broadhurst
djbroadhurst
Oct 1, 2002 10:29 pm
... On July 2 he told that he had done 2173 bits of something big: http://groups.yahoo.com/group/primeform/message/2621 On Sept 21 he told that he had recently...
9022
Milton Brown
miltbrown@...
Oct 2, 2002 12:37 am
I have done some work with RSA numbers, You are welcome to look at my WEB-site www.csulb.edu/~mbrown10 Milton L. Brown ... From: mohamed saleh...
9023
David Broadhurst
djbroadhurst
Oct 2, 2002 1:24 am
... There is good reason to suppose from http://groups.yahoo.com/group/primeform/message/2719 that Hans's target was posted by him as a PrP at ...
9026
paulunderwooduk
Oct 2, 2002 8:47 am
... and ... That's now changed and Primo/ECPP is again #1 general prime proving program/method. By "ordinary"; I thought the number does not have any special...
9027
Jon Perry
jon_perryuk
Oct 2, 2002 11:42 am
gcd(x,x+1)=1 Therefore gcd(kx,x+1)=gcd(k,x+1) Jon Perry perry@... http://www.users.globalnet.co.uk/~perry/maths BrainBench MVP for HTML and...
9028
Jon Perry
jon_perryuk
Oct 2, 2002 5:17 pm
Thanks. My proof went along these lines: If gcd(kx,x+1)>gcd(k,x+1), then gcd(x,x+1)>1 The theorem is extensible into: gcd(kx,y)=gcd(k,y).gcd(x,y) if gcd(x,y)=1...
9029
Jon Perry
jon_perryuk
Oct 2, 2002 7:49 pm
The general case: gcd(kx,y)=gcd(x,y).gcd(k,y/gcd(x,y)) Jon Perry perry@... http://www.users.globalnet.co.uk/~perry/maths BrainBench MVP for HTML...
9030
Max B
zen_ghost_floating@...
Oct 2, 2002 8:27 pm
"If we look at the prime factors of a Fibonacci number, there will be at least one of them that has never before appeared as a factor in any earlier Fibonacci...
9031
David Broadhurst
djbroadhurst
Oct 2, 2002 10:34 pm
Carmichael's theorem is _far_ more general than that. Knott gave only the Fibonacci case p=1,q=-1. In general consider U(p,q,n) with p>0 and d=p^2-4*q>0. Then...
9032
Shane
divineprime
Oct 3, 2002 8:43 am
... The proof is in this form: F(n)/x = characteristic factors. Observing: F(2^n)/L(2^ n-2)= all characteristic factors. x may have something to do with...
F(pp) / F(p)*F(p)= all characteristic factors. There must be more?...
9035
Milton Brown
miltbrown@...
Oct 3, 2002 12:56 pm
... From: Milton Brown [mailto:miltbrown@...] Sent: Tuesday, October 01, 2002 3:07 PM To: 'mohamed saleh'; primenumbers@yahoogroups.com Cc:...
9036
Phil Carmody
thefatphil
Oct 3, 2002 4:23 pm
I was wandering around Andy Steward's incredible generalised repunit resource, http://www.users.globalnet.co.uk/~aads/ and something regarding KP proofs...
9037
Jon Perry
jon_perryuk
Oct 3, 2002 5:49 pm
Anyone know why 4ab+3(a+b)+2 is never a square? It comes from looking at the 'alternative' 5-squares conjecture, find a set a,b,c,d such that ab-1=u^2 ac-1=v^2...
9038
jbrennen
Oct 3, 2002 6:11 pm
... Rewrite this as: N = (4*a+3) * b + (3*a+2) In order for N to be a square, we must have (3*a+2) be a quadratic residue modulo (4*a+3). Note that (3*a+2) is...
9040
Shane
divineprime
Oct 3, 2002 10:09 pm
F(3p)/ 2F(p) ___________ ~ 2 L(3p)/ 4L(p) Sweet! Product of F(3p)'s characteristic factors divided by the product of L(3p)'s characteristic factors....
9041
David Broadhurst
djbroadhurst
Oct 4, 2002 1:12 am
Phil: To find the 5 top triplets at http://www.ltkz.demon.co.uk/ktuplets.htm#largest3 I had recourse to sub-BLS square testing: ...
9042
Andy Steward
repunituk
Oct 4, 2002 6:34 am
Firstly, congratulations to Paul, Paul, David and Bouk. Secondly, DAMN. I had this great idea while I was on holiday about ten days ago. It appears to be much...
