Hi, Let p2 be a large prime. We find another large random prime p1, such that y = (2*p1*p2) + 1. Is there any way, one could quantify the probability of y...
... Well, it's just like any arbitrary number of the same size except that it's even, it's not divisible by p1, and it's not divisible by p2. Therefore there's...
Hello group, after about 140 days on an Athlon XP 2600 with the use of a combined sieving and prp program written by Jens Kruse Andersen, I found a new ...
To all interested: Sometimes there's a multiple of 210 that's symmetrically surrounded by twelve primes, six on both sides, by distances of 1, 11, 13, 17, 19,...
Group,... know that this is a modest effort to explain why only so few Fermat numbers are 'prime'. Overview: It can be shown that F(1)...F(4) are the only...
... And y is not congruent to 1 mod 3, or to 1 mod 5, or to 1 mod 7, etc. Which should make y somewhat more likely to be divisible by these small numbers......
... Phil meant y is odd, but another factor must also be considered. If q is a random odd prime other than p1 and p2, then q does not divide y-1. q must divide...
Hi Gerald, I m not a mathguy, but i would like to show you a different approach, which brings up rather similar findings regarding to multiples of 210, ...
... All 4 above posts were mailed September 9 but it took 5 days to deliver the posts by Jack and I. We are far apart and the posts showed up the same minute...
edited from earlier e-mail... no comments? the 'pair-wise effort has to be continuous to produce more Fer-mat primes. ... should say... has been once removed,...
I'm making progress, but I know I'm not yet at the frontier of factoring capability. Here is an example of my current program output. The coefficients of the...
The only pertinent web reference I could find is Green and Tao's theorem which says that for every integer k=1 or greater, one can find an arithmetic...
w_sindelar@...
Sep 24, 2007 1:07 pm
19094
... This follows from the widely believed conjecture that all admissible prime constellations have infinitely many occurrences. If the conjecture is true then...
Group, Q: Why are the number of them... finite? A: They are bounded by two 'modulo' conditions, namely... G(x): 2^(2^(x+1)) +1 == 2^q (mod (2^x +1)) H(x):...
Thank you once again, Jens. I value your responses. I would like to add a few comments in reply. ... I'm lost here. Seems like a convoluted approach. Here's...
w_sindelar@...
Sep 26, 2007 7:48 pm
19097
... You asked for comments on your statement (which is an unproven guess). I briefly showed that it would follow from a well-known and trusted conjecture,...
Group, I believe that (O)dd (P)erfect (N)umbers do NOT exist in the same manner as (E)ven (P)erfect (N)umbers do! If we allow the generic formula of...
... I'll try. Are you aware of the concept of "admissible constellations?" For example, twin primes, x and x+2, are believed to occur infinitely often. A...
Bill, Your analysis appears to be restricted to consideration of odd perfect numbers with only two distinct prime divisors. It's known that any odd perfect...
An odd perfect number, N, would have to be of the form N = p1^a1 p2^a2 . . . p_k^a_k where each of the p's are odd primes. The sum of the divisors of N,...
... Jens, I think I may have offended you by writing "I'm lost here. Seems like a convoluted approach." Looking back at this, I can see that it can be taken as...
w_sindelar@...
Sep 29, 2007 1:20 pm
19103
On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth" ... I'm glad you did, and I thank you. You must be a mind reader. You somehow sensed why I got lost trying to...
w_sindelar@...
Sep 29, 2007 1:20 pm
19104
... Jens, I think I may have offended you by writing "I'm lost here. Seems like a convoluted approach." Looking back at this, I can see that it can be taken as...
w_sindelar@...
Sep 29, 2007 1:25 pm
19105
On Thu, 27 Sep 2007 01:55:00 -0000 "elevensmooth" ... I'm glad you did, and I thank you. You must be a mind reader. You somehow sensed why I got lost trying to...
w_sindelar@...
Sep 29, 2007 1:25 pm
19106
... No problem. You can search more information about admissible constellations with a search engine. If a prime p <= k does not divide the common difference...
I am not satisfied that Li(x) describes the number of primes < x I have a new calculation that gives... x= 1000 Ki(1000)= 167.9 and... x= 7919 Ki(7919)=...
... http://en.wikipedia.org/w/index.php?title=Prime-counting_function#Formulas_for_prime-counting_functions See the paragraph about the Riemann's R-function. ...
