... of ... *m)) ... a /whole ... and ... & ... ((would imply that 'n' > 1 (now, I think my argument is preserved.)) ... from ... F4=65537. ... ((would imply...
I have to say, I have a lot of trouble following any argument that uses a modulus of 1. Could you use more words, more generalized equations, describe more ...
Hello, the k-tuplets page is down since several days. In addition Tony Forbes' e-mail adress @ltkz.demon.co.uk is not working. Does anyone know an actual...
... See http://mersenneforum.org/showthread.php?t=8832 The page is archived at http://web.archive.org/web/20070405115024/www.ltkz.demon.co.uk/ktuplets.htm Tony...
... http://primes.utm.edu/glossary/page.php?sort=CunninghamChain says: "Note that some authors extend the definition of Cunningham Chain to all sequences of...
Does it even converge? If gap size at p is about ln(p) then (with all equalities being "approximately equal to") q=p+ln(p) so q-p*ln(q)=p+ln(p)-p*ln(p+ln (p))...
That's right. One could also say: q ~ p, for p/q ~ 1. Then 1/(q-p*ln q) ~ 1/(p-p*ln p) and int(1/(x-x*ln x)) = -ln (ln x - 1) -> -inf. Thanks. Can you...
Hi all, I just realixzed that the 40th and largest Carol Prime n=253987, was reported on the prime database on Carol's birthday, May 7, 2007... ... Moody...
Suppose you are trying to approximate sum over large primes p of a(p) where a(p) is some monotonically decreasing to 0 function of p. Instead of writing...
Powers of Two Numbers are 1 21 421 8421 168421 32168421 6432168421 1286432168421 ... The conjecture is that 421 is the only Power of Two Prime. The other...
miltbrown@...
Aug 11, 2007 8:26 pm
19066
... Proof: Sum(i=0..N, 10^something(i)*2^i) mod 3 = Sum(i=0..N, 2^i) mod 3 = 2^(N+1)-1 mod 3 = 0 for odd N, 1 for even N. Q.E.D. Peter -- [Name] Peter Kosinar...
... Chain) ... http://ourworld.compuserve.com/homepages/hlifchitz/Henri/us/CunnGenus.htm, ... I can feel another project coming on: The longest chains...
Hi all, I am now ready to publish Carol/Kynea numbers. How do i go about doing that/ What journal should I submit to/ Is there a special format? Any help...
No prime p is (p-1)modulo 4, as primes that are(p-1)modulo 2 are [(p-1)/2]modulo4, and if the order of pmodulo2 is less than p-1, then the modulo4 of p is...
... Hmm, must have been drunk when I wrote this. For modulo read modulo order, or multiplicative order. No base that is square produces a p-1 multiplicative...
Hello all: I have obtained this result: DivisorSigma[2k,m]/DivisorSigma[k,m] is integer if and only if m is a perfect square DivisorSigma[k,m]=sum[d^k, d...
Hello all: I have obtained this result: DivisorSigma[2k,m]/DivisorSigma[k,m] is integer if and only if m is a perfect square DivisorSigma[k,m]=sum[d^k, d...
... I obtain a lot of small counterexamples with k=1, some with k=2, and ... 4 (3) (7) ... q1 := 2726235765168410 q2 := 7247255655544865674860411008810 ...
7203 is NOT a perfect square The result is true only if only m is a perfect square: m=n^2 m= 1, 4, 9, 16, 25, 36, ...... ... I obtain a lot of small...
7203 is not a perfect square! Maybe you didn't mean "if and only if." Part of "if and only if" is: sigma[2k](m)/sigma[k](m) is integer only if m is a square I...
... As Adam already disproved the "only if" part, let's concentrate on the "if" one. First, there are two simple observations to make: 1) If p is a prime, ...
Main page ( http://www.opertech.com/primes/k-tuples.html ) The latest updates have been posted, with improvements to more than 2200 widths. Currently there are...
... This page mentions the incompatibility of the 2 conjectures here http://en.wikipedia.org/wiki/Second_Hardy-Littlewood_conjecture and references your...
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 ...
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