[Sorry for the repeat message folks.] RMA.NET is now available for beta testing on Windows Vista. The source files are located here: ...
19019
Werner D. Sand
theo2357
Jul 5, 2007 2:25 pm
Who can help me calculating up to 10 exact decimal places sum (1/n^(1-1/n) - 1/n) sum (1/p^(1-1/p) - 1/p) sum (ln(n) / n^2) sum (ln(p) / p^2) n=positive...
19020
miltbrown@...
Jul 5, 2007 2:26 pm
An interesting extension of Patrick De Geest's Table is where 10^n+p and p are both prime (the first instance) The table begins: n---------p ============= ...
19021
Dirk Augustin
trex400
Jul 5, 2007 8:23 pm
... Patrick already has a list where 10^n-p1, 10^n+p2, n, p1 and p2 are all prime on his page http://www.worldofnumbers.com/index.html . Just search for...
19022
leavemsg1
Jul 7, 2007 5:09 pm
maybe... let Z = 2^(2^(p+1))+1 ; p is prime Z is prime iff [Z (mod (2^p+1))] == 2^q ; for some q for p = 2, 3,..., next??? eg. p=2, 2^8+1 mod 5 == 2^1 and... ...
19023
Bill Bouris
leavemsg1
Jul 8, 2007 11:02 pm
... maybe if q can only be odd???, then a statement that all Fermat primes greater than F4 are composite may arise??? just tinkering with the idea!...
19024
Werner D. Sand
theo2357
Jul 9, 2007 11:29 am
Thanks for evaluations received. What I mean is: Can sum(ln(n)/n^2 ~ 0.93 reach 1? Can sum(ln(p)/p^2 ~ 0.49 reach 0.5? WDS ... [Non-text portions of this...
19025
N.L.
nluhn
Jul 9, 2007 5:15 pm
Hello members ! Is it possible that exist a new prime-k-tuplet page ? Since some days, I can't reach http://www.ltkz.demon.co.uk/ktuplets.htm best -- Norman ...
19026
Peter Kosinar
pkosinar
Jul 9, 2007 9:49 pm
Hello, ... Consider the inequality sum(ln(k)/k^2, k=A+1..infinity) < integral(ln(k)/k^2, k=A..infinity) The right-hand side can be evaluated as (ln(A)+1)/A....
19027
Sudarshan Iyengar
sudarshansr
Jul 10, 2007 9:29 am
What is the probability of a number being square free... it seems that the more square free numbers than the non-square free numbers for big numbers. Would be...
19028
Jose Ramón Brox
ambroxius
Jul 10, 2007 10:48 am
... The density of numbers being nth-powerfree is 1/zeta(n), where zeta is the Riemann Zeta Function. So, for squarefree numbers, we have your probability is...
19029
Johannes Z.
joz1977
Jul 11, 2007 5:50 pm
I´d like to share a little concept for ACCOUSTICAL FACTORIZATION by using the relation between tones and overtones on strings, many guitarplayers might be...
19030
Adam
a_math_guy
Jul 12, 2007 3:28 pm
Using a similar trick, one can verify the sum over the primes is strictly less than 0.5. One overestimates the tail end of the sum: sum(log(t)/t^2,t=p to 00,...
19031
John W. Nicholson
reddwarf2956
Jul 12, 2007 3:37 pm
Is there a list of the known of Ramanujan prime? What is the largest value?...
19032
scolnik
scolnik2003
Jul 12, 2007 5:45 pm
PHYSICS: Factoring Numbers with Waves Zubairy Science 27 April 2007: 554-555 DOI: 10.1126/science.1140915 Hugo Scolnik A wise man hears one word and...
19033
Johannes Z.
joz1977
Jul 12, 2007 6:40 pm
Hugo Thank You for the hint, I´ll have a look at it soon as I get free access somewhere......
19035
scolnik
scolnik2003
Jul 13, 2007 11:56 am
I was searching for an interesting note I've read years ago. Here it is: http://www.mathpages.com/home/kmath222.htm Hugo Scolnik [Non-text portions of this...
19036
Adam
a_math_guy
Jul 13, 2007 6:17 pm
Whoops! Of course those inequalities are the other way around!! One overestimates the tail end of the sum: sum(log(t)/t^2,t=p to 00, t prime) <...
19037
Werner D. Sand
theo2357
Jul 14, 2007 11:53 am
Fine trick, but what makes you sure the integral is not e.g. = 0.008? Didn't you only shift the problem upon the integral? Werner ... [Non-text portions of...
19038
Peter Kosinar
pkosinar
Jul 14, 2007 2:25 pm
... The integral can be evaluated explicitly, exactly as I did in my original post. I just didn't find this method "nice trick"-y enugh, as it required adding...
19039
Werner D. Sand
theo2357
Jul 16, 2007 7:36 am
... 0.008? ... original ... required ... sum). ... is ... 1000, ... terms ... 0.5. ... You are quite right. I integrated only numerically. Everything clear. ...
19040
Phil Carmody
thefatphil
Jul 18, 2007 1:58 pm
Let p be a large prime, and g a generator of Z/pZ. Let u<p be non-zero modulo small prime q. From g^u (mod p), is it ever possible to tell anything more about...
19041
snudco
Jul 18, 2007 2:43 pm
Hello All, I generated the following fractal plot: http://www.isenbek.com based on the first 10,000 primes. I was wondering if anyone has seen this before, and...
19042
Shane
divineprime
Jul 18, 2007 5:23 pm
The new version of RMA.NET, has been extensively tested, and is now ready for release. This windows program is by far the fastest prime number software ...
19043
Werner D. Sand
theo2357
Jul 21, 2007 1:17 pm
The divergence of sum(1/2^ln n) and sum(1/2^ln p) can easily be proven. How can be proven that sum(1/3^ln n) and sum(1/3^ln p) converge? (IF they do. What are...
19044
Phil Carmody
thefatphil
Jul 21, 2007 2:16 pm
... 3^ln(n) = 3^(log_3(n)*ln(3)) = n^ln(3) sum(1/n^ln(3)) converges as n^ln(3) dominates n*ln(n)^2 whose sum of reciprocals converges. => sum(1/p^ln(3)) also...
19045
Werner D. Sand
theo2357
Jul 22, 2007 9:44 am
... proven. ... the tail ... Thank you. Remains adding that the convergence radius r of sum(1/a^ln n) and sum(1/a^ln p) is r=a>e corresponding to the unity in...
19046
Kermit Rose
kermit1941
Jul 24, 2007 4:34 am
We fill the array with it's index up to the length of the table. For example if we wish to use the Fermat prime number sieve to find the primes < 250, we set...
19047
andrew_j_walker
Jul 29, 2007 1:00 am
Hi, I'm not sure if my email got through to you so I'll post here. I put a message on usenet a few years back ...
19048
Kermit Rose
kermit1941
Jul 29, 2007 12:17 pm
2^(2^m) + 1 mod prime 3 cannot divide any number in this series because (2 -1)^2 + 1 = 2 5 cannot divide any number in this series except 5 because (2 -...
