This article is rated Start-class on Wikipedia's
content assessment scale. It is of interest to the following WikiProjects: | |||||||||||
|
Daily pageviews of this article
A graph should have been displayed here but
graphs are temporarily disabled. Until they are enabled again, visit the interactive graph at
pageviews.wmcloud.org |
I do not know where the claim that a Riesel number is the same as a Woodall number comes from.
A Riesel number is a number k such that the sequence k*2n-1 contains no primes. They are analogous to Sierpinski numbers with the plus sign replaced by a minus sign. —Preceding unsigned comment added by 203.194.48.245 ( talk) 15:52, 12 June 2005
Who is the Suyama who submitted a proof? Is it the videogame and anime voice actor, or the Australian company that distributes generators and motors? If it's a mathematician, someone who knows could add his first name and link to the article yet to be created. —Preceding unsigned comment added by 70.50.160.69 ( talk) 15:55, 31 May 2006
This launches straight into the history with little elaboration. Can it be further explained or linked to articles which will help the reader better understand the concept? —Preceding unsigned comment added by 168.158.220.3 ( talk • contribs) 22:38, 16 September 2010
"...is a natural number of the form n × 2n − 1 (written Wn)."
As a non-expert, this raises several questions for me.
I understand that math concepts can't always be explained to an elementary level, but I feel like this article leaves no room to research its meaning. 168.158.220.3 ( talk) 20:23, 17 September 2010 (UTC)
sligocki had made some recent changes which help to clarify the concepts. To work in his changes, perhaps the following collaboration?
I was also thinking that an example would help. Perhaps on the next line:
I don't know, just a thought. Otherwise, I think the changes are very helpful. 168.158.220.3 ( talk) 00:55, 20 September 2010 (UTC)
... Woodall's or Riesel's theorem (as it should be named): ... let R= k*2^n -1, n >2 is a natural number and k <= 2^n-1. If for some 'b', b^((R-1)/2) == +1 (mod R), then 'R' is prime; (similar to Proth's theorem, but gcd(b, 3)= 1 is necessary!) ... proof: if 'm' is from the set of natural numbers, then every odd prime divisor 'r' of a^(2^m) +1 implies that q == +1(mod a^(m+1)) [concluded from generalized Fermat-number 'proofs' by Proth, and with me examining Proth's theorem]. ... now, if 'q' is any prime divisor of 'S', then b^((R-1)/2)= (b^k)^(2^(n-1)) == +1 (mod q) implies that q == +1 (mod 2^n). ... thus, if 'S' is composite, 'S' will be the product of at least two primes each of which may have a maximum value of (2^n -1), and it follows that... ... k*2^n +1 = (2^n)*(2^n) -1 = (2^n +1)*(2^n -1) <= (2^n)*(2^n) -2*(2^n) +1; implies that -2 <= -2*(2^n) and dividing by 2^n and multiplying by -1, we have.. 1 > 2^n, contradiction! ... hence, for some 'b', if k <= 2^n -1 and b^((R-1)/2) == +1 (mod R), then 'R' is prime.
... I didn't violate any copyrights; it's MY modification that a high school student could verify. Thanks,Bill; my e-mail is leavemsg1@yahoo.com
The numerator of the number (−n)*2−n − 1 is 2n + n, the first few of them are
Numbers n such that 2n + n is prime are
The dual Woodall prime themselves are
(−n)*2−n − 1 = (2n + n)/(2n), and the numerator of this number is 2n + n
In the section titled Woodall primes: "In 1976 Christopher Hooley showed that almost all Cullen numbers are composite. Hooley's proof was reworked by Hiromi Suyama to show that it works for any sequence of numbers n · 2n+a + b where a and b are integers, and in particular also for Woodall numbers. Nonetheless, it is conjectured that there are infinitely many Woodall primes."
I performed a Google search on "Hiromi Suyama" and I can't seem to find the proof anywhere. Most of the search results I could find are just duplicates of the Wikipedia page or an older version of it. Has it really been proven that most Woodall numbers are composite, and if so, was it proven recently (within the last ten years)? Notice that this edit modified the sentence, "It is conjectured that almost all Woodall numbers are composite; a proof has been submitted by H. Suyama, but it has not been verified yet" to what it reads today. We may need to delete this sentence from the article if we cannot find sources to verify it. 173.76.101.2 ( talk) 22:29, 14 October 2018 (UTC)
In the mentioned section, the article states
and in a later sentence that
That is a contradiction—except if almost all doesn't mean all but finitely many in the natural numbers; then that must be mentioned. I'm going ahead and remove the second sentence since the first one has a credible source that I cannot easily check. Spezialist( talk) 19:21, 12 July 2021 (UTC)