Talk:Woodall number

From Wikipedia, the free encyclopedia
Jump to: navigation, search
WikiProject Mathematics (Rated Stub-class, Mid-priority)
WikiProject Mathematics
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
Stub Class
Mid Priority
 Field: Number theory

Riesel number[edit]

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

Suyama[edit]

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

I don't know either but the result appears to follow by the method used in Hooley's book (pages 115-119) for the same result on Cullen numbers. Richard Pinch (talk) 18:57, 30 July 2008 (UTC)
I found a freely available copy of the full article by Wilfrid Keller, "New Cullen Primes" at ams.org (the one originally linked to required subscription). It explains the basics of the proof by Christopher Hooley which shows that almost all Cullen numbers are composite, and a rework of this proof by Hiromi Suyama which (among other things) shows that this is also true for Woodall numbers. I updated the article accordingly, and changed the reference to the new URL. Adcoon (talk) 07:54, 5 September 2010 (UTC)

Huh[edit]

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 (talkcontribs) 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.

  • 'n x 2n - 1' appears to be an expression. How is it the form of a number?
  • 'written Wn'. Is 'written' an adjective or a verb, in this case? Can this be reworded "written as 'Wn'"?
  • Is "(written Wn)" referring to the expression or the Woodall number? Can this parenthesis be moved to the introduction?

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)

I think you will see this kind of thing for a lot of similar articles. An expression like n × 2n − 1 is usually thought to be pretty self-exlanatory, for example that n is some arbitrary natural number. But if I were to suggest an improvement, perhaps the following is better?
In number theory, a Woodall number is a natural number of the form
Wn=n × 2n − 1
where n is a natural number. The first few Woodall numbers are:
1, 7, 23, 63, 159, 383, 895, … (sequence A003261 in OEIS).
Woodall numbers were first studied by Allan J. C. Cunningham and H. J. Woodall in 1917, inspired by James Cullen's earlier study of the similarly-defined Cullen numbers. Woodall numbers curiously arise in Goodstein's theorem.[citation needed]
It would replace the first paragraph of the current article. Also, the article for Cullen numbers has pretty much the same style, so if this article is changed then Cullen numbers should likewise be changed. Adcoon (talk) 12:27, 18 September 2010 (UTC)
That's actually very helpful, thank you. I understand the concepts now.
I have a few suggestions for clarity, please let me know if they work:
In number theory, a Woodall number(Wn) is any natural number which fits the form
Wn=n × 2n − 1
where n is also a natural number.
Otherwise, I am in full support of the change. I think it is much more clear to a lay person. Thanks :) 168.158.220.3 (talk) 00:48, 20 September 2010 (UTC)


sligocki had made some recent changes which help to clarify the concepts. To work in his changes, perhaps the following collaboration?

In number theory, a Woodall number (Wn) is any natural number equal to
n × 2n − 1
for some natural number n.

I was also thinking that an example would help. Perhaps on the next line:

For example, 7 is a Woodall number because it fits the equation
2 x 2(2) - 1

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)

Fine by me, please WP:BE BOLD and edit the page to make it clearer :) Then other people can critique your change by changing it further and if we run into a conflict where we cannot agree on a way to say things, we can discuss more here. — sligocki (talk) 01:19, 20 September 2010 (UTC)
Went ahead and changed the page. Tried to add in the various suggestions. I don't think it is necessary to give a specific example since the list of the first Woodall numbers serves that purpose. But if you think it needs to be that specific, then just add it in. Also switched around two paragraphs; the one about almost all woodall numbers being composite seemed to fit better after the one about woodall primes. Adcoon (talk) 07:58, 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.

  • QED

... 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