mersenneforum.org Riesel primes
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 2009-07-18, 02:52 #1 Primeinator     "Kyle" Feb 2005 Somewhere near M50..sshh! 15768 Posts Riesel primes For Riesel primes of the form k*2^n -1, I know K is an odd integer. What are the restrictions for the value of n?
 2009-07-18, 02:54 #2 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17·251 Posts I'm pretty sure it's only that n is such that 2^n>k (otherwise every number would be a Riesel number, and every prime a Riesel prime). Last fiddled with by Mini-Geek on 2009-07-18 at 02:55
2009-07-18, 05:54   #3
Primeinator

"Kyle"
Feb 2005
Somewhere near M50..sshh!

2×3×149 Posts

Quote:
 Originally Posted by Mini-Geek I'm pretty sure it's only that n is such that 2^n>k (otherwise every number would be a Riesel number, and every prime a Riesel prime).
How would 2^n < k in of itself make a number riesel? I thought that any number of the form k*2^n -1 is a riesel number.

2009-07-18, 12:34   #4
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts

Quote:
 Originally Posted by Primeinator How would 2^n < k in of itself make a number riesel? I thought that any number of the form k*2^n -1 is a riesel number.
I was mistaken about every number being represented by some k*2^n-1. Every odd number can be represented by k*2^n-1 if 2^n can be less than k. Let's choose 15648132147819545 for example.
15648132147819545=k*2^n-1
15648132147819546=k*2^n
15648132147819546 has only one factor of 2, so we set n to 1
2*7824066073909773=k*2
7824066073909773=k
So 15648132147819545=7824066073909773*2^1-1

In case you're wondering, there is an identical requirement for Proth (k*2^n+1) numbers.
http://en.wikipedia.org/wiki/Proth_number

2009-07-18, 15:23   #5
Primeinator

"Kyle"
Feb 2005
Somewhere near M50..sshh!

89410 Posts

Quote:
 Originally Posted by Mini-Geek I was mistaken about every number being represented by some k*2^n-1. Every odd number can be represented by k*2^n-1 if 2^n can be less than k. In case you're wondering, there is an identical requirement for Proth (k*2^n+1) numbers. http://en.wikipedia.org/wiki/Proth_number
Now that makes sense. And of course, with the exception of two, being odd is a necessity for any number x being a Riesel prime.

However, you can still get an odd number by having n < k. A simple example:

3*2^2 -1 = 11 = ! prime.
15*2^3 - 1 = 119 = ! prime

Any number of k*2^n -1 will be odd, given k is odd. For n > k a prime can turn up. The first is k =3 and n = 6 which yields the following
3*2^6 - 1 = 191 = ! prime.

Finally, n = k is also a possibility. 3*2^3 -1 = 23 = ! prime.

Apparently, all three scenarios are possible?

 2009-07-18, 15:50 #6 kar_bon     Mar 2006 Germany B3216 Posts if you search for a comprehensive collection of Riesel primes have a look at www.rieselprime.de BTW: n=k are called Woodall primes (see above link also) Last fiddled with by kar_bon on 2009-07-18 at 15:51
2009-07-18, 16:05   #7
Primeinator

"Kyle"
Feb 2005
Somewhere near M50..sshh!

11011111102 Posts

Quote:
 Originally Posted by kar_bon if you search for a comprehensive collection of Riesel primes have a look at www.rieselprime.de BTW: n=k are called Woodall primes (see above link also)
From this site it appears that N is always bigger than k. However, even though the opposite can still produce a prime, is it considered a Riesel prime?

2009-07-18, 16:19   #8
kar_bon

Mar 2006
Germany

54628 Posts

Quote:
 Originally Posted by Primeinator From this site it appears that N is always bigger than k. However, even though the opposite can still produce a prime, is it considered a Riesel prime?
mhhh?

have a look at the range for example 2000<k<4000 in the Data Section.
only k=2001 got 24 primes with n<k!!!

2009-07-18, 16:49   #9
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts

Remember that the restriction is 2^n>k, not n>k, so e.g. n=4 k=15 is allowed (2^4=16, 16>15) 15*2^4-1=239 which is prime. www.rieselprime.de lists primes when 2^n<k, even though these aren't technically Riesel numbers.
Quote:
 Originally Posted by Primeinator Finally, n = k is also a possibility. 3*2^3 -1 = 23 = ! prime.
This might interest you: n*2^n+1 is a Cullen Number. (essentially a Proth number with k=n and no 2^n>k restriction; the Proth side equivalent of a Woodall number)
Quote:
 Originally Posted by Primeinator Apparently, all three scenarios are possible?
Apparently.

So...
When k*2^n+1 is prime, k*2^n+1 is called a Proth prime
When k*2^n-1 is prime, k*2^n-1 is called a Riesel prime
When k*2^n+1 is composite for every n with this specific k, k is called a Sierpinski number
When k*2^n-1 is composite for every n with this specific k, k is called a Riesel number
When k*2^n+1 with odd k, positive integer n, and 2^n>k, k*2^n+1 is called a Proth number
When k*2^n-1 with odd k, positive integer n, and 2^n>k, k*2^n-1 is called ...what? (we're referring to it as Riesel number here, but that's technically incorrect since that refers to the equiv. of a Sierpinski number)

or in text: "Riesel number" technically refers to a k such that all k*2^n-1 are composite, and "Riesel prime" refers to primes of the form k*2^n-1, right? Is there any name for numbers of the form k*2^n-1, analogous to "Proth number" for numbers of the form k*2^n+1? I know there is rarely confusion, at least in projects that aren't searching for Riesel numbers, but it is still an incorrect and vague reference.

2009-07-19, 19:14   #10
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by Primeinator How would 2^n < k in of itself make a number riesel? I thought that any number of the form k*2^n -1 is a riesel number.
Every single integer is of the form k*2^n - 1.

2009-07-19, 23:14   #11
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17×251 Posts

Quote:
 Originally Posted by R.D. Silverman Every single integer is of the form k*2^n - 1.
Is this assuming k and n are integers? If so, I don't see how this could be.

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