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k*2^n-1 and k*2^n+1 are twins   Message List  
Reply | Forward Message #16665 of 21093 |
Re: k*2^n-1 and k*2^n+1 are twins

--- In primenumbers@yahoogroups.com, "jbrennen" <jack@b...> wrote:
> Can anybody find a value of k which yields more twin primes than
> k=202507305 (3*5*7*11*13*13487) ?
>
> When k=202507305, k*2^n+/-1 are twin primes for n:
>
> 2, 12, 17, 28, 31, 33, 42, 55, 62, 86, 89, 91
>
> (and most likely for no other values of n)
>
> At the time that I found this one, I remember searching far and
> wide for a "better" k with more than 12 twin primes, with no luck.
>
> If you find a value of k with more than 12 twin primes, please
> let me know!

Jack

This goes to show it sometimes takes a while to reply to posts, in
this case 2 1/2 years!

In any case the following k has 13 twins to n=10000

7985650262654529465

And the twins are n=1,3,17,37,38,39,70,97,485,556,561,1082,1086

So there is a nice bitwin length 3 in there are well.

This resulted in a new search I have started to determine highest
scoring k.2^n+ & - 1 series, where 1 point is awarded for each twin
or cunningham chain length 2. (a CC length 3 gets two points, etc).

The k quoted has 13 twins, and 15 points for cunningham chains (up
to n=10000), so 28 is a nice easy target to beat.

Regards

Robert Smith







Sat May 21, 2005 2:42 pm

robert44444uk
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Message #16665 of 21093 |
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Hello! More than a year ago there were some messages about "Proth weight" and "Nash weight", i.e. the weight of k for k*2^n+1 and k*2^n-1 respectively. As for...
Andrey Kulsha
andrey_601
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Nov 18, 2002
7:47 pm

... [SNIP] ... Your observation is correct. The reasons why the ks have various Proth/Nash weights is because various subsets of ns are forbidden from being...
Phil Carmody
thefatphil
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Nov 18, 2002
10:48 pm

... 'PSieve', a program that Chris Nash and I worked on quite a while ago, enabled the weights of many things to be investigated, including twins. In fact, at ...
Paul Jobling
paul_joblinguk
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Nov 19, 2002
11:31 am

Can anybody find a value of k which yields more twin primes than k=202507305 (3*5*7*11*13*13487) ? When k=202507305, k*2^n+/-1 are twin primes for n: 2, 12,...
jbrennen
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Nov 19, 2002
6:04 pm

... Jack This goes to show it sometimes takes a while to reply to posts, in this case 2 1/2 years! In any case the following k has 13 twins to n=10000 ...
Robert
robert44444uk
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May 21, 2005
2:43 pm

... That's a pretty good one, and the last 5 of those exponents... wow! I would have expected my "record" to be beaten by a k value with an abundance of small...
Jack Brennen
jbrennen
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May 22, 2005
1:30 pm

... in ... wow! I know, extraordinary to get 5 after n=484 ... I agree, but they are not so easy to find - 10 twins to n=100 is not too hard, but getting those...
Robert
robert44444uk
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May 22, 2005
4:10 pm

... See also my announcement on NMBRTHRY a couple of years back. I believe I injected the 15-twin k into a usenet post at about the same time, if you really...
thefatphil
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May 23, 2005
11:02 am

... Ah, I smell a Sierpinski-with-Riesel problem. Solution: k=5, with a covering set {3}, not so? David...
David Broadhurst
djbroadhurst
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Nov 18, 2002
11:55 pm

... Yes, 5*2^n-1 and 5*2^n+1 are never both prime. :-) The same is true of all k>3 which are not divisible by 3. The interesting value of k is k=111, which is...
Jack Brennen
jbrennen
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Nov 19, 2002
3:06 am

... I should clarify that when extending the Sierpinski problem to twin primes, we must impose the additional requirement that k be divisible by 3 -- without...
Jack Brennen
jbrennen
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Nov 19, 2002
3:35 am

... Elizabeth Regina I might have protested at that "must", but I bow to the good humour of ... But my word, your non-trivial case ABC2 111*2^(36*$a-25)+1 |...
David Broadhurst
djbroadhurst
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Nov 19, 2002
5:38 am

... Doing the same with k=1341 (where n must be equal to 21 mod 36): ABC2 1341*2^(36*$a-15)+1 | 1341*2^(36*$a-15)-1 a: from 1 to 200 1341*2^(36*14-15)+1 ...
Paul Jobling
paul_joblinguk
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Nov 19, 2002
11:51 am

Has anybody thought of taking this problem sideways in a dual Sierpinski problem manner? i.e. is there a k such that 2^n+k and 2^n+k+2 (or 2^n+k-2 if you want...
Gary Chaffey
garychaffey
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Nov 19, 2002
12:32 pm

... Surely the dual would be k such that 2^n+k and 2^n-k are never prime? Paul. __________________________________________________ Virus checked by MessageLabs...
Paul Jobling
paul_joblinguk
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Nov 19, 2002
12:39 pm
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