## Re: k*2^n-1 and k*2^n+1 are twins

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• ... 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
Message 1 of 15 , May 21, 2005
--- 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
• ... 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
Message 2 of 15 , May 22, 2005
Robert wrote:
> 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

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 twin primes (n <= 100) -- just because it's so
much easier to selectively prune the search tree. I'm sure that my
hypothetical search algorithm would have given up on this k value
once it found only 8 twins with n <= 100.

The factorization of this k:

3 5 7 11 13 19 29 37 251 1009 103007

Any significance to that? I'm not surprised to see the abundance
of small primes, but the product of the "larger" primes in the list
(251 * 1009 * 103007 == 26087449813) seems pretty large for you to
have done a brute force search for a large number of twins...
Maybe you were sieving for bi-twin chains and came across this
lucky coefficient???

Good find, regardless of how or why you found it. :)
• ... 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
Message 3 of 15 , May 22, 2005
--- In primenumbers@yahoogroups.com, Jack Brennen <jb@b...> wrote:
> Robert wrote:
> > 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
>
> That's a pretty good one, and the last 5 of those exponents...
wow!

I know, extraordinary to get 5 after n=484

> I would have expected my "record" to be beaten by a k value with an
> abundance of small twin primes (n <= 100) -- just because it's so
> much easier to selectively prune the search tree. I'm sure that my
> hypothetical search algorithm would have given up on this k value
> once it found only 8 twins with n <= 100.
>
I agree, but they are not so easy to find - 10 twins to n=100 is not
too hard, but getting those to be twins after 100 is a pig.

> The factorization of this k:
>
> 3 5 7 11 13 19 29 37 251 1009 103007
>
> Any significance to that?

Yes, this is a Payam M(37-) E(51-) number, meaning it is
3*5*11*13*19*29*37 = M(37), and in fact has no factors with order
base 2 of less than 52, for the k.2^n-1 series. In this case it is
also not that divisible on the + side either.

But there are zillions of M(37-) numbers, they will always be really
prime, but sometimes they are also prime on the + side, as in this
case, because no small factors appear there either. When this is the
case, a happy coincidence will occasionally occur, and the twins and
cunninghams line up.

My score of 28 (twins +cunninghams) is really a good target to beat.
Been going all day and only got one 24.

Regards

Robert Smith
• ... 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
Message 4 of 15 , May 23, 2005
--- In primenumbers@yahoogroups.com, Jack Brennen <jb@b...> wrote:
> Robert wrote:
> > 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
>
> 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 twin primes (n <= 100) -- just because it's so
> much easier to selectively prune the search tree. I'm sure that my
> hypothetical search algorithm would have given up on this k value
> once it found only 8 twins with n <= 100.
>
> The factorization of this k:
>
> 3 5 7 11 13 19 29 37 251 1009 103007
>
> Any significance to that? I'm not surprised to see the abundance
> of small primes, but the product of the "larger" primes in the list
> (251 * 1009 * 103007 == 26087449813) seems pretty large for you to
> have done a brute force search for a large number of twins...
> Maybe you were sieving for bi-twin chains and came across this
> lucky coefficient???
>
> Good find, regardless of how or why you found it. :)