- 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! - --- In primenumbers@yahoogroups.com, "jbrennen" <jack@b...> wrote:
> Can anybody find a value of k which yields more twin primes than

Jack

> 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!

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 - Robert wrote:
> This goes to show it sometimes takes a while to reply to posts, in

That's a pretty good one, and the last 5 of those exponents... wow!

> 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

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 primenumbers@yahoogroups.com, Jack Brennen <jb@b...> wrote:
> Robert wrote:

in

> > This goes to show it sometimes takes a while to reply to posts,

> > this case 2 1/2 years!

wow!

> >

> > 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...

I know, extraordinary to get 5 after n=484

> I would have expected my "record" to be beaten by a k value with an

I agree, but they are not so easy to find - 10 twins to n=100 is not

> 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.

>

too hard, but getting those to be twins after 100 is a pig.

> The factorization of this k:

Yes, this is a Payam M(37-) E(51-) number, meaning it is

>

> 3 5 7 11 13 19 29 37 251 1009 103007

>

> Any significance to that?

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 - --- In primenumbers@yahoogroups.com, Jack Brennen <jb@b...> wrote:
> Robert wrote:

See also my announcement on NMBRTHRY a couple of years back.

> > 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. :)

I believe I injected the 15-twin k into a usenet post at about the

same time, if you really want its numerical value.

Phil