There cannot be another:

The first cluster (3,5,7,11,13,17) cannot recur because the

initial triple of primes (3,5,7) = (n, n+2, n+4) must

contain a multiple of 3. The only way to arrange 6 primes

into such a dense cluster without having such a triple is

the second cluster (5,7,11,13,17,19) = (n, n+2, n+6, n+8,

n+12, n+14) which must contain a multiple of 5. To see why,

take each term mod 5.

Mark Underwood wrote:

>

> Hi Tavai

>

> That seems intuitive to me as well, so I was surprised that for 6

> consecutive primes it *seems* that (3,5,7,11,13,17) and

> (5,7,11,13,17,19) are tied for second densest and that there are no

> others. One would think that if there are an infinite number of six

> primes at this density then another could be found fairly easily. Do

> you know of another in this case?

>

> Mark

>

> --- In primenumbers@y..., "Tavai A. Mitrytu" <jtpk@y...> wrote:

> > I have a similar running conjecture which states:

> > "Any finite arrangement of primes which occurs at

> > least twice repeats infinitely." You have to have a

> > repeat to get a minimum period.

>

>

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>

>

>

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