David wrote:

> Here is the first set of 9:

>

> {v=[

> 206710003, 206710013, 206710019,

> 206710027, 206710037, 206710043,

> 206710051, 206710061, 206710067];

> print(vector(#v-1,k,v[k+1]-v[k]));}

>

> [10, 6, 8, 10, 6, 8, 10, 6]

A gap cycle of 6, 8, 10 in some order can only get one longer than

this before becoming inadmissible modulo 5.

If we want more than 10 primes then we need other gaps.

Puzzle:

If (a, b, c) and (b, c, d) are Pythagorean triples then is a=d?

I assume so in the following.

This means all solutions must have a gap cycle of 3 numbers.

If 5 divides the sum of the 3 numbers then the cycle never

becomes inadmissible modulo 5.

(10, 24, 26) is the smallest even Pythagorean triple with sum

divisible by 5.

The first case of 12 consecutive primes for that cycle in some order:

490537270893409 + d,

for d = 0, 10, 34, 60, 70, 94, 120, 130, 154, 180, 190, 214.

The gaps are 10, 24, 26, 10, 24, 26, 10, 24, 26, 10, 24.

A few hundred cases of 12 non-consecutive primes were found

during the search.

I suspect the above is the first case of 12 consecutive primes

with any gaps.

A cycle with (10, 24, 26) cannot give more than 12 primes because

it becomes inadmissible modulo 7.

In order to get an admissible pattern with 13 primes I think we

would need gaps of a size which makes it very hard to find

consecutive primes.

If the k-tuple conjecture (*) is true then there are arbitrarily

long solutions.

For example, (3*n#, 4*n#, 5*n#) is a Pythagorean triple for all n,

and the gaps never become inadmissible modulo primes up to n.

(*) There are variations of the k-tuple conjecture.

We only need one saying that all admissible patterns have at least

one occurrence.

Then it follows that they also have occurrences with consecutive primes.

--

Jens Kruse Andersen