Phil Carmody wrote:

> I specifically wanted to offer the choice. I wanted the chains

> to be longer, and also less simple to find.

With the choice we quickly get a chain with 12 sums:

281 + 283 + 293 = 857

857 + 859 + 863 = 2579

2551 + 2557 + 2579 = 7687

7681 + 7687 + 7691 = 23059

23059 + 23063 + 23071 = 69193

69193 + 69197 + 69203 = 207593

207569 + 207589 + 207593 = 622751

622751 + 622777 + 622781 = 1868309

1868287 + 1868291 + 1868309 = 5604887

5604881 + 5604887 + 5604901 = 16814669

16814669 + 16814671 + 16814701 = 50444041

50444027 + 50444041 + 50444059 = 151332127

The longest chain with initial primes below 10^11 is probably 14 sums:

2878090951 + 2878090961 + 2878090967 = 8634272879

8634272839 + 8634272879 + 8634272891 = 25902818609

25902818609 + 25902818629 + 25902818663 = 77708455901

77708455819 + 77708455823 + 77708455901 = 233125367543

233125367543 + 233125367599 + 233125367609 = 699376102751

699376102751 + 699376102757 + 699376102783 = 2098128308291

2098128308239 + 2098128308273 + 2098128308291 = 6294384924803

6294384924787 + 6294384924803 + 6294384924823 = 18883154774413

18883154774413 + 18883154774441 + 18883154774447 = 56649464323301

56649464323301 + 56649464323303 + 56649464323307 = 169948392969911

169948392969877 + 169948392969881 + 169948392969911 = 509845178909669

509845178909611 + 509845178909669 + 509845178909687 = 1529535536728967

1529535536728967 + 1529535536728987 + 1529535536728999 = 4588606610186953

4588606610186927 + 4588606610186947 + 4588606610186953 = 13765819830560827

There are also 14 starting with

44453980303 + 44453980309 + 44453980379 = 133361940991

> I presume you just ran a prime sieve and then probed for each one

> with no deduplication?

I didn't bother with deduplication but I ran two prime sieves in

parallel, around p and 3p. This meant the first sum never had to be prp

tested, and its surrounding primes were also generated without prp tests.

It could be extended to a third sieve around 9p and so on, but each

extra sieve has to sieve a larger interval and avoids fewer prp tests.

> I hadn't considered Dickson. It's the big gun, certainly.

> I wondered initially if it would fail on guaranteeing that the

> positioned primes are consecutive, but if you ask for a triplet

> {p,p+3+/-1,p+6}, then that comes for free.

Dickson's conjecture can be manipulated to ensure consecutive primes

in this problem for any admissible pattern. We can just demand extra

primes chosen so specific prime factors are forced to divide certain

numbers in a gap.

For example, we can ensure consecutive {p, p+6} by demanding four primes

{p, p+6, p+8, p+14}. This forces 3 to divide p+4, and 5 to divide p+2.

--

Jens Kruse Andersen