- Here is a set of 11 consecutive primes consisting of 8 contiguous subsets

of 4 primes. Each subset has 3 distinct differences (gaps) between

adjacent primes and 4 distinct rightmost digits.

(2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287).

For example, the first subset of 4 primes is (2213, 2221, 2237, 2239).

The 3 distinct gaps are (8, 6, 2). The 4 distinct rightmost digits are

(3, 1, 7, 9). The next contiguous subset is (2221, 2237, 2239, 2243) with

distinct gaps (6, 2, 4) and distinct rightmost digits (1, 7, 9, 3). And

so on for 8 subsets.

Can theory hint where on the number line one is most likely to find sets

like this? Anyone care to try to find a bigger one? Thanks folks.

Bill Sindelar

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w_sindelar@... wrote:

> Here is a set of 11 consecutive primes consisting

Why stop at 11? Here is the same sort of thing,

> of 8 contiguous subsets of 4 primes. Each subset

> has 3 distinct differences (gaps) between adjacent primes

> and 4 distinct rightmost digits.

> (2213, 2221, 2237, 2239,

> 2243, 2251, 2267, 2269,

> 2273, 2281, 2287).

but with 12 consecutive primes:

[2123971, 2123983, 2123999, 2124007,

2124011, 2124013, 2124019, 2124037,

2124041, 2124043, 2124049, 2124127]

David - Bill Sindelar wrote:
> Here is a set of 11 consecutive primes consisting of 8 contiguous subsets

The ending digit must go through a cycle of length 4, for example the above

> of 4 primes. Each subset has 3 distinct differences (gaps) between

> adjacent primes and 4 distinct rightmost digits.

> (2213, 2221, 2237, 2239, 2243, 2251, 2267, 2269, 2273, 2281, 2287).

> Can theory hint where on the number line one is most likely to find sets

> like this? Anyone care to try to find a bigger one? Thanks folks.

3179 3179 ...

If the gap requirement is satisfied modulo 10 for the cycle then

it will automatically be satisfied for the whole sequence.

The record setters up to 10^12 except length 19 are all of this type.

12:

2123971

2123983

2123999

2124007

2124011

2124013

2124019

2124037

2124041

2124043

2124049

2124127

13:

66945301

66945313

66945349

66945407

66945421

66945443

66945449

66945467

66945491

66945503

66945509

66945517

66945521

14:

66945301

66945313

66945349

66945407

66945421

66945443

66945449

66945467

66945491

66945503

66945509

66945517

66945521

66945553

15:

184171621

184171643

184171679

184171697

184171711

184171733

184171739

184171747

184171751

184171753

184171759

184171777

184171781

184171843

184171849

16:

2493412841

2493412847

2493412849

2493412853

2493412891

2493412937

2493412939

2493413023

2493413051

2493413057

2493413059

2493413063

2493413071

2493413077

2493413099

2493413113

17:

2732087209

2732087227

2732087231

2732087263

2732087279

2732087317

2732087321

2732087333

2732087359

2732087447

2732087461

2732087483

2732087489

2732087537

2732087551

2732087563

2732087569

18:

10206934519

10206934523

10206934541

10206934567

10206934579

10206934583

10206934601

10206934607

10206934609

10206934613

10206934631

10206934697

10206934709

10206934723

10206934741

10206934757

10206934769

10206934783

19:

554804585227

554804585233

554804585299

554804585311

554804585347

554804585393

554804585449

554804585471

554804585507

554804585513

554804585579

554804585651

554804585737

554804585743

554804585779

554804585831

554804585837

554804585873

554804585899

20:

692245399661

692245399687

692245399709

692245399723

692245399751

692245399787

692245399789

692245399793

692245399801

692245399837

692245399849

692245399873

692245399891

692245399907

692245399919

692245399943

692245399991

692245399997

692245400009

692245400023

--

Jens Kruse Andersen - Thank you David, and thank you Jens.You never cease to astound me. In the 9 examples you so quickly calculated, the sets 12, 13, 14 and 15 have the pattern (1, 3, 9, 7) for the beginning set of 4 consecutive primes. For sets 16 and 20 the pattern is (1, 7, 9, 3). For set 11 the pattern is (3, 1, 7, 9). For set 19 the pattern is ((7, 3, 9, 1). For set 18 the pattern is (9, 3, 1, 7). For set 17 the pattern is (9, 7, 1, 3).

There are 24 possible patterns into which the 4 digits 1, 3, 7 and 9 can be arranged, so that each pattern contains distinct digits. The sets 11 to 20 account for six.

The obvious question comes up. Is it possible to find additional sets 21, 22, 23, 24...N, such that for the sets 11 to N all 24 possible patterns are used in the begining sets of 4 consecutive primes?

Bill Sindelar

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"Jens Kruse Andersen" <jens.k.a@...> wrote:

> The ending digit must go through a cycle of length 4 ...

That was indeed my methodology, adopted

> If the gap requirement is satisfied modulo 10 for the cycle then

> it will automatically be satisfied for the whole sequence.

in order quickly to attain length 12,

with gaps in Z/Z10 given by

[2, 6, 8, 4, 2, 6, 8, 4, 2, 6, 8]

It's truly impressive that Jens attained length 20.

Moreover, he had the foresight to study gaps in Z.

Videlicit:

> 19:

where the gaps in Z/Z10 are

> 554804585227

> 554804585233

> 554804585299

> 554804585311

> 554804585347

> 554804585393

> 554804585449

> 554804585471

> 554804585507

> 554804585513

> 554804585579

> 554804585651

> 554804585737

> 554804585743

> 554804585779

> 554804585831

> 554804585837

> 554804585873

> 554804585899

[6, 6, 2, 6, 6, 6, 2, 6, 6, 6, 2, 6, 6, 6, 2, 6, 6, 6]

but are still OK in Z:

[6, 66, 12, 36, 46, 56, 22, 36, 6, 66, 72, 86, 6, 36, 52, 6, 36, 26]

David (well and truly outdone by Jens, as so often before)