## Re: Special Set of Eleven Consecutive Primes

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• ... Why stop at 11? Here is the same sort of thing, but with 12 consecutive primes: [2123971, 2123983, 2123999, 2124007, 2124011, 2124013, 2124019, 2124037,
Message 1 of 5 , Jul 8 8:16 AM
w_sindelar@... wrote:

> 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).

Why stop at 11? Here is the same sort of thing,
but with 12 consecutive primes:

[2123971, 2123983, 2123999, 2124007,
2124011, 2124013, 2124019, 2124037,
2124041, 2124043, 2124049, 2124127]

David
• ... The ending digit must go through a cycle of length 4, for example the above 3179 3179 ... If the gap requirement is satisfied modulo 10 for the cycle then
Message 2 of 5 , Jul 8 9:10 AM
Bill Sindelar wrote:
> 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).

> 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.

The ending digit must go through a cycle of length 4, for example the above
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,
Message 3 of 5 , Jul 9 6:50 AM
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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• ... That was indeed my methodology, adopted 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
Message 4 of 5 , Jul 9 3:57 PM
"Jens Kruse Andersen" <jens.k.a@...> wrote:

> The ending digit must go through a cycle of length 4 ...
> If the gap requirement is satisfied modulo 10 for the cycle then
> it will automatically be satisfied for the whole sequence.

That was indeed my methodology, adopted
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:
> 554804585227
> 554804585233
> 554804585299
> 554804585311
> 554804585347
> 554804585393
> 554804585449
> 554804585471
> 554804585507
> 554804585513
> 554804585579
> 554804585651
> 554804585737
> 554804585743
> 554804585779
> 554804585831
> 554804585837
> 554804585873
> 554804585899

where the gaps in Z/Z10 are

[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)
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