## Re: Special Set of Eleven Consecutive Primes

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• 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 1 of 5 , Jul 9, 2011
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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 2 of 5 , Jul 9, 2011
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"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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