Mersenne Digest Tuesday, April 10 2001 Volume 01 : Number 838
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Date: Sat, 7 Apr 2001 21:51:36 +0200 (MET DST)
From: [EMAIL PROTECTED]
Subject: Re: Mersenne: arithmetic progression of consecutive primes
> From: [EMAIL PROTECTED]
> In a message dated 05/04/2001 05:19:38 GMT Daylight Time, "Gary Untermeyer"
> <[EMAIL PROTECTED]> writes:
>
> > Let x be a prime number. Consider the series of numbers that take the
> > following form:
> >
> > x, x + n, x + 2n, x + 3n, x + 4n, x + 5n, x + 6n, where n is an
> > even positive whole number.
> >
> > In this series of seven numbers, can anyone tell me why, if ALL of these
> > numbers are prime, that the minimum value of n is 210 if all the terms
> > are _consecutive_ prime numbers?
>
> Inverting the argument, if the sequence has N terms, then n must be a
> multiple of all the primes <= N. So in your example (N = 7), n must be a
> multiple of 2, 3, 5 and 7, i.e. of 210. This is still the case for N = 8
> (your next case), 9, and 10. For N=11, n must be a multiple of 210*11=2310.
>
> Note that this is true for _all_ arithmetic progressions of primes, not just
> for progressions of _consecutive_ primes.
We need the additional hypothesis that the least prime exceeds N.
The primes 5, 53, 101, 149, 197 are in arithmetic progression,
but the difference (48) is not divisible by 5.
Likewise for 7, 157, 307, 457, 607, 757, 907.
Peter Montgomery
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Date: Tue, 10 Apr 2001 17:42:35 +0200
From: "Tobias" <[EMAIL PROTECTED]>
Subject: Mersenne: Test
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This is only a test to see if it works
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End of Mersenne Digest V1 #838
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