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For example

3 consecutive primes in arithmetic progression

3,5,7  distance 2
151,157,163 distance 6

4 consecutive primes in arithmetic progression

251,257,263,269 distance 6
1741,1747,1753,1753 distance 6
76543,76561,76579,76597 distance 18

Are there five consecutive primes in arithmetic progression?

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2 Answers 2

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Yes for the lengths you ask for this is known to exist.

The minimal example is $9843019+ 30 n$ for $n=0,1,2,3,4$ (taken from the page at the end).

A more common way to phrase this would be to ask about (five) consecutive primes in arithmetic progression.

Indeed, it is conjectured that there are arbitrarily long (finite) arithmetic progressions of consecutive primes, however this is open. (Without the restriction of the primes being consecutive primes this is known by a well-known result of Green and Tao.)

The longest known arithmetic progression of consecutive primes has length ten.

For further details one could start at http://en.wikipedia.org/wiki/Primes_in_arithmetic_progression (see the section towards the end on consecutive primes in AP) for record data related to this see http://users.cybercity.dk/~dsl522332/math/cpap.htm

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  • $\begingroup$ 4 consecutive primes in arithmetic progression is 251+6n for n=0,1,2,3. So much difference between 251 and 9843019 ! $\endgroup$ May 1, 2013 at 13:25
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    $\begingroup$ A reason why a relatively big increase here is not so surprsing is that the difference in a k term progression must be a multiple of all the primes up to k (ignoring isolated phenomena if the starting value should be very small). So from k=4 to k=5 the minimal possible distance increases from 6=2.3 to 30=2.3.5. And, this is a big problem if one wants consecutive primes. Since the 'typical' gap between primes of size about x is about log x, so say for x a million it is (only) about 14. If one yere to drop the consecutive requirement things also change: eg 5,11,17,23,29 or $\endgroup$
    – user9072
    May 1, 2013 at 14:02
  • $\begingroup$ with difference 30 one has 7,37,67,97,127,157 and also length ten appears much much earlier 199+210n for n=0,...,9 By contrast the known consecutive progs lengths ten have 93 digits (note that here one needs difference 210 as explained above) and for ln x to be about that size one needs beyond ninety digits. And for 11 the difference would jump to 2310 so that chances are one will need slightly beyond thousand digits numbers $\endgroup$
    – user9072
    May 1, 2013 at 14:15
  • $\begingroup$ The common difference of CPAP is always a multiple of 6 for prime numbers above 47 ? $\endgroup$ May 1, 2013 at 15:20
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    $\begingroup$ Yes. Note the difference must be even (as all primes except 2 are odd). And if the difference d is 1 mod 3, then one of p, p+d, p+2d will be 0 mod 3, so that it cannot be prime except if it is 3. Same for d= -1 mod 3. So 3|d, so 6|d. More generally: Assume p,p+d,...,p+(k-1)d are all prime. Let q be a prime <=k. If d is not 0 mod q, then 0,d,...,(k-1)d covers all the classes modulo q, so we have some j such that jd = -p mod q and thus p+jd = 0 mod q, a contradiction to p+jd being prime, except p+jd =q (which is what I meant with the isolated phenomena for small starting value above). $\endgroup$
    – user9072
    May 1, 2013 at 15:53
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Yes. In fact, there is a set of 10 consecutive primes that are in arithmetic progression. I believe this is the longest known set. See here.

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