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A031217 Number of terms in longest arithmetic progression of consecutive primes starting at n-th prime (conjectured to be unbounded). 1

%I

%S 2,3,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,

%T 2,3,2,2,3,2,2,2,2,2,2,3,2,2,2,2,2,2,2,4,3,2,2,2,2,2,2,2,2,2,2,2,2,2,

%U 2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2

%N Number of terms in longest arithmetic progression of consecutive primes starting at n-th prime (conjectured to be unbounded).

%C a(n) <= 4 for n <= 10^5. - _Reinhard Zumkeller_, Feb 02 2007

%C The first instance of 4 consecutive primes in an arithmetic progression is (251, 257, 263, 269), which starts with the 54th prime. The first instance of 5 consecutive primes in an arithmetic progression is (9843019, 9843049, 9843079, 9843109, 9843139), which starts with the 654926th prime. [From Harvey P. Dale, Jul 13 2011]

%D R. K. Guy, Unsolved Problems in Number Theory, A6.

%H R. Zumkeller, <a href="/A031217/b031217.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Pri#primes_AP">Index entries for sequences related to primes in arithmetic progressions</a>

%e At 47 there are 3 consecutive primes in A.P., 47 53 59.

%t max = 5; a[n_] := Catch[pp = NestList[ NextPrime, Prime[n], max-1]; Do[ If[ Length[ Union[ Differences[pp[[1 ;; -k]] ] ] ] == 1, Throw[max-k+1]], {k, 1, max-1}]]; Table[a[n], {n, 1, 105}] (* _Jean-Fran├žois Alcover_, Jul 17 2012 *)

%o (PARI) a(n)=my(p=prime(n),q=nextprime(p+1),g=q-p,k=2); while(nextprime(q+1)==q+g, q+=g; k++); k \\ _Charles R Greathouse IV_, Jun 20 2013

%Y Cf. A001223.

%K nonn,easy,nice

%O 1,1

%A _N. J. A. Sloane_.

%E More terms from _James A. Sellers_

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Last modified September 18 19:54 EDT 2014. Contains 246924 sequences.