

A031217


Number of terms in longest arithmetic progression of consecutive primes starting at nth prime (conjectured to be unbounded).


1



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, 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, 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
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OFFSET

1,1


COMMENTS

a(n) <= 4 for n <= 10^5.  Reinhard Zumkeller, Feb 02 2007
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]


REFERENCES

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


LINKS

R. Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to primes in arithmetic progressions


EXAMPLE

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


MATHEMATICA

max = 5; a[n_] := Catch[pp = NestList[ NextPrime, Prime[n], max1]; Do[ If[ Length[ Union[ Differences[pp[[1 ;; k]] ] ] ] == 1, Throw[maxk+1]], {k, 1, max1}]]; Table[a[n], {n, 1, 105}] (* JeanFrançois Alcover, Jul 17 2012 *)


PROG

(PARI) a(n)=my(p=prime(n), q=nextprime(p+1), g=qp, k=2); while(nextprime(q+1)==q+g, q+=g; k++); k \\ Charles R Greathouse IV, Jun 20 2013


CROSSREFS

Cf. A001223.
Sequence in context: A090387 A030329 A120881 * A078545 A111497 A220554
Adjacent sequences: A031214 A031215 A031216 * A031218 A031219 A031220


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane.


EXTENSIONS

More terms from James A. Sellers


STATUS

approved



