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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
 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 (list; graph; refs; listen; history; text; internal format)
 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. [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 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], 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-François Alcover, Jul 17 2012 *) PROG (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 CROSSREFS Cf. A001223. Sequence in context: A090387 A030329 A120881 * A111497 A220554 A208243 Adjacent sequences:  A031214 A031215 A031216 * A031218 A031219 A031220 KEYWORD nonn,easy,nice AUTHOR EXTENSIONS More terms from James A. Sellers STATUS approved

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