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

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

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

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]

Examples

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

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

Crossrefs

Cf. A001223.

Programs

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 *)
Length[Split[Differences[#]][[1]]]&/@Partition[Prime[Range[120]],10,1]+1 (* Harvey P. Dale, Mar 17 2024 *)

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

Extensions

More terms from James Sellers