A105470 a(n)=1 if there is number of the form 6k+3 with prime(n) <= 6k+3 <= prime(n+1), otherwise 0.

1, 1, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1

Offset: 1

Author

Giovanni Teofilatto, May 02 2005

Keywords

• easy
• nonn

Comments

Except for the first pair of primes and for twin primes there is always at least one number of the form 6n+3 between two successive primes.

Examples

			a(3)=0 because between prime(3) and prime(4) there are no numbers of the form 6k+3;
a(4)=1 because between prime(4) and prime(5) there is one number of the form 6k+3: 9.
		

Crossrefs

Cf. A100810, A106002.

Programs

Mathematica

f[n_] := Count[Table[Mod[k, 6], {k, Prime[n], Prime[n + 1]}], 3];Table[If[f[n] == 0, 0, 1], {n, 120}] (* Ray Chandler, Oct 17 2006 *)
Join[{1,1},If[Last[#]-First[#]==2,0,1]&/@Partition[Prime[Range[ 3,200]],2,1]] (* Harvey P. Dale, Nov 27 2013 *)

Extensions

Edited by Ray Chandler, Oct 17 2006