A098726 Take three consecutive primes starting with the n-th prime. Calculate d(i,j) = abs(prime(i) - prime(j)), for all {i,j}, i.e., all possible differences. a(n) is the number of distinct differences (which can be either 3 or 2).

3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 2, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3

Offset: 1

Author

Labos Elemer, Oct 05 2004

Keywords

• nonn

Comments

a(n) = 2 iff the consecutive prime differences are equal.

It appears that a(n) = 2 for n in A064113. - Michel Marcus, Jul 27 2017

Crossrefs

Cf. A080370, A054643, A064113.

Programs

Mathematica

k=3;t=Table[Abs[Prime[n+i]-Prime[n+j]], {i, 0, k-1}, {j, 0, k-1}]; u=Delete[Union[Flatten[t]], 1];a(n)=Length[u]

Extensions

Name edited by Michel Marcus, Jul 27 2017