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%I #10 Jul 27 2017 04:19:31
%S 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,
%T 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,
%U 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
%N 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).
%C a(n) = 2 iff the consecutive prime differences are equal.
%C It appears that a(n) = 2 for n in A064113. - _Michel Marcus_, Jul 27 2017
%t 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]
%Y Cf. A080370, A054643, A064113.
%K nonn
%O 1,1
%A _Labos Elemer_, Oct 05 2004
%E Name edited by _Michel Marcus_, Jul 27 2017
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