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Sum(mu(m) : prime(n) < m < prime(n+1)), where mu is the Moebius function (A008683).
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%I #3 Mar 30 2012 18:50:30

%S 0,0,1,1,0,2,0,2,1,-1,3,2,-1,1,1,3,0,1,0,0,1,1,3,4,0,-1,0,0,0,4,0,2,

%T -1,5,0,0,3,0,-1,1,0,0,0,0,0,7,6,1,0,-2,1,0,1,1,1,1,0,0,1,-1,-1,7,0,0,

%U 1,5,2,2,0,0,0,1,0,0,2,0,5,0,0,2,0,-1,0,-2,-1,3,2,1,1,0,2,2,1,0,2,0,5,0,4,4,2,1,2,1,0,3

%N Sum(mu(m) : prime(n) < m < prime(n+1)), where mu is the Moebius function (A008683).

%C a(n) = A002321(A000040(n+1)-1) - A002321(A000040(n)).

%e n=16: adding mu(m) for A000040(16)=53 < m < A000040(17)=59: a(16) = mu(54)+mu(55)+mu(56)+mu(57)+mu(58) = mu(2*3^3)+mu(5*11)+mu(7*2^3)+mu(3*19)+mu(2*29) = 0+(-1)*(-1)+0+(-1)*(-1)+(-1)*(-1) = 1+1+1 = 3.

%K sign

%O 1,6

%A _Reinhard Zumkeller_, Jan 30 2003