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A279817 a(1) = -1; for n>1, Sum_{d|n} a(n-d+1) = 0.

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%I A279817 #16 Dec 21 2016 13:54:58
%S A279817 -1,1,1,0,1,0,1,-1,0,1,1,-2,1,1,-1,1,1,-3,1,-3,1,2,1,-6,0,0,0,0,1,-2,
%T A279817 1,-2,-1,5,-1,-4,1,3,0,3,1,-3,1,-7,-3,10,1,-9,0,-10,2,-4,1,-7,2,6,-1,
%U A279817 4,1,-25,1,2,-2,4,-1,-11,1,6,-1,13,1,-20,1,-3,-4,0
%N A279817 a(1) = -1; for n>1, Sum_{d|n} a(n-d+1) = 0.
%F A279817 For primes p and q:
%F A279817 a(p) = 1.
%F A279817 If p^2 - p + 1 is prime then a(p^2) = 0.
%F A279817 If p*q - p + 1 and p*q - q + 1 are primes then a(p*q) = -1.
%e A279817 When n is any prime p, we have Sum_{d|p} a(p-d+1) = 0, so a(p-1+1) + a(p-p+1) = 0, hence a(p)=1.
%e A279817 For n=4, we have a(4-1+1) + a(4-2+1) + a(4-4+1) = 0, so a(4) + a(3) + a(1) = 0, hence a(4)=0.
%p A279817 a := proc (n) option remember; -add(a(n-d+1), d = `minus`(numtheory:-divisors(n), {1})) end proc; a(1) := -1; seq(simplify(a(i)), i = 1 .. 1000)
%K A279817 sign
%O A279817 1,12
%A A279817 _Gevorg Hmayakyan_, Dec 19 2016