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%I #14 Mar 01 2016 18:46:29
%S 1,1,1,0,1,-1,1,1,0,-1,1,3,1,-1,-1,-1,1,3,1,3,-1,-1,1,-7,0,-1,1,3,1,7,
%T 1,3,-1,-1,-1,-12,1,-1,-1,-7,1,7,1,3,3,-1,1,19,0,3,-1,3,1,-7,-1,-7,-1,
%U -1,1,-27,1,-1,3,-6,-1,7,1,3,-1,7,1,45,1,-1,3,3,-1,7,1,19,-1,-1,1,-27,-1,-1,-1,-7,1,-27,-1,3,-1,-1,-1,-51,1,3,3,-12,1,7
%N Sum{k=1 to oo} a(k)/k^r = sqrt(zeta(r) -3/4) +1/2.
%C The "+ 1/2" in the Dirichlet series generating function was added so the first term of the sequence is an integer. We could have added/subtracted any other integer+1/2 instead and then had the first term equal another integer. "zeta(r)" refers to sum{k=1 to oo} 1/k^r.
%H Robert Israel, <a href="/A097558/b097558.txt">Table of n, a(n) for n = 1..10000</a>
%F a(1)=1; for n>=2, a(n) = 1 - sum{k|n, 2<=k<=n-1} a(n/k) a(k).
%F From _Robert Israel_, Mar 01 2016: (Start)
%F a(n) depends only on the prime signature of n.
%F If p is prime, a(p^k) = (-1)^(k+1)*A005043(k-1).
%F If n is squarefree, a(n) = (-1)^(A001222(n)-1)*A048287(A001222(n)).
%F (End)
%p A[1]:= 1:
%p for n from 2 to 100 do
%p A[n]:= 1 - add(A[n/k]*A[k], k= numtheory:-divisors(n) minus {1,n})
%p od:
%p seq(A[n],n=1..100); # _Robert Israel_, Mar 01 2016
%Y Cf. A001222, A005043, A048287.
%K sign
%O 1,12
%A _Leroy Quet_, Aug 27 2004
%E More terms from _David Wasserman_, Dec 27 2007
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