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a(1) = 1; a(n+1) = Sum_{d|n} (-1)^(d+1)*a(d).
5

%I #8 Apr 28 2019 20:09:44

%S 1,1,0,1,-1,0,0,1,-2,-1,0,1,-2,-1,1,1,-3,-2,0,1,-2,-1,1,2,-5,-5,3,2,

%T -2,-1,2,3,-6,-5,2,2,-4,-3,3,2,-5,-4,3,4,-4,-5,6,7,-13,-12,7,5,-3,-2,

%U 5,5,-8,-7,5,6,-7,-6,8,5,-11,-13,8,9,-8,-6,9,10,-17,-16,12,8,-6

%N a(1) = 1; a(n+1) = Sum_{d|n} (-1)^(d+1)*a(d).

%F G.f.: x * (1 + Sum_{n>=1} (-1)^(n+1)*a(n)*x^n/(1 - x^n)).

%F L.g.f.: log(Product_{n>=1} (1 - x^n)^((-1)^n*a(n)/n)) = Sum_{n>=1} a(n+1)*x^n/n.

%t a[n_] := a[n] = Sum[(-1)^(d + 1) a[d], {d, Divisors[n - 1]}]; a[1] = 1; Table[a[n], {n, 1, 77}]

%t a[n_] := a[n] = SeriesCoefficient[x (1 + Sum[(-1)^(k + 1) a[k] x^k/(1 - x^k), {k, 1, n - 1}]), {x, 0, n}]; Table[a[n], {n, 1, 77}]

%Y Cf. A003238, A281487, A307776, A307777, A307779.

%K sign

%O 1,9

%A _Ilya Gutkovskiy_, Apr 28 2019