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%I #32 Nov 23 2021 18:18:41
%S 1,-2,-3,2,-5,6,-7,-2,6,10,-11,-6,-13,14,15,2,-17,-12,-19,-10,21,22,
%T -23,6,20,26,-12,-14,-29,-30,-31,-2,33,34,35,12,-37,38,39,10,-41,-42,
%U -43,-22,-30,46,-47,-6,42,-40,51,-26,-53,24,55,14,57,58,-59,30
%N a(1) = 1; a(n) = -Sum_{d|n, d<n} rad(n/d) * a(d), where rad = A007947.
%C Dirichlet inverse of A007947.
%C Moebius transform of A125131.
%H Antti Karttunen, <a href="/A325126/b325126.txt">Table of n, a(n) for n = 1..20000</a>
%F G.f. A(x) satisfies: A(x) = x - Sum_{k>=2} rad(k) * A(x^k).
%F From _Isaac Saffold_, May 30 2020: (Start)
%F a(n) = A008836(n)*A326297(n)*A007947(n).
%F Proof:
%F Define lambda(n) := A008836(n); h(n) := A326297(n); rad(n) := A007947(n).
%F As lambda(n), h(n), and rad(n) are multiplicative, the identity needs only to be proved for prime power n.
%F It is clear that the identity holds for n = 1 = p^0. For a given nonnegative integer k, assume the identity holds for all v such that 0 <= v <= k. Then, by the recursive formula for Dirichlet inverses,
%F a(p^(k+1)) = -Sum_{v=0..k} lambda(p^v)*h(p^v)*rad(p^v)*rad(p^(k+1-v))
%F = -p * (1 + p*Sum_{v=1..k}((-1)^v * (p-1)^(v-1)))
%F = -p * (1 - p*Sum_{v=0..(k-1)}((1 - p)^v))
%F = -p * (1 - p*(((1-p)^k - 1) / -p))
%F = -p * (1-p)^k
%F = (-1)^(k+1) * (p-1)^k * p
%F = lambda(p^(k+1)) * h(p^(k+1)) * rad(p^(k+1))
%F Thus the identity holds for p^(k+1), k >= 0.
%F As k is arbitrary and the identity holds for p^0, it holds for the prime powers, and thus for all positive integers. Q.E.D. (End)
%t a[n_] := If[n == 1, n, -Sum[If[d < n, Last[Select[Divisors[n/d], SquareFreeQ]] a[d], 0], {d, Divisors[n]}]]; Table[a[n], {n, 1, 60}]
%t f[p_, e_] := -p*(1 - p)^(e - 1); a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* _Amiram Eldar_, Oct 14 2020 *)
%o (PARI) rad(n) = factorback(factorint(n)[, 1]); \\ A007947
%o lista(nn) = {my(va=vector(nn)); va[1] = 1; for (n=2, nn, va[n] = -sumdiv(n, d, if (d<n, rad(n/d) * va[d]));); va;} \\ _Michel Marcus_, Jun 01 2020
%Y Cf. A007947, A008836, A125131, A326297.
%Y Cf. also A327564, A332732, A349340, A349612, A349618.
%K sign,mult,easy
%O 1,2
%A _Ilya Gutkovskiy_, Sep 04 2019
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