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a(n) = Sum_{d|n} mu(d)*d^n.
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%I #22 Nov 07 2018 10:03:42

%S 1,-3,-26,-15,-3124,45864,-823542,-255,-19682,9990233352,

%T -285311670610,2176246800,-302875106592252,11111328602468784,

%U 437893859848932344,-65535,-827240261886336764176,101559568985784,-1978419655660313589123978,99999904632567310800

%N a(n) = Sum_{d|n} mu(d)*d^n.

%H Seiichi Manyama, <a href="/A321222/b321222.txt">Table of n, a(n) for n = 1..388</a>

%F G.f.: Sum_{k>=1} mu(k)*(k*x)^k/(1 - (k*x)^k).

%F a(n) = Product_{p|n, p prime} (1 - p^n).

%t Table[Sum[MoebiusMu[d] d^n, {d, Divisors[n]}], {n, 20}]

%t nmax = 20; Rest[CoefficientList[Series[Sum[MoebiusMu[k] (k x)^k/(1 - (k x)^k), {k, 1, nmax}], {x, 0, nmax}], x]]

%t Table[Product[1 - Boole[PrimeQ[d]] d^n, {d, Divisors[n]}], {n, 20}]

%o (PARI) a(n) = sumdiv(n, d, moebius(d)*d^n) \\ _Andrew Howroyd_, Nov 06 2018

%Y Cf. A008683, A023900, A046970, A063453, A067858, A189922, A189923.

%K sign

%O 1,2

%A _Ilya Gutkovskiy_, Nov 06 2018