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%I #36 Dec 11 2023 15:10:54
%S 1,1,1,0,1,1,1,-1,0,1,1,0,1,1,1,-1,1,0,1,0,1,1,1,-1,0,1,-1,0,1,1,1,0,
%T 1,1,1,0,1,1,1,-1,1,1,1,0,0,1,1,-1,0,0,1,0,1,-1,1,-1,1,1,1,0,1,1,0,1,
%U 1,1,1,0,1,1,1,0,1,1,0,0,1,1,1,-1,-1,1,1,0,1,1,1,-1,1,0,1,0,1,1
%N Dirichlet g.f.: Sum_{n>0} a(n)/n^s = (zeta(s)*zeta(6*s))/(zeta(2*s)*zeta(3*s)).
%H Antti Karttunen, <a href="/A299406/b299406.txt">Table of n, a(n) for n = 1..65537</a>
%H Vaclav Kotesovec, <a href="/A299406/a299406.jpg">Plot of Sum_{k=1..n} a(k) / (2*n*Pi^4/(315*Zeta(3))) for n = 1..1000000</a>
%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>
%F a(n) is multiplicative with a(p^e)=(-1)^(e mod 3 + e mod 6) if e mod 3 < 2, otherwise 0, p prime and e >= 0.
%F Dirichlet inverse b(n) is multiplicative with b(p^e) = (-1)^e if e < 3, otherwise 0, p prime and e >= 0.
%F Sum_{d|n} a(d)*A005361(n/d) = A000005(n).
%F Conjecture: a(n) = A008836(n) * A210826(n).
%F Sum_{k=1..n} a(k) ~ 2*n*Pi^4/(315*Zeta(3)). - _Vaclav Kotesovec_, Feb 08 2019
%t f[e_] := If[Mod[e, 3] < 2, (-1)^(Mod[e, 3] + Mod[e, 6]), 0];
%t a[n_] := a[n] = Times @@ (f /@ FactorInteger[n][[All, 2]]);
%t Array[a, 100] (* _Jean-François Alcover_, Feb 26 2018 *)
%o (PARI) A299406(n) = { my(es = factor(n)[, 2]); factorback(apply(e -> if(2==(e%3),0,(-1)^((e%3)+(e%6))), es)); }; \\ _Antti Karttunen_, Jul 29 2018
%Y Cf. A000005, A005361, A008836, A210826.
%K sign,mult
%O 1,1
%A _Werner Schulte_, Feb 20 2018
%E Name clarified by _Andrey Zabolotskiy_, Dec 11 2023