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A362028 a(n) = Sum_{k=1..n} (-1)^(n-k) * mu(k)^2, where mu(k) is the Moebius function.

%I A362028 #22 Mar 05 2024 11:52:29
%S A362028 1,0,1,-1,2,-1,2,-2,2,-1,2,-2,3,-2,3,-3,4,-4,5,-5,6,-5,6,-6,6,-5,5,-5,
%T A362028 6,-5,6,-6,7,-6,7,-7,8,-7,8,-8,9,-8,9,-9,9,-8,9,-9,9,-9,10,-10,11,-11,
%U A362028 12,-12,13,-12,13,-13,14,-13,13,-13,14,-13,14,-14,15,-14,15,-15,16,-15,15,-15,16,-15,16,-16,16,-15,16,-16,17
%N A362028 a(n) = Sum_{k=1..n} (-1)^(n-k) * mu(k)^2, where mu(k) is the Moebius function.
%H A362028 Seiichi Manyama, <a href="/A362028/b362028.txt">Table of n, a(n) for n = 1..10000</a>
%H A362028 László Tóth, <a href="https://www.emis.de/journals/JIS/VOL20/Toth/toth25.html">Alternating Sums Concerning Multiplicative Arithmetic Functions</a>, Journal of Integer Sequences, Vol. 20 (2017), Article 17.2.1.
%F A362028 G.f.: (Sum_{k>=1} mu(k)^2 * x^k) / (1 + x).
%F A362028 a(n) = -a(n-1) + A008966(n) for n > 1.
%F A362028 abs(a(n)) = (2/Pi^2) * n + O(R(n)), where R(n) = x^(1/2)*exp(-c * log(n)^(3/5) / log(log(n))^(1/5)) and c is a positive constant, unconditionally, or x^(11/35+eps) assuming the Riemann hypothesis (Tóth, 2017). - _Amiram Eldar_, Mar 05 2024
%t A362028 With[{m = 100}, -(-1)^Range[m] * Accumulate[Array[(-1)^(# + 1) * MoebiusMu[#]^2 &, m]]] (* _Amiram Eldar_, Mar 05 2024 *)
%o A362028 (PARI) a(n) = sum(k=1, n, (-1)^(n-k)*moebius(k)^2);
%o A362028 (PARI) lista(kmax) = {my(s = 0); for(k = 1, kmax, s += (-1)^(k+1)*moebius(k)^2; print1((-1)^(k+1)*s, ", "))}; \\ _Amiram Eldar_, Mar 05 2024
%Y A362028 Cf. A008683, A008966, A013928, A185197, A362029.
%K A362028 sign,easy
%O A362028 1,5
%A A362028 _Seiichi Manyama_, Apr 05 2023