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a(n) = n^3 * Sum_{d|n} (-1)^(n/d + 1) * mu(d) / d^3.
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%I #14 Nov 16 2022 06:55:44

%S 1,-9,26,-56,124,-234,342,-448,702,-1116,1330,-1456,2196,-3078,3224,

%T -3584,4912,-6318,6858,-6944,8892,-11970,12166,-11648,15500,-19764,

%U 18954,-19152,24388,-29016,29790,-28672,34580,-44208,42408,-39312,50652,-61722,57096,-55552,68920,-80028

%N a(n) = n^3 * Sum_{d|n} (-1)^(n/d + 1) * mu(d) / d^3.

%H Amiram Eldar, <a href="/A338548/b338548.txt">Table of n, a(n) for n = 1..10000</a>

%F G.f.: Sum_{k>=1} mu(k) * x^k * (1 - 4*x^k + x^(2*k)) / (1 + x^k)^4.

%F G.f. A(x) satisfies: A(x) = x * (1 - 4*x + x^2) / (1 + x)^4 - Sum_{k>=2} A(x^k).

%F Dirichlet g.f.: (1 - 2^(4 - s)) * zeta(s - 3) / zeta(s).

%F a(n) = J_3(n) if n odd, J_3(n) - 16 * J_3(n/2) if n even, where J_3 = A059376 (Jordan function J_3).

%F Multiplicative with a(2) = -9, a(2^e) = -7*2^(3*(e-1)) for e > 1, and a(p^e) = (p^3-1)*p^(3*(e-1)) for p > 2. - _Amiram Eldar_, Nov 15 2022

%t Table[n^3 Sum[(-1)^(n/d + 1) MoebiusMu[d]/d^3, {d, Divisors[n]}], {n, 1, 42}]

%t nmax = 42; CoefficientList[Series[Sum[MoebiusMu[k] x^k (1 - 4 x^k + x^(2 k))/(1 + x^k)^4, {k, 1, nmax}], {x, 0, nmax}], x] // Rest

%t f[p_, e_] := (p^3 - 1)*p^(3*(e - 1)); f[2, 1] = -9; f[2, e_] := -7*2^(3*(e - 1)); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* _Amiram Eldar_, Nov 15 2022 *)

%o (PARI) a(n) = n^3 * sumdiv(n, d, (-1)^(n/d+1)*moebius(d)/d^3); \\ _Michel Marcus_, Nov 02 2020

%Y Cf. A008683, A059376, A138503, A325596, A338547, A338549.

%K sign,mult

%O 1,2

%A _Ilya Gutkovskiy_, Nov 02 2020