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- $$\mu(n) = \underset{n = 1}{1} - \underset{a = n}{\sum_{a \geq 2}} 1 + \underset{ab = n}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd = n}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots$$
- $$M(x) = 1 - \sum_{2 \leq a \leq x} 1 + \underset{ab \leq x}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots $$
- ===As a sum of the number of points under n-dimensional hyperboloids{{citation needed|date=July 2018}}===
- :<math>M(x) = 1 - \sum_{2 \leq a \leq x} 1 + \underset{ab \leq x}{\sum_{a \geq 2} \sum_{b \geq 2}} 1 - \underset{abc \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2}} 1 + \underset{abcd \leq x}{\sum_{a \geq 2} \sum_{b \geq 2} \sum_{c \geq 2} \sum_{d \geq 2}} 1 - \cdots</math>
- This formulation expanding the Mertens function suggests asymptotic bounds obtained by considering the [[Divisor_summatory_function#Piltz_divisor_problem|Piltz divisor problem]] which generalizes the [[Divisor_summatory_function|Dirichlet divisor problem]] of computing [[asymptotic estimate]]s for the summatory function of the [[divisor function]].
- nn = 15;
- Table[Sum[If[a^2 == n, 1, 0], {a, 1, nn}] -
- Sum[Sum[If[a^2*b == n, 1, 0], {a, 1, nn}], {b, 2, nn}] +
- Sum[Sum[Sum[If[a^2*b*c == n, 1, 0], {a, 1, nn}], {b, 2, nn}], {c, 2,
- nn}] - +Sum[
- Sum[Sum[Sum[If[a^2*b*c*d == n, 1, 0], {a, 1, nn}], {b, 2,
- nn}], {c, 2, nn}], {d, 2, nn}], {n, 1, nn}]
- LiouvilleLambda[Range[nn]]
- %% - %
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