Odd Mertens function
Let $M^*(n)$ be the "odd Mertens function", defined by $M^*(n) = sum mu(k)$ for odd $k$, $1 le k le n$.
Let $r$ be an odd number. Since $mu(r)$ is multiplicative, $mu(2r) = -mu(r)$ and $mu(4r) = mu(8r) = cdots = 0$.
So splitting $M(n) = sum_{k=1}^n mu(k)$, the standard Mertens function, by even and odd $k$, we only need to consider odd $k$ and even $k$ not divisible by 4. This gives the identity $M(n) = M^*(n) - M^*(n/2)$ or equivalently $M^*(n) = M(n) + M^*(n/2)$, so with a sublinear algorithm to calculate the Mertens function we have our odd Mertens function too.
My question: can we calculate $M^*(n)$ directly (without recursion) in sub-linear time?
number-theory algorithms mobius-function
asked yesterday
qwr
6,55542654
add a comment |
Let $M^*(n)$ be the "odd Mertens function", defined by $M^*(n) = sum mu(k)$ for odd $k$, $1 le k le n$.
Let $r$ be an odd number. Since $mu(r)$ is multiplicative, $mu(2r) = -mu(r)$ and $mu(4r) = mu(8r) = cdots = 0$.
So splitting $M(n) = sum_{k=1}^n mu(k)$, the standard Mertens function, by even and odd $k$, we only need to consider odd $k$ and even $k$ not divisible by 4. This gives the identity $M(n) = M^*(n) - M^*(n/2)$ or equivalently $M^*(n) = M(n) + M^*(n/2)$, so with a sublinear algorithm to calculate the Mertens function we have our odd Mertens function too.
My question: can we calculate $M^*(n)$ directly (without recursion) in sub-linear time?
number-theory algorithms mobius-function
asked yesterday
qwr
6,55542654
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
– Peter
yesterday
add a comment |
Let $M^*(n)$ be the "odd Mertens function", defined by $M^*(n) = sum mu(k)$ for odd $k$, $1 le k le n$.
Let $r$ be an odd number. Since $mu(r)$ is multiplicative, $mu(2r) = -mu(r)$ and $mu(4r) = mu(8r) = cdots = 0$.
So splitting $M(n) = sum_{k=1}^n mu(k)$, the standard Mertens function, by even and odd $k$, we only need to consider odd $k$ and even $k$ not divisible by 4. This gives the identity $M(n) = M^*(n) - M^*(n/2)$ or equivalently $M^*(n) = M(n) + M^*(n/2)$, so with a sublinear algorithm to calculate the Mertens function we have our odd Mertens function too.
My question: can we calculate $M^*(n)$ directly (without recursion) in sub-linear time?
number-theory algorithms mobius-function
asked yesterday
qwr
6,55542654
Let $M^*(n)$ be the "odd Mertens function", defined by $M^*(n) = sum mu(k)$ for odd $k$, $1 le k le n$.
Let $r$ be an odd number. Since $mu(r)$ is multiplicative, $mu(2r) = -mu(r)$ and $mu(4r) = mu(8r) = cdots = 0$.
So splitting $M(n) = sum_{k=1}^n mu(k)$, the standard Mertens function, by even and odd $k$, we only need to consider odd $k$ and even $k$ not divisible by 4. This gives the identity $M(n) = M^*(n) - M^*(n/2)$ or equivalently $M^*(n) = M(n) + M^*(n/2)$, so with a sublinear algorithm to calculate the Mertens function we have our odd Mertens function too.
My question: can we calculate $M^*(n)$ directly (without recursion) in sub-linear time?
number-theory algorithms mobius-function
number-theory algorithms mobius-function
asked yesterday
qwr
6,55542654
asked yesterday
qwr
6,55542654
asked yesterday
qwr
6,55542654
asked yesterday
qwr
6,55542654
asked yesterday
qwr
6,55542654
6,55542654
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
– Peter
yesterday
add a comment |
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
– Peter
yesterday
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
– Peter
yesterday
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
– Peter
yesterday
add a comment |
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The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
– Peter
yesterday