Odd Mertens function
$begingroup$
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 Jan 5 at 5:43
qwrqwr
6,56542655
$endgroup$
add a comment |
$begingroup$
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 Jan 5 at 5:43
qwrqwr
6,56542655
$endgroup$
$begingroup$
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
$endgroup$
– Peter
Jan 5 at 9:16
add a comment |
$begingroup$
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 Jan 5 at 5:43
qwrqwr
6,56542655
$endgroup$
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 Jan 5 at 5:43
qwrqwr
6,56542655
asked Jan 5 at 5:43
qwrqwr
6,56542655
asked Jan 5 at 5:43
qwrqwr
6,56542655
asked Jan 5 at 5:43
qwrqwr
6,56542655
asked Jan 5 at 5:43
qwrqwr
6,56542655
6,56542655
$begingroup$
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
$endgroup$
– Peter
Jan 5 at 9:16
add a comment |
$begingroup$
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
$endgroup$
– Peter
Jan 5 at 9:16
$begingroup$
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
$endgroup$
– Peter
Jan 5 at 9:16
$begingroup$
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
$endgroup$
– Peter
Jan 5 at 9:16
add a comment |
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$begingroup$
The effort to calculate the sum is only halved this way, this does not make the calculation sublinear , if it was not before.
$endgroup$
– Peter
Jan 5 at 9:16