Sum of the number of relatively prime integers up to $x$, $x-1$, $ldots$, $1$Counting elements of reduced residue systems modulo one number which are smaller than anotherEuler's totient function Proof that $nmid varphi(2^n-1)$Euler's totient function for large numbersNumber theory question about Euler's totient functionDoes the totient function reach all its values when restricted to odd numbers?Find all values of N that satisfy ϕ(N) = x for any given x valueOn factoring and integer given the value of its Euler's totient function.How to count the number of perfect square greater than $N$ and less than $N^2$ that are relatively prime to $N$?How many numbers less than $m$ and relatively prime to $n$, where $m>n$?Showing that two binomials are relatively prime for all positive integers (Euclidean ?)
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Transcription Beats per minute
Sum of the number of relatively prime integers up to $x$, $x-1$, $ldots$, $1$
Counting elements of reduced residue systems modulo one number which are smaller than anotherEuler's totient function Proof that $nmid varphi(2^n-1)$Euler's totient function for large numbersNumber theory question about Euler's totient functionDoes the totient function reach all its values when restricted to odd numbers?Find all values of N that satisfy ϕ(N) = x for any given x valueOn factoring and integer given the value of its Euler's totient function.How to count the number of perfect square greater than $N$ and less than $N^2$ that are relatively prime to $N$?How many numbers less than $m$ and relatively prime to $n$, where $m>n$?Showing that two binomials are relatively prime for all positive integers (Euclidean ?)
$begingroup$
If there is a number $x$, and we want to find the sum of the number of relatively prime integers up to $x$, $x-1$, $dots$ until $1$, is there a formula for this or any way to solve it? Like if the number is $6$, you add up $0$ (for numbers relatively prime to $1$), $1$ (for numbers relatively prime to $2$), $2$ (for $3$), $2$ (for $4$), $4$ (for $5$), and $2$ (for $6$) for a total of $11$ numbers. I tried using Euler's Totient Function, but with a high number, that would require far too many computations. Is there any way to compute this for a high number, let's say $2019$, without taking a lot of time?
totient-function coprime
$endgroup$
|
show 3 more comments
$begingroup$
If there is a number $x$, and we want to find the sum of the number of relatively prime integers up to $x$, $x-1$, $dots$ until $1$, is there a formula for this or any way to solve it? Like if the number is $6$, you add up $0$ (for numbers relatively prime to $1$), $1$ (for numbers relatively prime to $2$), $2$ (for $3$), $2$ (for $4$), $4$ (for $5$), and $2$ (for $6$) for a total of $11$ numbers. I tried using Euler's Totient Function, but with a high number, that would require far too many computations. Is there any way to compute this for a high number, let's say $2019$, without taking a lot of time?
totient-function coprime
$endgroup$
1
$begingroup$
mathworld.wolfram.com/TotientSummatoryFunction.html
$endgroup$
– saulspatz
Mar 15 at 23:30
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
Mar 15 at 23:46
$begingroup$
So basically you want to compute $sum_ileq n varphi(i)$?
$endgroup$
– tomasz
Mar 15 at 23:50
$begingroup$
Yes but how do you compute it?
$endgroup$
– Jaemin Kim
Mar 16 at 0:17
$begingroup$
"let's say 2019" That's oddly specific. Where is this problem from, exactly? And are you supposed to calculate it by hand?
$endgroup$
– Arthur
Mar 16 at 0:50
|
show 3 more comments
$begingroup$
If there is a number $x$, and we want to find the sum of the number of relatively prime integers up to $x$, $x-1$, $dots$ until $1$, is there a formula for this or any way to solve it? Like if the number is $6$, you add up $0$ (for numbers relatively prime to $1$), $1$ (for numbers relatively prime to $2$), $2$ (for $3$), $2$ (for $4$), $4$ (for $5$), and $2$ (for $6$) for a total of $11$ numbers. I tried using Euler's Totient Function, but with a high number, that would require far too many computations. Is there any way to compute this for a high number, let's say $2019$, without taking a lot of time?
totient-function coprime
$endgroup$
If there is a number $x$, and we want to find the sum of the number of relatively prime integers up to $x$, $x-1$, $dots$ until $1$, is there a formula for this or any way to solve it? Like if the number is $6$, you add up $0$ (for numbers relatively prime to $1$), $1$ (for numbers relatively prime to $2$), $2$ (for $3$), $2$ (for $4$), $4$ (for $5$), and $2$ (for $6$) for a total of $11$ numbers. I tried using Euler's Totient Function, but with a high number, that would require far too many computations. Is there any way to compute this for a high number, let's say $2019$, without taking a lot of time?
totient-function coprime
totient-function coprime
edited Mar 17 at 11:19
Rócherz
3,0013821
3,0013821
asked Mar 15 at 23:25
Jaemin KimJaemin Kim
31
31
1
$begingroup$
mathworld.wolfram.com/TotientSummatoryFunction.html
$endgroup$
– saulspatz
Mar 15 at 23:30
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
Mar 15 at 23:46
$begingroup$
So basically you want to compute $sum_ileq n varphi(i)$?
$endgroup$
– tomasz
Mar 15 at 23:50
$begingroup$
Yes but how do you compute it?
$endgroup$
– Jaemin Kim
Mar 16 at 0:17
$begingroup$
"let's say 2019" That's oddly specific. Where is this problem from, exactly? And are you supposed to calculate it by hand?
$endgroup$
– Arthur
Mar 16 at 0:50
|
show 3 more comments
1
$begingroup$
mathworld.wolfram.com/TotientSummatoryFunction.html
$endgroup$
– saulspatz
Mar 15 at 23:30
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
Mar 15 at 23:46
$begingroup$
So basically you want to compute $sum_ileq n varphi(i)$?
$endgroup$
– tomasz
Mar 15 at 23:50
$begingroup$
Yes but how do you compute it?
$endgroup$
– Jaemin Kim
Mar 16 at 0:17
$begingroup$
"let's say 2019" That's oddly specific. Where is this problem from, exactly? And are you supposed to calculate it by hand?
$endgroup$
– Arthur
Mar 16 at 0:50
1
1
$begingroup$
mathworld.wolfram.com/TotientSummatoryFunction.html
$endgroup$
– saulspatz
Mar 15 at 23:30
$begingroup$
mathworld.wolfram.com/TotientSummatoryFunction.html
$endgroup$
– saulspatz
Mar 15 at 23:30
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
Mar 15 at 23:46
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
Mar 15 at 23:46
$begingroup$
So basically you want to compute $sum_ileq n varphi(i)$?
$endgroup$
– tomasz
Mar 15 at 23:50
$begingroup$
So basically you want to compute $sum_ileq n varphi(i)$?
$endgroup$
– tomasz
Mar 15 at 23:50
$begingroup$
Yes but how do you compute it?
$endgroup$
– Jaemin Kim
Mar 16 at 0:17
$begingroup$
Yes but how do you compute it?
$endgroup$
– Jaemin Kim
Mar 16 at 0:17
$begingroup$
"let's say 2019" That's oddly specific. Where is this problem from, exactly? And are you supposed to calculate it by hand?
$endgroup$
– Arthur
Mar 16 at 0:50
$begingroup$
"let's say 2019" That's oddly specific. Where is this problem from, exactly? And are you supposed to calculate it by hand?
$endgroup$
– Arthur
Mar 16 at 0:50
|
show 3 more comments
1 Answer
1
active
oldest
votes
$begingroup$
There is a well-known expression for the Euler function using the Möbius function:
$$ varphi(n) = nsum_dmid nfracmu(d)d. $$
Consequently,
beginalign
sum_n=1^N varphi(n)
&= sum_n=1^N n sum_dmid n fracmu(d)d \
&= sum_d=1^N fracmu(d)d sum_kle N/d kd \
&= frac12,sum_d=1^N mu(d) leftlfloor frac Ndright rfloor left(leftlfloor frac Ndright rfloor +1right).
endalign
This expression is standartly used to give an asymptotic for the sum in the LHS, but it also can be used to efficiently calculate this sum.
$endgroup$
add a comment |
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$begingroup$
There is a well-known expression for the Euler function using the Möbius function:
$$ varphi(n) = nsum_dmid nfracmu(d)d. $$
Consequently,
beginalign
sum_n=1^N varphi(n)
&= sum_n=1^N n sum_dmid n fracmu(d)d \
&= sum_d=1^N fracmu(d)d sum_kle N/d kd \
&= frac12,sum_d=1^N mu(d) leftlfloor frac Ndright rfloor left(leftlfloor frac Ndright rfloor +1right).
endalign
This expression is standartly used to give an asymptotic for the sum in the LHS, but it also can be used to efficiently calculate this sum.
$endgroup$
add a comment |
$begingroup$
There is a well-known expression for the Euler function using the Möbius function:
$$ varphi(n) = nsum_dmid nfracmu(d)d. $$
Consequently,
beginalign
sum_n=1^N varphi(n)
&= sum_n=1^N n sum_dmid n fracmu(d)d \
&= sum_d=1^N fracmu(d)d sum_kle N/d kd \
&= frac12,sum_d=1^N mu(d) leftlfloor frac Ndright rfloor left(leftlfloor frac Ndright rfloor +1right).
endalign
This expression is standartly used to give an asymptotic for the sum in the LHS, but it also can be used to efficiently calculate this sum.
$endgroup$
add a comment |
$begingroup$
There is a well-known expression for the Euler function using the Möbius function:
$$ varphi(n) = nsum_dmid nfracmu(d)d. $$
Consequently,
beginalign
sum_n=1^N varphi(n)
&= sum_n=1^N n sum_dmid n fracmu(d)d \
&= sum_d=1^N fracmu(d)d sum_kle N/d kd \
&= frac12,sum_d=1^N mu(d) leftlfloor frac Ndright rfloor left(leftlfloor frac Ndright rfloor +1right).
endalign
This expression is standartly used to give an asymptotic for the sum in the LHS, but it also can be used to efficiently calculate this sum.
$endgroup$
There is a well-known expression for the Euler function using the Möbius function:
$$ varphi(n) = nsum_dmid nfracmu(d)d. $$
Consequently,
beginalign
sum_n=1^N varphi(n)
&= sum_n=1^N n sum_dmid n fracmu(d)d \
&= sum_d=1^N fracmu(d)d sum_kle N/d kd \
&= frac12,sum_d=1^N mu(d) leftlfloor frac Ndright rfloor left(leftlfloor frac Ndright rfloor +1right).
endalign
This expression is standartly used to give an asymptotic for the sum in the LHS, but it also can be used to efficiently calculate this sum.
answered Mar 17 at 11:37
W-t-PW-t-P
1,529612
1,529612
add a comment |
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$begingroup$
mathworld.wolfram.com/TotientSummatoryFunction.html
$endgroup$
– saulspatz
Mar 15 at 23:30
$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
Mar 15 at 23:46
$begingroup$
So basically you want to compute $sum_ileq n varphi(i)$?
$endgroup$
– tomasz
Mar 15 at 23:50
$begingroup$
Yes but how do you compute it?
$endgroup$
– Jaemin Kim
Mar 16 at 0:17
$begingroup$
"let's say 2019" That's oddly specific. Where is this problem from, exactly? And are you supposed to calculate it by hand?
$endgroup$
– Arthur
Mar 16 at 0:50