Count Integers Not Greater Than $a$ Coprime To $b$number of coprimes to a less than bEuler's totient function and complex numbersCounting elements of reduced residue systems modulo one number which are smaller than anotherIn a given sequence of consecutive integers, finding the count of integers with a least prime factor greater than $p$Eulers totient function divided by $n$, counting numbers in the set [1,m] that are coprime to nEuler's theorem (modular arithmetic) for non-coprime integersFor $n > 2, n in mathbbZ$, show the sum of integers coprime to $n$ in the range $[1,n-1]$ is equal to $frac12n phi(n)$Proof any arithmetic progression coprime count same as toting function of nHow 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 $m>n$?Are there any known methods for finding Upper/Lower bounds on the number of Totients of x less than another number y?
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Count Integers Not Greater Than $a$ Coprime To $b$
number of coprimes to a less than bEuler's totient function and complex numbersCounting elements of reduced residue systems modulo one number which are smaller than anotherIn a given sequence of consecutive integers, finding the count of integers with a least prime factor greater than $p$Eulers totient function divided by $n$, counting numbers in the set [1,m] that are coprime to nEuler's theorem (modular arithmetic) for non-coprime integersFor $n > 2, n in mathbbZ$, show the sum of integers coprime to $n$ in the range $[1,n-1]$ is equal to $frac12n phi(n)$Proof any arithmetic progression coprime count same as toting function of nHow 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 $m>n$?Are there any known methods for finding Upper/Lower bounds on the number of Totients of x less than another number y?
$begingroup$
I'd like to ask how to count $f(a,b)$, the number of integers not greater than $a$ which are coprime to a given number $b$. Can $f$ be expressed using Euler's totient function?
combinatorics elementary-number-theory totient-function coprime
asked Mar 22 at 11:02
Hang WuHang Wu
551312
$endgroup$
add a comment |
$begingroup$
I'd like to ask how to count $f(a,b)$, the number of integers not greater than $a$ which are coprime to a given number $b$. Can $f$ be expressed using Euler's totient function?
combinatorics elementary-number-theory totient-function coprime
asked Mar 22 at 11:02
Hang WuHang Wu
551312
$endgroup$
add a comment |
$begingroup$
I'd like to ask how to count $f(a,b)$, the number of integers not greater than $a$ which are coprime to a given number $b$. Can $f$ be expressed using Euler's totient function?
combinatorics elementary-number-theory totient-function coprime
asked Mar 22 at 11:02
Hang WuHang Wu
551312
$endgroup$
I'd like to ask how to count $f(a,b)$, the number of integers not greater than $a$ which are coprime to a given number $b$. Can $f$ be expressed using Euler's totient function?
combinatorics elementary-number-theory totient-function coprime
combinatorics elementary-number-theory totient-function coprime
asked Mar 22 at 11:02
Hang WuHang Wu
551312
asked Mar 22 at 11:02
Hang WuHang Wu
551312
asked Mar 22 at 11:02
Hang WuHang Wu
551312
asked Mar 22 at 11:02
Hang WuHang Wu
551312
asked Mar 22 at 11:02
Hang WuHang Wu
551312
551312
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can do this using the basic property of the Mobius function $mu$, which is
$$ sum_dmid m mu(d) = begincases 1, m=1 \ 0, m>1 endcases $$
(where $m$ is a positive integer, and summation extends over all positive divisors $d$ of $m$). Namely, we have
$$ f(a,b) = sum_n=1^a sum_dmid(n,b)mu(d) = sum_dmid b mu(d) sum_substack1le nle a\dmid n 1. $$
The inner sum counts all integers $nin[1,a]$ divisible by $d$; hence, is equal to $lfloor a/drfloor$. This gives
$$ f(a,b) = sum_dmid b mu(d) lfloor a/drfloor. $$
This is an exact formula which can be used to efficiently compute your $f(a,b)$. You can also use it to get a good approximation: since $lfloor a/drfloor=a/d-theta$, where $|theta|<1$,
$$ f(a,b) = asum_dmid b fracmu(d)d + R = afracvarphi(b)b + R, $$
where $|R|$ does not exceed the number of divisors of $b$.
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
$endgroup$
add a comment |
$begingroup$
In some cases yes $f$ can be expressed in terms of Eulers totient function. Any time b divides a, $f(a,b)$ is simply $fracabphi(b)$. Euler totient function can also be written in terms of $f$ though. Euler totient function is $$prod_pf(p,p)$$ in general $f(a,b)$ is $lfloorfracabrfloorphi(b)$ plus the cardinality of the set of numbers (via inclusion-exclusion principle or f itself) not divisible by a prime factor of b less than $abmod b$
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You can do this using the basic property of the Mobius function $mu$, which is
$$ sum_dmid m mu(d) = begincases 1, m=1 \ 0, m>1 endcases $$
(where $m$ is a positive integer, and summation extends over all positive divisors $d$ of $m$). Namely, we have
$$ f(a,b) = sum_n=1^a sum_dmid(n,b)mu(d) = sum_dmid b mu(d) sum_substack1le nle a\dmid n 1. $$
The inner sum counts all integers $nin[1,a]$ divisible by $d$; hence, is equal to $lfloor a/drfloor$. This gives
$$ f(a,b) = sum_dmid b mu(d) lfloor a/drfloor. $$
This is an exact formula which can be used to efficiently compute your $f(a,b)$. You can also use it to get a good approximation: since $lfloor a/drfloor=a/d-theta$, where $|theta|<1$,
$$ f(a,b) = asum_dmid b fracmu(d)d + R = afracvarphi(b)b + R, $$
where $|R|$ does not exceed the number of divisors of $b$.
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
$endgroup$
add a comment |
$begingroup$
You can do this using the basic property of the Mobius function $mu$, which is
$$ sum_dmid m mu(d) = begincases 1, m=1 \ 0, m>1 endcases $$
(where $m$ is a positive integer, and summation extends over all positive divisors $d$ of $m$). Namely, we have
$$ f(a,b) = sum_n=1^a sum_dmid(n,b)mu(d) = sum_dmid b mu(d) sum_substack1le nle a\dmid n 1. $$
The inner sum counts all integers $nin[1,a]$ divisible by $d$; hence, is equal to $lfloor a/drfloor$. This gives
$$ f(a,b) = sum_dmid b mu(d) lfloor a/drfloor. $$
This is an exact formula which can be used to efficiently compute your $f(a,b)$. You can also use it to get a good approximation: since $lfloor a/drfloor=a/d-theta$, where $|theta|<1$,
$$ f(a,b) = asum_dmid b fracmu(d)d + R = afracvarphi(b)b + R, $$
where $|R|$ does not exceed the number of divisors of $b$.
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
$endgroup$
add a comment |
$begingroup$
You can do this using the basic property of the Mobius function $mu$, which is
$$ sum_dmid m mu(d) = begincases 1, m=1 \ 0, m>1 endcases $$
(where $m$ is a positive integer, and summation extends over all positive divisors $d$ of $m$). Namely, we have
$$ f(a,b) = sum_n=1^a sum_dmid(n,b)mu(d) = sum_dmid b mu(d) sum_substack1le nle a\dmid n 1. $$
The inner sum counts all integers $nin[1,a]$ divisible by $d$; hence, is equal to $lfloor a/drfloor$. This gives
$$ f(a,b) = sum_dmid b mu(d) lfloor a/drfloor. $$
This is an exact formula which can be used to efficiently compute your $f(a,b)$. You can also use it to get a good approximation: since $lfloor a/drfloor=a/d-theta$, where $|theta|<1$,
$$ f(a,b) = asum_dmid b fracmu(d)d + R = afracvarphi(b)b + R, $$
where $|R|$ does not exceed the number of divisors of $b$.
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
$endgroup$
You can do this using the basic property of the Mobius function $mu$, which is
$$ sum_dmid m mu(d) = begincases 1, m=1 \ 0, m>1 endcases $$
(where $m$ is a positive integer, and summation extends over all positive divisors $d$ of $m$). Namely, we have
$$ f(a,b) = sum_n=1^a sum_dmid(n,b)mu(d) = sum_dmid b mu(d) sum_substack1le nle a\dmid n 1. $$
The inner sum counts all integers $nin[1,a]$ divisible by $d$; hence, is equal to $lfloor a/drfloor$. This gives
$$ f(a,b) = sum_dmid b mu(d) lfloor a/drfloor. $$
This is an exact formula which can be used to efficiently compute your $f(a,b)$. You can also use it to get a good approximation: since $lfloor a/drfloor=a/d-theta$, where $|theta|<1$,
$$ f(a,b) = asum_dmid b fracmu(d)d + R = afracvarphi(b)b + R, $$
where $|R|$ does not exceed the number of divisors of $b$.
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
answered Mar 22 at 11:36
W-t-PW-t-P
1,614612
1,614612
add a comment |
add a comment |
$begingroup$
In some cases yes $f$ can be expressed in terms of Eulers totient function. Any time b divides a, $f(a,b)$ is simply $fracabphi(b)$. Euler totient function can also be written in terms of $f$ though. Euler totient function is $$prod_pf(p,p)$$ in general $f(a,b)$ is $lfloorfracabrfloorphi(b)$ plus the cardinality of the set of numbers (via inclusion-exclusion principle or f itself) not divisible by a prime factor of b less than $abmod b$
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
add a comment |
$begingroup$
In some cases yes $f$ can be expressed in terms of Eulers totient function. Any time b divides a, $f(a,b)$ is simply $fracabphi(b)$. Euler totient function can also be written in terms of $f$ though. Euler totient function is $$prod_pf(p,p)$$ in general $f(a,b)$ is $lfloorfracabrfloorphi(b)$ plus the cardinality of the set of numbers (via inclusion-exclusion principle or f itself) not divisible by a prime factor of b less than $abmod b$
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
add a comment |
$begingroup$
In some cases yes $f$ can be expressed in terms of Eulers totient function. Any time b divides a, $f(a,b)$ is simply $fracabphi(b)$. Euler totient function can also be written in terms of $f$ though. Euler totient function is $$prod_pf(p,p)$$ in general $f(a,b)$ is $lfloorfracabrfloorphi(b)$ plus the cardinality of the set of numbers (via inclusion-exclusion principle or f itself) not divisible by a prime factor of b less than $abmod b$
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
$endgroup$
In some cases yes $f$ can be expressed in terms of Eulers totient function. Any time b divides a, $f(a,b)$ is simply $fracabphi(b)$. Euler totient function can also be written in terms of $f$ though. Euler totient function is $$prod_pf(p,p)$$ in general $f(a,b)$ is $lfloorfracabrfloorphi(b)$ plus the cardinality of the set of numbers (via inclusion-exclusion principle or f itself) not divisible by a prime factor of b less than $abmod b$
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
edited Mar 23 at 0:58
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
answered Mar 23 at 0:48
Roddy MacPheeRoddy MacPhee
722118
722118
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
add a comment |
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
So you mean $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq a\ forall p < a mod b bigwedge p1$?
$endgroup$
– Hang Wu
Mar 23 at 3:16
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
But I think the formula should be $f(a,b)=lfloorfracabrfloorphi(b)+sum_substack1leq i leq (a mod b) \ gcd(i,b)=11$.
$endgroup$
– Hang Wu
Mar 23 at 3:35
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
$begingroup$
$a=1005,b=200; f(a,b)= 5(80)+1(4)-2(1)$ because 2 and 5 are less than or equal to 5. 5 leaves 4 coprime to it, 2 takes away 2 of them. this leaves 5(80)+2= 402 numbers coprime to 200.
$endgroup$
– Roddy MacPhee
Mar 23 at 12:57
add a comment |
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