Formula for product of sums of pairs of coprime divisors of $n$.
Can we develop a formula for
$$
r(n)=prod_{
begin{array}{c}
x,ymid n \
(x,y)=1
end{array}}
(x+y)
$$
In words this is the product of sums of all coprime pairs of divisors of $n$.
For example
$$r(12)=(1+2)(1+3)(1+4)(1+6)(1+12)(2+3)(3+4)=191100$$
There is some information on the number of coprime pairs of divisors (A063647).
In answer to the comment, this problem comes from wanting to know whether a certain function of $n$ divides at least one of the $(x+y)$ terms. This is true iff the function divides $r(n)$. I have no idea if a nice form exists, but I would certainly accept a form in terms of the prime factorization of $n$.
combinatorics number-theory divisor-counting-function coprime
asked Nov 30 at 0:59
tyobrien
1,171513
add a comment |
Can we develop a formula for
$$
r(n)=prod_{
begin{array}{c}
x,ymid n \
(x,y)=1
end{array}}
(x+y)
$$
In words this is the product of sums of all coprime pairs of divisors of $n$.
For example
$$r(12)=(1+2)(1+3)(1+4)(1+6)(1+12)(2+3)(3+4)=191100$$
There is some information on the number of coprime pairs of divisors (A063647).
In answer to the comment, this problem comes from wanting to know whether a certain function of $n$ divides at least one of the $(x+y)$ terms. This is true iff the function divides $r(n)$. I have no idea if a nice form exists, but I would certainly accept a form in terms of the prime factorization of $n$.
combinatorics number-theory divisor-counting-function coprime
asked Nov 30 at 0:59
tyobrien
1,171513
4
Where did this problem come from and/or do you have a reason to believe there's a nice closed form expression for this?
– Carl Schildkraut
Nov 30 at 1:01
I mean you can write the prime factorization of $n$ and get an expression.
– mathworker21
Nov 30 at 1:07
"This is true iff the function divides r(n)." That's not true in general. f(n) = n+1 comes to mind...
– Todor Markov
Nov 30 at 18:36
True @TodorMarkov I realized this recently.
– tyobrien
Dec 1 at 16:08
However I may be able to show that my $f(n)$ is a prime...
– tyobrien
Dec 3 at 22:16
add a comment |
Can we develop a formula for
$$
r(n)=prod_{
begin{array}{c}
x,ymid n \
(x,y)=1
end{array}}
(x+y)
$$
In words this is the product of sums of all coprime pairs of divisors of $n$.
For example
$$r(12)=(1+2)(1+3)(1+4)(1+6)(1+12)(2+3)(3+4)=191100$$
There is some information on the number of coprime pairs of divisors (A063647).
In answer to the comment, this problem comes from wanting to know whether a certain function of $n$ divides at least one of the $(x+y)$ terms. This is true iff the function divides $r(n)$. I have no idea if a nice form exists, but I would certainly accept a form in terms of the prime factorization of $n$.
combinatorics number-theory divisor-counting-function coprime
asked Nov 30 at 0:59
tyobrien
1,171513
Can we develop a formula for
$$
r(n)=prod_{
begin{array}{c}
x,ymid n \
(x,y)=1
end{array}}
(x+y)
$$
In words this is the product of sums of all coprime pairs of divisors of $n$.
For example
$$r(12)=(1+2)(1+3)(1+4)(1+6)(1+12)(2+3)(3+4)=191100$$
There is some information on the number of coprime pairs of divisors (A063647).
In answer to the comment, this problem comes from wanting to know whether a certain function of $n$ divides at least one of the $(x+y)$ terms. This is true iff the function divides $r(n)$. I have no idea if a nice form exists, but I would certainly accept a form in terms of the prime factorization of $n$.
combinatorics number-theory divisor-counting-function coprime
combinatorics number-theory divisor-counting-function coprime
asked Nov 30 at 0:59
tyobrien
1,171513
asked Nov 30 at 0:59
tyobrien
1,171513
edited Nov 30 at 1:54
asked Nov 30 at 0:59
tyobrien
1,171513
asked Nov 30 at 0:59
tyobrien
1,171513
asked Nov 30 at 0:59
tyobrien
1,171513
1,171513
4
Where did this problem come from and/or do you have a reason to believe there's a nice closed form expression for this?
– Carl Schildkraut
Nov 30 at 1:01
I mean you can write the prime factorization of $n$ and get an expression.
– mathworker21
Nov 30 at 1:07
"This is true iff the function divides r(n)." That's not true in general. f(n) = n+1 comes to mind...
– Todor Markov
Nov 30 at 18:36
True @TodorMarkov I realized this recently.
– tyobrien
Dec 1 at 16:08
However I may be able to show that my $f(n)$ is a prime...
– tyobrien
Dec 3 at 22:16
add a comment |
4
Where did this problem come from and/or do you have a reason to believe there's a nice closed form expression for this?
– Carl Schildkraut
Nov 30 at 1:01
I mean you can write the prime factorization of $n$ and get an expression.
– mathworker21
Nov 30 at 1:07
"This is true iff the function divides r(n)." That's not true in general. f(n) = n+1 comes to mind...
– Todor Markov
Nov 30 at 18:36
True @TodorMarkov I realized this recently.
– tyobrien
Dec 1 at 16:08
However I may be able to show that my $f(n)$ is a prime...
– tyobrien
Dec 3 at 22:16
4
4
Where did this problem come from and/or do you have a reason to believe there's a nice closed form expression for this?
– Carl Schildkraut
Nov 30 at 1:01
Where did this problem come from and/or do you have a reason to believe there's a nice closed form expression for this?
– Carl Schildkraut
Nov 30 at 1:01
I mean you can write the prime factorization of $n$ and get an expression.
– mathworker21
Nov 30 at 1:07
I mean you can write the prime factorization of $n$ and get an expression.
– mathworker21
Nov 30 at 1:07
"This is true iff the function divides r(n)." That's not true in general. f(n) = n+1 comes to mind...
– Todor Markov
Nov 30 at 18:36
"This is true iff the function divides r(n)." That's not true in general. f(n) = n+1 comes to mind...
– Todor Markov
Nov 30 at 18:36
True @TodorMarkov I realized this recently.
– tyobrien
Dec 1 at 16:08
True @TodorMarkov I realized this recently.
– tyobrien
Dec 1 at 16:08
However I may be able to show that my $f(n)$ is a prime...
– tyobrien
Dec 3 at 22:16
However I may be able to show that my $f(n)$ is a prime...
– tyobrien
Dec 3 at 22:16
add a comment |
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4
Where did this problem come from and/or do you have a reason to believe there's a nice closed form expression for this?
– Carl Schildkraut
Nov 30 at 1:01
I mean you can write the prime factorization of $n$ and get an expression.
– mathworker21
Nov 30 at 1:07
"This is true iff the function divides r(n)." That's not true in general. f(n) = n+1 comes to mind...
– Todor Markov
Nov 30 at 18:36
True @TodorMarkov I realized this recently.
– tyobrien
Dec 1 at 16:08
However I may be able to show that my $f(n)$ is a prime...
– tyobrien
Dec 3 at 22:16