Number of ways a natural number can be written as a sum of naturals that are all coprime to it.
$begingroup$
let $X: mathbb{N}^2 to mathbb{N}$
Let $X(a ,b)$ be the number of unique ways we can write $a$ as the sum of $b$ many numbers, where each of the $b$ numbers are co-prime to $a$. Where $a$ $in mathbb{N}$ and $b$ $in mathbb{N}$
Example:
$X(a ,2) = |{(x, y): x + y = a$, $gcd(a, x) = gcd(a, y) = 1}|$
I can easily show that $X(a, 2) = frac{phi(a)}{2}$, where $phi$ is Euler's totient function, when $a > 2$.
Proof:
Let $Phi_{a} = {k: gcd(a, k) = 1, k in mathbb{N}}$
let $k in mathbb{N}$, then it is easy to see $forall k < a$, $exists n in mathbb{N}$ such that $a = k + n$
Specifically, this is $n = a - k$
Now if we only consider $k in Phi_{a}$ we can see that $n in Phi_{a}$. Why is this so? Here is why:
Let $k in Phi_{a}$ i.e. $gcd(a, k) = 1$, now if we assume $n notin Phi_{a}$ i.e. $gcd(a, n) neq 1$ it implies that $exists m in mathbb{N}$ such that $m | a$ and $m | n$. But this means that $m | k$ (because $n = a - k$) which contradicts $gcd(a, k) = 1$. Thus our assumption was wrong and therefore $n in Phi_{a}$.
Hence it follows that, $forall x in Phi_{a}$ $exists y in Phi_{a}$ such that $x + y = a$. We also know that $|Phi_{a}| = phi(a)$ thus once we pair our numbers together we have exactly $frac{phi(a)}{2}$ unique pairs $(x, y)$. Here unique means if we have counted the pair $(x, y)$ then we do not count the pair $(y, x)$ as we consider them to be the same pair.
End of proof
Now to my actual question: Can we find the value of $X(a, b)$ in terms of $a$ and $b$ when $b > 2$? So far I have only defined the trivial cases:
$X(a ,b) = begin{cases} 0, & b gt a &OR& b + 1equiv aequiv 0pmod 2 &OR& b = 1, agt 1\ 1, & b = a\ frac{phi(a)}{2},& b = 2, a gt 2\ ?,& Otherwiseend{cases}$
Just so that my question is clear, for $b = 3$ we have:
$X(a ,3) = |{(x, y, z): x + y + z = a$, $gcd(a, x) = gcd(a, y) = gcd(a, z) = 1}|$
This question is purely out of interest, thanks in advance for any answers.
elementary-number-theory
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
$endgroup$
|
show 2 more comments
$begingroup$
let $X: mathbb{N}^2 to mathbb{N}$
Let $X(a ,b)$ be the number of unique ways we can write $a$ as the sum of $b$ many numbers, where each of the $b$ numbers are co-prime to $a$. Where $a$ $in mathbb{N}$ and $b$ $in mathbb{N}$
Example:
$X(a ,2) = |{(x, y): x + y = a$, $gcd(a, x) = gcd(a, y) = 1}|$
I can easily show that $X(a, 2) = frac{phi(a)}{2}$, where $phi$ is Euler's totient function, when $a > 2$.
Proof:
Let $Phi_{a} = {k: gcd(a, k) = 1, k in mathbb{N}}$
let $k in mathbb{N}$, then it is easy to see $forall k < a$, $exists n in mathbb{N}$ such that $a = k + n$
Specifically, this is $n = a - k$
Now if we only consider $k in Phi_{a}$ we can see that $n in Phi_{a}$. Why is this so? Here is why:
Let $k in Phi_{a}$ i.e. $gcd(a, k) = 1$, now if we assume $n notin Phi_{a}$ i.e. $gcd(a, n) neq 1$ it implies that $exists m in mathbb{N}$ such that $m | a$ and $m | n$. But this means that $m | k$ (because $n = a - k$) which contradicts $gcd(a, k) = 1$. Thus our assumption was wrong and therefore $n in Phi_{a}$.
Hence it follows that, $forall x in Phi_{a}$ $exists y in Phi_{a}$ such that $x + y = a$. We also know that $|Phi_{a}| = phi(a)$ thus once we pair our numbers together we have exactly $frac{phi(a)}{2}$ unique pairs $(x, y)$. Here unique means if we have counted the pair $(x, y)$ then we do not count the pair $(y, x)$ as we consider them to be the same pair.
End of proof
Now to my actual question: Can we find the value of $X(a, b)$ in terms of $a$ and $b$ when $b > 2$? So far I have only defined the trivial cases:
$X(a ,b) = begin{cases} 0, & b gt a &OR& b + 1equiv aequiv 0pmod 2 &OR& b = 1, agt 1\ 1, & b = a\ frac{phi(a)}{2},& b = 2, a gt 2\ ?,& Otherwiseend{cases}$
Just so that my question is clear, for $b = 3$ we have:
$X(a ,3) = |{(x, y, z): x + y + z = a$, $gcd(a, x) = gcd(a, y) = gcd(a, z) = 1}|$
This question is purely out of interest, thanks in advance for any answers.
elementary-number-theory
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
$endgroup$
1
$begingroup$
Since $phi(1)=phi(2)=1$, $phi(a)$ isn't always even.
$endgroup$
– J.G.
Dec 6 '18 at 9:12
$begingroup$
Here are the values of $X(a,3)$ for $3leq aleq27$: 1, 0, 6, 0, 15, 0, 9, 0, 45, 0, 66, 0, 12, 0, 120, 0, 153, 0, 30, 0, 231, 0, 150, 0, 81. This is enough to show that the sequence isn't known to OEIS or Google.
$endgroup$
– Chris Culter
Dec 27 '18 at 10:19
1
$begingroup$
@ChrisCulter You must be counting all the tuples $(x,y,z)$ as different (that is integer combinations). I think the question asks for integer partitions (see the last sentence of op's proof). Although then the notation $(x,y,dots)$ isn't correct. We should demand $xleq y leqdots$. For this partition definition I got the values for $X(a, 3)$ for $ageq 0$: $$0, 0, 0, 1, 0, 2, 0, 4, 0, 3, 0, 10, 0, 14, 0, 4, 0, 24, 0, 30, 0, 8, 0, 44,dots$$. That isn't found either.
$endgroup$
– ploosu2
Dec 29 '18 at 9:41
$begingroup$
If you allow repetitions (that is, $(2,3)$ and $(3,2)$ are both counted in the sum $X(5,2)$ ) then you can write a nice recursive relation between $X(a,b)$ and some $X(?,b-1)$s. It's unlikely that you can find something better than that (as others have checked on OEIS).
$endgroup$
– Breakfastisready
Dec 30 '18 at 2:19
2
$begingroup$
If you sum over all $b$ for a fixed $a$, and then index a sequence by $a$, you get A057562
$endgroup$
– alex.jordan
Jan 1 at 6:53
|
show 2 more comments
$begingroup$
let $X: mathbb{N}^2 to mathbb{N}$
Let $X(a ,b)$ be the number of unique ways we can write $a$ as the sum of $b$ many numbers, where each of the $b$ numbers are co-prime to $a$. Where $a$ $in mathbb{N}$ and $b$ $in mathbb{N}$
Example:
$X(a ,2) = |{(x, y): x + y = a$, $gcd(a, x) = gcd(a, y) = 1}|$
I can easily show that $X(a, 2) = frac{phi(a)}{2}$, where $phi$ is Euler's totient function, when $a > 2$.
Proof:
Let $Phi_{a} = {k: gcd(a, k) = 1, k in mathbb{N}}$
let $k in mathbb{N}$, then it is easy to see $forall k < a$, $exists n in mathbb{N}$ such that $a = k + n$
Specifically, this is $n = a - k$
Now if we only consider $k in Phi_{a}$ we can see that $n in Phi_{a}$. Why is this so? Here is why:
Let $k in Phi_{a}$ i.e. $gcd(a, k) = 1$, now if we assume $n notin Phi_{a}$ i.e. $gcd(a, n) neq 1$ it implies that $exists m in mathbb{N}$ such that $m | a$ and $m | n$. But this means that $m | k$ (because $n = a - k$) which contradicts $gcd(a, k) = 1$. Thus our assumption was wrong and therefore $n in Phi_{a}$.
Hence it follows that, $forall x in Phi_{a}$ $exists y in Phi_{a}$ such that $x + y = a$. We also know that $|Phi_{a}| = phi(a)$ thus once we pair our numbers together we have exactly $frac{phi(a)}{2}$ unique pairs $(x, y)$. Here unique means if we have counted the pair $(x, y)$ then we do not count the pair $(y, x)$ as we consider them to be the same pair.
End of proof
Now to my actual question: Can we find the value of $X(a, b)$ in terms of $a$ and $b$ when $b > 2$? So far I have only defined the trivial cases:
$X(a ,b) = begin{cases} 0, & b gt a &OR& b + 1equiv aequiv 0pmod 2 &OR& b = 1, agt 1\ 1, & b = a\ frac{phi(a)}{2},& b = 2, a gt 2\ ?,& Otherwiseend{cases}$
Just so that my question is clear, for $b = 3$ we have:
$X(a ,3) = |{(x, y, z): x + y + z = a$, $gcd(a, x) = gcd(a, y) = gcd(a, z) = 1}|$
This question is purely out of interest, thanks in advance for any answers.
elementary-number-theory
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
$endgroup$
let $X: mathbb{N}^2 to mathbb{N}$
Let $X(a ,b)$ be the number of unique ways we can write $a$ as the sum of $b$ many numbers, where each of the $b$ numbers are co-prime to $a$. Where $a$ $in mathbb{N}$ and $b$ $in mathbb{N}$
Example:
$X(a ,2) = |{(x, y): x + y = a$, $gcd(a, x) = gcd(a, y) = 1}|$
I can easily show that $X(a, 2) = frac{phi(a)}{2}$, where $phi$ is Euler's totient function, when $a > 2$.
Proof:
Let $Phi_{a} = {k: gcd(a, k) = 1, k in mathbb{N}}$
let $k in mathbb{N}$, then it is easy to see $forall k < a$, $exists n in mathbb{N}$ such that $a = k + n$
Specifically, this is $n = a - k$
Now if we only consider $k in Phi_{a}$ we can see that $n in Phi_{a}$. Why is this so? Here is why:
Let $k in Phi_{a}$ i.e. $gcd(a, k) = 1$, now if we assume $n notin Phi_{a}$ i.e. $gcd(a, n) neq 1$ it implies that $exists m in mathbb{N}$ such that $m | a$ and $m | n$. But this means that $m | k$ (because $n = a - k$) which contradicts $gcd(a, k) = 1$. Thus our assumption was wrong and therefore $n in Phi_{a}$.
Hence it follows that, $forall x in Phi_{a}$ $exists y in Phi_{a}$ such that $x + y = a$. We also know that $|Phi_{a}| = phi(a)$ thus once we pair our numbers together we have exactly $frac{phi(a)}{2}$ unique pairs $(x, y)$. Here unique means if we have counted the pair $(x, y)$ then we do not count the pair $(y, x)$ as we consider them to be the same pair.
End of proof
Now to my actual question: Can we find the value of $X(a, b)$ in terms of $a$ and $b$ when $b > 2$? So far I have only defined the trivial cases:
$X(a ,b) = begin{cases} 0, & b gt a &OR& b + 1equiv aequiv 0pmod 2 &OR& b = 1, agt 1\ 1, & b = a\ frac{phi(a)}{2},& b = 2, a gt 2\ ?,& Otherwiseend{cases}$
Just so that my question is clear, for $b = 3$ we have:
$X(a ,3) = |{(x, y, z): x + y + z = a$, $gcd(a, x) = gcd(a, y) = gcd(a, z) = 1}|$
This question is purely out of interest, thanks in advance for any answers.
elementary-number-theory
elementary-number-theory
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
edited Jan 1 at 8:44
DanielOnMSE
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
asked Dec 6 '18 at 9:03
DanielOnMSEDanielOnMSE
966
966
1
$begingroup$
Since $phi(1)=phi(2)=1$, $phi(a)$ isn't always even.
$endgroup$
– J.G.
Dec 6 '18 at 9:12
$begingroup$
Here are the values of $X(a,3)$ for $3leq aleq27$: 1, 0, 6, 0, 15, 0, 9, 0, 45, 0, 66, 0, 12, 0, 120, 0, 153, 0, 30, 0, 231, 0, 150, 0, 81. This is enough to show that the sequence isn't known to OEIS or Google.
$endgroup$
– Chris Culter
Dec 27 '18 at 10:19
1
$begingroup$
@ChrisCulter You must be counting all the tuples $(x,y,z)$ as different (that is integer combinations). I think the question asks for integer partitions (see the last sentence of op's proof). Although then the notation $(x,y,dots)$ isn't correct. We should demand $xleq y leqdots$. For this partition definition I got the values for $X(a, 3)$ for $ageq 0$: $$0, 0, 0, 1, 0, 2, 0, 4, 0, 3, 0, 10, 0, 14, 0, 4, 0, 24, 0, 30, 0, 8, 0, 44,dots$$. That isn't found either.
$endgroup$
– ploosu2
Dec 29 '18 at 9:41
$begingroup$
If you allow repetitions (that is, $(2,3)$ and $(3,2)$ are both counted in the sum $X(5,2)$ ) then you can write a nice recursive relation between $X(a,b)$ and some $X(?,b-1)$s. It's unlikely that you can find something better than that (as others have checked on OEIS).
$endgroup$
– Breakfastisready
Dec 30 '18 at 2:19
2
$begingroup$
If you sum over all $b$ for a fixed $a$, and then index a sequence by $a$, you get A057562
$endgroup$
– alex.jordan
Jan 1 at 6:53
|
show 2 more comments
1
$begingroup$
Since $phi(1)=phi(2)=1$, $phi(a)$ isn't always even.
$endgroup$
– J.G.
Dec 6 '18 at 9:12
$begingroup$
Here are the values of $X(a,3)$ for $3leq aleq27$: 1, 0, 6, 0, 15, 0, 9, 0, 45, 0, 66, 0, 12, 0, 120, 0, 153, 0, 30, 0, 231, 0, 150, 0, 81. This is enough to show that the sequence isn't known to OEIS or Google.
$endgroup$
– Chris Culter
Dec 27 '18 at 10:19
1
$begingroup$
@ChrisCulter You must be counting all the tuples $(x,y,z)$ as different (that is integer combinations). I think the question asks for integer partitions (see the last sentence of op's proof). Although then the notation $(x,y,dots)$ isn't correct. We should demand $xleq y leqdots$. For this partition definition I got the values for $X(a, 3)$ for $ageq 0$: $$0, 0, 0, 1, 0, 2, 0, 4, 0, 3, 0, 10, 0, 14, 0, 4, 0, 24, 0, 30, 0, 8, 0, 44,dots$$. That isn't found either.
$endgroup$
– ploosu2
Dec 29 '18 at 9:41
$begingroup$
If you allow repetitions (that is, $(2,3)$ and $(3,2)$ are both counted in the sum $X(5,2)$ ) then you can write a nice recursive relation between $X(a,b)$ and some $X(?,b-1)$s. It's unlikely that you can find something better than that (as others have checked on OEIS).
$endgroup$
– Breakfastisready
Dec 30 '18 at 2:19
2
$begingroup$
If you sum over all $b$ for a fixed $a$, and then index a sequence by $a$, you get A057562
$endgroup$
– alex.jordan
Jan 1 at 6:53
1
1
$begingroup$
Since $phi(1)=phi(2)=1$, $phi(a)$ isn't always even.
$endgroup$
– J.G.
Dec 6 '18 at 9:12
$begingroup$
Since $phi(1)=phi(2)=1$, $phi(a)$ isn't always even.
$endgroup$
– J.G.
Dec 6 '18 at 9:12
$begingroup$
Here are the values of $X(a,3)$ for $3leq aleq27$: 1, 0, 6, 0, 15, 0, 9, 0, 45, 0, 66, 0, 12, 0, 120, 0, 153, 0, 30, 0, 231, 0, 150, 0, 81. This is enough to show that the sequence isn't known to OEIS or Google.
$endgroup$
– Chris Culter
Dec 27 '18 at 10:19
$begingroup$
Here are the values of $X(a,3)$ for $3leq aleq27$: 1, 0, 6, 0, 15, 0, 9, 0, 45, 0, 66, 0, 12, 0, 120, 0, 153, 0, 30, 0, 231, 0, 150, 0, 81. This is enough to show that the sequence isn't known to OEIS or Google.
$endgroup$
– Chris Culter
Dec 27 '18 at 10:19
1
1
$begingroup$
@ChrisCulter You must be counting all the tuples $(x,y,z)$ as different (that is integer combinations). I think the question asks for integer partitions (see the last sentence of op's proof). Although then the notation $(x,y,dots)$ isn't correct. We should demand $xleq y leqdots$. For this partition definition I got the values for $X(a, 3)$ for $ageq 0$: $$0, 0, 0, 1, 0, 2, 0, 4, 0, 3, 0, 10, 0, 14, 0, 4, 0, 24, 0, 30, 0, 8, 0, 44,dots$$. That isn't found either.
$endgroup$
– ploosu2
Dec 29 '18 at 9:41
$begingroup$
@ChrisCulter You must be counting all the tuples $(x,y,z)$ as different (that is integer combinations). I think the question asks for integer partitions (see the last sentence of op's proof). Although then the notation $(x,y,dots)$ isn't correct. We should demand $xleq y leqdots$. For this partition definition I got the values for $X(a, 3)$ for $ageq 0$: $$0, 0, 0, 1, 0, 2, 0, 4, 0, 3, 0, 10, 0, 14, 0, 4, 0, 24, 0, 30, 0, 8, 0, 44,dots$$. That isn't found either.
$endgroup$
– ploosu2
Dec 29 '18 at 9:41
$begingroup$
If you allow repetitions (that is, $(2,3)$ and $(3,2)$ are both counted in the sum $X(5,2)$ ) then you can write a nice recursive relation between $X(a,b)$ and some $X(?,b-1)$s. It's unlikely that you can find something better than that (as others have checked on OEIS).
$endgroup$
– Breakfastisready
Dec 30 '18 at 2:19
$begingroup$
If you allow repetitions (that is, $(2,3)$ and $(3,2)$ are both counted in the sum $X(5,2)$ ) then you can write a nice recursive relation between $X(a,b)$ and some $X(?,b-1)$s. It's unlikely that you can find something better than that (as others have checked on OEIS).
$endgroup$
– Breakfastisready
Dec 30 '18 at 2:19
2
2
$begingroup$
If you sum over all $b$ for a fixed $a$, and then index a sequence by $a$, you get A057562
$endgroup$
– alex.jordan
Jan 1 at 6:53
$begingroup$
If you sum over all $b$ for a fixed $a$, and then index a sequence by $a$, you get A057562
$endgroup$
– alex.jordan
Jan 1 at 6:53
|
show 2 more comments
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1
$begingroup$
Since $phi(1)=phi(2)=1$, $phi(a)$ isn't always even.
$endgroup$
– J.G.
Dec 6 '18 at 9:12
$begingroup$
Here are the values of $X(a,3)$ for $3leq aleq27$: 1, 0, 6, 0, 15, 0, 9, 0, 45, 0, 66, 0, 12, 0, 120, 0, 153, 0, 30, 0, 231, 0, 150, 0, 81. This is enough to show that the sequence isn't known to OEIS or Google.
$endgroup$
– Chris Culter
Dec 27 '18 at 10:19
1
$begingroup$
@ChrisCulter You must be counting all the tuples $(x,y,z)$ as different (that is integer combinations). I think the question asks for integer partitions (see the last sentence of op's proof). Although then the notation $(x,y,dots)$ isn't correct. We should demand $xleq y leqdots$. For this partition definition I got the values for $X(a, 3)$ for $ageq 0$: $$0, 0, 0, 1, 0, 2, 0, 4, 0, 3, 0, 10, 0, 14, 0, 4, 0, 24, 0, 30, 0, 8, 0, 44,dots$$. That isn't found either.
$endgroup$
– ploosu2
Dec 29 '18 at 9:41
$begingroup$
If you allow repetitions (that is, $(2,3)$ and $(3,2)$ are both counted in the sum $X(5,2)$ ) then you can write a nice recursive relation between $X(a,b)$ and some $X(?,b-1)$s. It's unlikely that you can find something better than that (as others have checked on OEIS).
$endgroup$
– Breakfastisready
Dec 30 '18 at 2:19
2
$begingroup$
If you sum over all $b$ for a fixed $a$, and then index a sequence by $a$, you get A057562
$endgroup$
– alex.jordan
Jan 1 at 6:53