How to count the number of solutions to linear Diophantine equation with multiple variables recursively?
$begingroup$
Consider the diophantine equation of the form
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
where $n,kgeq 1$ and $a_i$ are non-negative integers. Then we are concerned $v(n)$ denote the number of solutions for a fixed tuple $(a_1,a_2,cdots ,a_k).$ Can $v(n)$ be expressed recursively, that is, in terms of $v(n-1),v(n-2),$ etc?
I considered the following example to see if I could generalize
$$x_1+2x_2 = n.$$
I noticed that if $(x_1,x_2)$ is a solution of $x_1+2x_2 = n-k$ then $(x_1+k,x_2)$ is a solution of $x_1+2x_2=n.$ Furthermore for $k=2p$ then if $x_1+2x_2 = n-k=n - 2p$ then $(x_1,x_2+p)$ is a solution of $x_1+2x_2=n.$ Therefore,
$$v(n) = sum_{k=1}^{n}v(n-k)+v(n-2)+v(n-4)+cdots $$
Is this correct? How do I get a general formula for $v(n)?$ Any ideas will be much appreciated?
combinatorics number-theory
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
$endgroup$
add a comment |
$begingroup$
Consider the diophantine equation of the form
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
where $n,kgeq 1$ and $a_i$ are non-negative integers. Then we are concerned $v(n)$ denote the number of solutions for a fixed tuple $(a_1,a_2,cdots ,a_k).$ Can $v(n)$ be expressed recursively, that is, in terms of $v(n-1),v(n-2),$ etc?
I considered the following example to see if I could generalize
$$x_1+2x_2 = n.$$
I noticed that if $(x_1,x_2)$ is a solution of $x_1+2x_2 = n-k$ then $(x_1+k,x_2)$ is a solution of $x_1+2x_2=n.$ Furthermore for $k=2p$ then if $x_1+2x_2 = n-k=n - 2p$ then $(x_1,x_2+p)$ is a solution of $x_1+2x_2=n.$ Therefore,
$$v(n) = sum_{k=1}^{n}v(n-k)+v(n-2)+v(n-4)+cdots $$
Is this correct? How do I get a general formula for $v(n)?$ Any ideas will be much appreciated?
combinatorics number-theory
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
$endgroup$
$begingroup$
Are the $x_k$ to be any integers? (then infinitely many solutions can happen), or are the $x_k$ to be non-negative?
$endgroup$
– coffeemath
Dec 29 '18 at 22:27
1
$begingroup$
$x_k$ are non-negative.
$endgroup$
– Hello_World
Dec 29 '18 at 22:48
$begingroup$
In your formula, in the sum part are many terms which coincide with the extra terms after the sum. Have you tried your formula in some cases of reasonably small $n$? It would be good to include them. [I assume the extra terms are restricted so no terms $v( rm{negative})$ occur.]
$endgroup$
– coffeemath
Dec 30 '18 at 0:37
1
$begingroup$
Yeah, I will try that. In the meantime I found this: nipne.ro/rjp/2013_58_9-10/1408_1417.pdf Refer to page 2. I think it is interesting.
$endgroup$
– Hello_World
Dec 30 '18 at 0:48
$begingroup$
Yes, that link of your comment seems to be to a proven recurrence.
$endgroup$
– coffeemath
Dec 30 '18 at 1:07
add a comment |
$begingroup$
Consider the diophantine equation of the form
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
where $n,kgeq 1$ and $a_i$ are non-negative integers. Then we are concerned $v(n)$ denote the number of solutions for a fixed tuple $(a_1,a_2,cdots ,a_k).$ Can $v(n)$ be expressed recursively, that is, in terms of $v(n-1),v(n-2),$ etc?
I considered the following example to see if I could generalize
$$x_1+2x_2 = n.$$
I noticed that if $(x_1,x_2)$ is a solution of $x_1+2x_2 = n-k$ then $(x_1+k,x_2)$ is a solution of $x_1+2x_2=n.$ Furthermore for $k=2p$ then if $x_1+2x_2 = n-k=n - 2p$ then $(x_1,x_2+p)$ is a solution of $x_1+2x_2=n.$ Therefore,
$$v(n) = sum_{k=1}^{n}v(n-k)+v(n-2)+v(n-4)+cdots $$
Is this correct? How do I get a general formula for $v(n)?$ Any ideas will be much appreciated?
combinatorics number-theory
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
$endgroup$
Consider the diophantine equation of the form
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
where $n,kgeq 1$ and $a_i$ are non-negative integers. Then we are concerned $v(n)$ denote the number of solutions for a fixed tuple $(a_1,a_2,cdots ,a_k).$ Can $v(n)$ be expressed recursively, that is, in terms of $v(n-1),v(n-2),$ etc?
I considered the following example to see if I could generalize
$$x_1+2x_2 = n.$$
I noticed that if $(x_1,x_2)$ is a solution of $x_1+2x_2 = n-k$ then $(x_1+k,x_2)$ is a solution of $x_1+2x_2=n.$ Furthermore for $k=2p$ then if $x_1+2x_2 = n-k=n - 2p$ then $(x_1,x_2+p)$ is a solution of $x_1+2x_2=n.$ Therefore,
$$v(n) = sum_{k=1}^{n}v(n-k)+v(n-2)+v(n-4)+cdots $$
Is this correct? How do I get a general formula for $v(n)?$ Any ideas will be much appreciated?
combinatorics number-theory
combinatorics number-theory
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
asked Dec 29 '18 at 22:15
Hello_WorldHello_World
4,14121931
4,14121931
$begingroup$
Are the $x_k$ to be any integers? (then infinitely many solutions can happen), or are the $x_k$ to be non-negative?
$endgroup$
– coffeemath
Dec 29 '18 at 22:27
1
$begingroup$
$x_k$ are non-negative.
$endgroup$
– Hello_World
Dec 29 '18 at 22:48
$begingroup$
In your formula, in the sum part are many terms which coincide with the extra terms after the sum. Have you tried your formula in some cases of reasonably small $n$? It would be good to include them. [I assume the extra terms are restricted so no terms $v( rm{negative})$ occur.]
$endgroup$
– coffeemath
Dec 30 '18 at 0:37
1
$begingroup$
Yeah, I will try that. In the meantime I found this: nipne.ro/rjp/2013_58_9-10/1408_1417.pdf Refer to page 2. I think it is interesting.
$endgroup$
– Hello_World
Dec 30 '18 at 0:48
$begingroup$
Yes, that link of your comment seems to be to a proven recurrence.
$endgroup$
– coffeemath
Dec 30 '18 at 1:07
add a comment |
$begingroup$
Are the $x_k$ to be any integers? (then infinitely many solutions can happen), or are the $x_k$ to be non-negative?
$endgroup$
– coffeemath
Dec 29 '18 at 22:27
1
$begingroup$
$x_k$ are non-negative.
$endgroup$
– Hello_World
Dec 29 '18 at 22:48
$begingroup$
In your formula, in the sum part are many terms which coincide with the extra terms after the sum. Have you tried your formula in some cases of reasonably small $n$? It would be good to include them. [I assume the extra terms are restricted so no terms $v( rm{negative})$ occur.]
$endgroup$
– coffeemath
Dec 30 '18 at 0:37
1
$begingroup$
Yeah, I will try that. In the meantime I found this: nipne.ro/rjp/2013_58_9-10/1408_1417.pdf Refer to page 2. I think it is interesting.
$endgroup$
– Hello_World
Dec 30 '18 at 0:48
$begingroup$
Yes, that link of your comment seems to be to a proven recurrence.
$endgroup$
– coffeemath
Dec 30 '18 at 1:07
$begingroup$
Are the $x_k$ to be any integers? (then infinitely many solutions can happen), or are the $x_k$ to be non-negative?
$endgroup$
– coffeemath
Dec 29 '18 at 22:27
$begingroup$
Are the $x_k$ to be any integers? (then infinitely many solutions can happen), or are the $x_k$ to be non-negative?
$endgroup$
– coffeemath
Dec 29 '18 at 22:27
1
1
$begingroup$
$x_k$ are non-negative.
$endgroup$
– Hello_World
Dec 29 '18 at 22:48
$begingroup$
$x_k$ are non-negative.
$endgroup$
– Hello_World
Dec 29 '18 at 22:48
$begingroup$
In your formula, in the sum part are many terms which coincide with the extra terms after the sum. Have you tried your formula in some cases of reasonably small $n$? It would be good to include them. [I assume the extra terms are restricted so no terms $v( rm{negative})$ occur.]
$endgroup$
– coffeemath
Dec 30 '18 at 0:37
$begingroup$
In your formula, in the sum part are many terms which coincide with the extra terms after the sum. Have you tried your formula in some cases of reasonably small $n$? It would be good to include them. [I assume the extra terms are restricted so no terms $v( rm{negative})$ occur.]
$endgroup$
– coffeemath
Dec 30 '18 at 0:37
1
1
$begingroup$
Yeah, I will try that. In the meantime I found this: nipne.ro/rjp/2013_58_9-10/1408_1417.pdf Refer to page 2. I think it is interesting.
$endgroup$
– Hello_World
Dec 30 '18 at 0:48
$begingroup$
Yeah, I will try that. In the meantime I found this: nipne.ro/rjp/2013_58_9-10/1408_1417.pdf Refer to page 2. I think it is interesting.
$endgroup$
– Hello_World
Dec 30 '18 at 0:48
$begingroup$
Yes, that link of your comment seems to be to a proven recurrence.
$endgroup$
– coffeemath
Dec 30 '18 at 1:07
$begingroup$
Yes, that link of your comment seems to be to a proven recurrence.
$endgroup$
– coffeemath
Dec 30 '18 at 1:07
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's denote with $v(k,n)$ the number of solutions of the following equation:
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
Take a look at the last item $x_k$. It can take any value from 0 to $lfloor{frac{n}{a_k}}rfloor$. So we have the following recurrence relation:
$$v(k,n)=sum_{i=0}^{lfloor{frac{n}{a_k}}rfloor} v(k-1,n-icdot a_k)tag{1}$$
...with the following exit criteria:
$$v(1,n)=begin{cases}
1, & text{if} a_1mid n \
0, & text{if} a_1nmid n
end{cases}$$
Here is an example: let's count the number of solutions for the following equation:
$$x_1+2x_2+3x_3+2x_4+6x_5+2x_6=100$$
You can do this in Java:
public class SolutionCounter {
public static int count(int a, int k, int n) {
if(k == 1) {
return (n % a[0] == 0)? 1: 0;
}
int cnt = 0;
for(int i = 0; i <= n / a[k - 1]; i++) {
cnt += count(a, k - 1, n - i * a[k - 1]);
}
return cnt;
}
public static void main(String args) {
int a = {1,2,3,2,6,2};
int sum = 100;
int cnt = count(a, a.length, sum);
System.out.println(cnt);
}
}
or with this ridiculously short Python script:
def cnt(a, k, n):
if k == 1:
return 1 if n % a[0] == 0 else 0
return sum(cnt(a, k - 1, n - i * a[k - 1]) for i in range(0, int(n / a[k - 1]) + 1))
print(cnt([1,2,3,2,6,2], 6, 100))
...and the result is: 847875.
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
$endgroup$
$begingroup$
Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
$endgroup$
– Oldboy
Jan 2 at 10:54
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
$endgroup$
– Hello_World
Jan 2 at 11:00
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
add a comment |
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let's denote with $v(k,n)$ the number of solutions of the following equation:
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
Take a look at the last item $x_k$. It can take any value from 0 to $lfloor{frac{n}{a_k}}rfloor$. So we have the following recurrence relation:
$$v(k,n)=sum_{i=0}^{lfloor{frac{n}{a_k}}rfloor} v(k-1,n-icdot a_k)tag{1}$$
...with the following exit criteria:
$$v(1,n)=begin{cases}
1, & text{if} a_1mid n \
0, & text{if} a_1nmid n
end{cases}$$
Here is an example: let's count the number of solutions for the following equation:
$$x_1+2x_2+3x_3+2x_4+6x_5+2x_6=100$$
You can do this in Java:
public class SolutionCounter {
public static int count(int a, int k, int n) {
if(k == 1) {
return (n % a[0] == 0)? 1: 0;
}
int cnt = 0;
for(int i = 0; i <= n / a[k - 1]; i++) {
cnt += count(a, k - 1, n - i * a[k - 1]);
}
return cnt;
}
public static void main(String args) {
int a = {1,2,3,2,6,2};
int sum = 100;
int cnt = count(a, a.length, sum);
System.out.println(cnt);
}
}
or with this ridiculously short Python script:
def cnt(a, k, n):
if k == 1:
return 1 if n % a[0] == 0 else 0
return sum(cnt(a, k - 1, n - i * a[k - 1]) for i in range(0, int(n / a[k - 1]) + 1))
print(cnt([1,2,3,2,6,2], 6, 100))
...and the result is: 847875.
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
$endgroup$
$begingroup$
Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
$endgroup$
– Oldboy
Jan 2 at 10:54
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
$endgroup$
– Hello_World
Jan 2 at 11:00
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
add a comment |
$begingroup$
Let's denote with $v(k,n)$ the number of solutions of the following equation:
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
Take a look at the last item $x_k$. It can take any value from 0 to $lfloor{frac{n}{a_k}}rfloor$. So we have the following recurrence relation:
$$v(k,n)=sum_{i=0}^{lfloor{frac{n}{a_k}}rfloor} v(k-1,n-icdot a_k)tag{1}$$
...with the following exit criteria:
$$v(1,n)=begin{cases}
1, & text{if} a_1mid n \
0, & text{if} a_1nmid n
end{cases}$$
Here is an example: let's count the number of solutions for the following equation:
$$x_1+2x_2+3x_3+2x_4+6x_5+2x_6=100$$
You can do this in Java:
public class SolutionCounter {
public static int count(int a, int k, int n) {
if(k == 1) {
return (n % a[0] == 0)? 1: 0;
}
int cnt = 0;
for(int i = 0; i <= n / a[k - 1]; i++) {
cnt += count(a, k - 1, n - i * a[k - 1]);
}
return cnt;
}
public static void main(String args) {
int a = {1,2,3,2,6,2};
int sum = 100;
int cnt = count(a, a.length, sum);
System.out.println(cnt);
}
}
or with this ridiculously short Python script:
def cnt(a, k, n):
if k == 1:
return 1 if n % a[0] == 0 else 0
return sum(cnt(a, k - 1, n - i * a[k - 1]) for i in range(0, int(n / a[k - 1]) + 1))
print(cnt([1,2,3,2,6,2], 6, 100))
...and the result is: 847875.
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
$endgroup$
$begingroup$
Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
$endgroup$
– Oldboy
Jan 2 at 10:54
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
$endgroup$
– Hello_World
Jan 2 at 11:00
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
add a comment |
$begingroup$
Let's denote with $v(k,n)$ the number of solutions of the following equation:
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
Take a look at the last item $x_k$. It can take any value from 0 to $lfloor{frac{n}{a_k}}rfloor$. So we have the following recurrence relation:
$$v(k,n)=sum_{i=0}^{lfloor{frac{n}{a_k}}rfloor} v(k-1,n-icdot a_k)tag{1}$$
...with the following exit criteria:
$$v(1,n)=begin{cases}
1, & text{if} a_1mid n \
0, & text{if} a_1nmid n
end{cases}$$
Here is an example: let's count the number of solutions for the following equation:
$$x_1+2x_2+3x_3+2x_4+6x_5+2x_6=100$$
You can do this in Java:
public class SolutionCounter {
public static int count(int a, int k, int n) {
if(k == 1) {
return (n % a[0] == 0)? 1: 0;
}
int cnt = 0;
for(int i = 0; i <= n / a[k - 1]; i++) {
cnt += count(a, k - 1, n - i * a[k - 1]);
}
return cnt;
}
public static void main(String args) {
int a = {1,2,3,2,6,2};
int sum = 100;
int cnt = count(a, a.length, sum);
System.out.println(cnt);
}
}
or with this ridiculously short Python script:
def cnt(a, k, n):
if k == 1:
return 1 if n % a[0] == 0 else 0
return sum(cnt(a, k - 1, n - i * a[k - 1]) for i in range(0, int(n / a[k - 1]) + 1))
print(cnt([1,2,3,2,6,2], 6, 100))
...and the result is: 847875.
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
$endgroup$
Let's denote with $v(k,n)$ the number of solutions of the following equation:
$$a_1x_1+a_2x_2+cdots +a_kx_k=n$$
Take a look at the last item $x_k$. It can take any value from 0 to $lfloor{frac{n}{a_k}}rfloor$. So we have the following recurrence relation:
$$v(k,n)=sum_{i=0}^{lfloor{frac{n}{a_k}}rfloor} v(k-1,n-icdot a_k)tag{1}$$
...with the following exit criteria:
$$v(1,n)=begin{cases}
1, & text{if} a_1mid n \
0, & text{if} a_1nmid n
end{cases}$$
Here is an example: let's count the number of solutions for the following equation:
$$x_1+2x_2+3x_3+2x_4+6x_5+2x_6=100$$
You can do this in Java:
public class SolutionCounter {
public static int count(int a, int k, int n) {
if(k == 1) {
return (n % a[0] == 0)? 1: 0;
}
int cnt = 0;
for(int i = 0; i <= n / a[k - 1]; i++) {
cnt += count(a, k - 1, n - i * a[k - 1]);
}
return cnt;
}
public static void main(String args) {
int a = {1,2,3,2,6,2};
int sum = 100;
int cnt = count(a, a.length, sum);
System.out.println(cnt);
}
}
or with this ridiculously short Python script:
def cnt(a, k, n):
if k == 1:
return 1 if n % a[0] == 0 else 0
return sum(cnt(a, k - 1, n - i * a[k - 1]) for i in range(0, int(n / a[k - 1]) + 1))
print(cnt([1,2,3,2,6,2], 6, 100))
...and the result is: 847875.
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
edited Jan 2 at 10:51
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
answered Dec 30 '18 at 9:44
OldboyOldboy
8,59011036
8,59011036
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Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
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– Oldboy
Jan 2 at 10:54
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
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– Hello_World
Jan 2 at 11:00
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
add a comment |
$begingroup$
Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
$endgroup$
– Oldboy
Jan 2 at 10:54
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
$endgroup$
– Hello_World
Jan 2 at 11:00
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
$begingroup$
Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
Could you give a recurrence in terms of $n$?
$endgroup$
– Hello_World
Jan 2 at 9:31
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
$endgroup$
– Oldboy
Jan 2 at 10:54
$begingroup$
(1) already gives recurrence formula for $v(n)$ in terms of $v(n-1)$, $v(n-2)$,... but you must also take values of $a_i$ into account. I doubt that simpler recurrence formula exists.
$endgroup$
– Oldboy
Jan 2 at 10:54
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
$endgroup$
– Hello_World
Jan 2 at 11:00
$begingroup$
Say you have $x_1+2x_2=n$ and the number of solutions to this is $v(n).$ Is there a way to express $v(n)$ in terms of $v(1),v(2),cdots,v(n-1)?$ In your recurrence, it like you are conditioning on one variable and therefore getting rid of a term. Whereas I want to condition on $n$. I hope this makes sense?
$endgroup$
– Hello_World
Jan 2 at 11:00
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
$begingroup$
@Hello_World Can you explain your request on the simplest possible example: $4x_1=n$. What is the recurrence formula for $v(n)$ in this simplest case? If we don't find a cute formula for one variable, chances are we won't find it for many more.
$endgroup$
– Oldboy
Jan 2 at 11:20
add a comment |
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$begingroup$
Are the $x_k$ to be any integers? (then infinitely many solutions can happen), or are the $x_k$ to be non-negative?
$endgroup$
– coffeemath
Dec 29 '18 at 22:27
1
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$x_k$ are non-negative.
$endgroup$
– Hello_World
Dec 29 '18 at 22:48
$begingroup$
In your formula, in the sum part are many terms which coincide with the extra terms after the sum. Have you tried your formula in some cases of reasonably small $n$? It would be good to include them. [I assume the extra terms are restricted so no terms $v( rm{negative})$ occur.]
$endgroup$
– coffeemath
Dec 30 '18 at 0:37
1
$begingroup$
Yeah, I will try that. In the meantime I found this: nipne.ro/rjp/2013_58_9-10/1408_1417.pdf Refer to page 2. I think it is interesting.
$endgroup$
– Hello_World
Dec 30 '18 at 0:48
$begingroup$
Yes, that link of your comment seems to be to a proven recurrence.
$endgroup$
– coffeemath
Dec 30 '18 at 1:07