Asymptotic Expression for the number of solutions to linear Diophantine equation.
$begingroup$
Consider the the general linear diophantine equation
$$sum_{i=1}^{k}a_ix_i =n $$
with $a_igeq 1, ngeq 1$ and $x_igeq 0.$
Then the generating function that counts the number of solutions to this equation is
$$F(x)=prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right) = 1+sum_{ngeq 1}a(n)x^{n},$$
where $a(n)$ is the number of solutions to the linear Diophantine equation. Due to Schur we have that if $gcd(a_1,a_2,cdots,a_k)=1$ then
$$a(n)sim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$
However, I am curious to know if there is a general asymptotic expression for $a(n)$ with perhaps no restrictions on the coefficients $a_i$ of the equation. I tried to see if I could mimic the proof of Schur's theorem to come up with something, but because in the general case we do not know which pole of $F$ has the highest multiplicity, it is hard to guess the answer.
Any ideas in this regard will be much appreciated.
number-theory asymptotics analytic-number-theory linear-diophantine-equations analytic-combinatorics
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
$endgroup$
add a comment |
$begingroup$
Consider the the general linear diophantine equation
$$sum_{i=1}^{k}a_ix_i =n $$
with $a_igeq 1, ngeq 1$ and $x_igeq 0.$
Then the generating function that counts the number of solutions to this equation is
$$F(x)=prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right) = 1+sum_{ngeq 1}a(n)x^{n},$$
where $a(n)$ is the number of solutions to the linear Diophantine equation. Due to Schur we have that if $gcd(a_1,a_2,cdots,a_k)=1$ then
$$a(n)sim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$
However, I am curious to know if there is a general asymptotic expression for $a(n)$ with perhaps no restrictions on the coefficients $a_i$ of the equation. I tried to see if I could mimic the proof of Schur's theorem to come up with something, but because in the general case we do not know which pole of $F$ has the highest multiplicity, it is hard to guess the answer.
Any ideas in this regard will be much appreciated.
number-theory asymptotics analytic-number-theory linear-diophantine-equations analytic-combinatorics
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
$endgroup$
1
$begingroup$
If $gcd(a_1,ldots,a_n)=d>1$, then we must have $dmid n$ to get $a(n)neq 0$, and division by $d$ brings you back to $gcd=1$. (The above is "almost" the general case.)
$endgroup$
– metamorphy
Jan 12 at 8:48
$begingroup$
Do you suggest that $a(n)sim frac{(n/d)^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$ where $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– model_checker
Jan 12 at 8:50
$begingroup$
$a(n)simfrac{(n/d)^{k-1}}{(k-1)!(a_1/d)ldots(a_k/d)}=frac{n^{k-1}d}{(k-1)!a_1ldots a_k}$, only when $dmid n$.
$endgroup$
– metamorphy
Jan 12 at 8:53
add a comment |
$begingroup$
Consider the the general linear diophantine equation
$$sum_{i=1}^{k}a_ix_i =n $$
with $a_igeq 1, ngeq 1$ and $x_igeq 0.$
Then the generating function that counts the number of solutions to this equation is
$$F(x)=prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right) = 1+sum_{ngeq 1}a(n)x^{n},$$
where $a(n)$ is the number of solutions to the linear Diophantine equation. Due to Schur we have that if $gcd(a_1,a_2,cdots,a_k)=1$ then
$$a(n)sim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$
However, I am curious to know if there is a general asymptotic expression for $a(n)$ with perhaps no restrictions on the coefficients $a_i$ of the equation. I tried to see if I could mimic the proof of Schur's theorem to come up with something, but because in the general case we do not know which pole of $F$ has the highest multiplicity, it is hard to guess the answer.
Any ideas in this regard will be much appreciated.
number-theory asymptotics analytic-number-theory linear-diophantine-equations analytic-combinatorics
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
$endgroup$
Consider the the general linear diophantine equation
$$sum_{i=1}^{k}a_ix_i =n $$
with $a_igeq 1, ngeq 1$ and $x_igeq 0.$
Then the generating function that counts the number of solutions to this equation is
$$F(x)=prod_{i=1}^{k}left(frac{1}{1-x^{a_i}}right) = 1+sum_{ngeq 1}a(n)x^{n},$$
where $a(n)$ is the number of solutions to the linear Diophantine equation. Due to Schur we have that if $gcd(a_1,a_2,cdots,a_k)=1$ then
$$a(n)sim frac{n^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$$
as $nto infty.$
However, I am curious to know if there is a general asymptotic expression for $a(n)$ with perhaps no restrictions on the coefficients $a_i$ of the equation. I tried to see if I could mimic the proof of Schur's theorem to come up with something, but because in the general case we do not know which pole of $F$ has the highest multiplicity, it is hard to guess the answer.
Any ideas in this regard will be much appreciated.
number-theory asymptotics analytic-number-theory linear-diophantine-equations analytic-combinatorics
number-theory asymptotics analytic-number-theory linear-diophantine-equations analytic-combinatorics
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
asked Jan 12 at 6:41
model_checkermodel_checker
4,45521931
4,45521931
1
$begingroup$
If $gcd(a_1,ldots,a_n)=d>1$, then we must have $dmid n$ to get $a(n)neq 0$, and division by $d$ brings you back to $gcd=1$. (The above is "almost" the general case.)
$endgroup$
– metamorphy
Jan 12 at 8:48
$begingroup$
Do you suggest that $a(n)sim frac{(n/d)^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$ where $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– model_checker
Jan 12 at 8:50
$begingroup$
$a(n)simfrac{(n/d)^{k-1}}{(k-1)!(a_1/d)ldots(a_k/d)}=frac{n^{k-1}d}{(k-1)!a_1ldots a_k}$, only when $dmid n$.
$endgroup$
– metamorphy
Jan 12 at 8:53
add a comment |
1
$begingroup$
If $gcd(a_1,ldots,a_n)=d>1$, then we must have $dmid n$ to get $a(n)neq 0$, and division by $d$ brings you back to $gcd=1$. (The above is "almost" the general case.)
$endgroup$
– metamorphy
Jan 12 at 8:48
$begingroup$
Do you suggest that $a(n)sim frac{(n/d)^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$ where $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– model_checker
Jan 12 at 8:50
$begingroup$
$a(n)simfrac{(n/d)^{k-1}}{(k-1)!(a_1/d)ldots(a_k/d)}=frac{n^{k-1}d}{(k-1)!a_1ldots a_k}$, only when $dmid n$.
$endgroup$
– metamorphy
Jan 12 at 8:53
1
1
$begingroup$
If $gcd(a_1,ldots,a_n)=d>1$, then we must have $dmid n$ to get $a(n)neq 0$, and division by $d$ brings you back to $gcd=1$. (The above is "almost" the general case.)
$endgroup$
– metamorphy
Jan 12 at 8:48
$begingroup$
If $gcd(a_1,ldots,a_n)=d>1$, then we must have $dmid n$ to get $a(n)neq 0$, and division by $d$ brings you back to $gcd=1$. (The above is "almost" the general case.)
$endgroup$
– metamorphy
Jan 12 at 8:48
$begingroup$
Do you suggest that $a(n)sim frac{(n/d)^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$ where $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– model_checker
Jan 12 at 8:50
$begingroup$
Do you suggest that $a(n)sim frac{(n/d)^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$ where $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– model_checker
Jan 12 at 8:50
$begingroup$
$a(n)simfrac{(n/d)^{k-1}}{(k-1)!(a_1/d)ldots(a_k/d)}=frac{n^{k-1}d}{(k-1)!a_1ldots a_k}$, only when $dmid n$.
$endgroup$
– metamorphy
Jan 12 at 8:53
$begingroup$
$a(n)simfrac{(n/d)^{k-1}}{(k-1)!(a_1/d)ldots(a_k/d)}=frac{n^{k-1}d}{(k-1)!a_1ldots a_k}$, only when $dmid n$.
$endgroup$
– metamorphy
Jan 12 at 8:53
add a comment |
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$begingroup$
If $gcd(a_1,ldots,a_n)=d>1$, then we must have $dmid n$ to get $a(n)neq 0$, and division by $d$ brings you back to $gcd=1$. (The above is "almost" the general case.)
$endgroup$
– metamorphy
Jan 12 at 8:48
$begingroup$
Do you suggest that $a(n)sim frac{(n/d)^{k−1}}{(k − 1)!a_1a_2 ··· a_k}$ where $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– model_checker
Jan 12 at 8:50
$begingroup$
$a(n)simfrac{(n/d)^{k-1}}{(k-1)!(a_1/d)ldots(a_k/d)}=frac{n^{k-1}d}{(k-1)!a_1ldots a_k}$, only when $dmid n$.
$endgroup$
– metamorphy
Jan 12 at 8:53