Asymptotic Expression for the number of solutions to linear Diophantine equation.












1












$begingroup$


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.










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$endgroup$








  • 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 $d=gcd(a_1,a_2,cdots,a_k)$?
    $endgroup$
    – Hello_World
    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












$begingroup$


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.










share|cite|improve this question









$endgroup$








  • 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 $d=gcd(a_1,a_2,cdots,a_k)$?
    $endgroup$
    – Hello_World
    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








1





$begingroup$


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.










share|cite|improve this question









$endgroup$




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






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asked Jan 12 at 6:41









Hello_WorldHello_World

4,13121731




4,13121731








  • 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 $d=gcd(a_1,a_2,cdots,a_k)$?
    $endgroup$
    – Hello_World
    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




    $begingroup$
    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 $d=gcd(a_1,a_2,cdots,a_k)$?
    $endgroup$
    – Hello_World
    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 $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– Hello_World
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 $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– Hello_World
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












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