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
$endgroup$
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
$endgroup$
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
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
$endgroup$
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
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
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 $d=gcd(a_1,a_2,cdots,a_k)$?
$endgroup$
– 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
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070652%2fasymptotic-expression-for-the-number-of-solutions-to-linear-diophantine-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070652%2fasymptotic-expression-for-the-number-of-solutions-to-linear-diophantine-equation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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