Estimate related to the Möbius functionAsymptotic density of k-almost primesWalsh Fourier Transform of the Möbius functionAsymptotics for the number of ways to sum primes such that the sum is <= nHow does Yitang Zhang use Cauchy's inequality and Theorem 2 to obtain the error term coming from the $S_2$ sumReferences to proofs of upper and lower bounds on the number of coprimes in an interval?Small quotients of smooth numbersAsymptotic estimate for a random model of primesChecking Mertens and the like in less than linear time or less than $sqrtx$ spaceAsymptotic formula for the average number of zeros of a polynomial modulo pIdelic/Hom representation of locally free class groupsQuestion on sums of multiplicative functions twisted by the Mobius function
Estimate related to the Möbius function
Asymptotic density of k-almost primesWalsh Fourier Transform of the Möbius functionAsymptotics for the number of ways to sum primes such that the sum is <= nHow does Yitang Zhang use Cauchy's inequality and Theorem 2 to obtain the error term coming from the $S_2$ sumReferences to proofs of upper and lower bounds on the number of coprimes in an interval?Small quotients of smooth numbersAsymptotic estimate for a random model of primesChecking Mertens and the like in less than linear time or less than $sqrtx$ spaceAsymptotic formula for the average number of zeros of a polynomial modulo pIdelic/Hom representation of locally free class groupsQuestion on sums of multiplicative functions twisted by the Mobius function
$begingroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
$endgroup$
add a comment |
$begingroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
$endgroup$
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_ n mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
add a comment |
$begingroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
$endgroup$
I need to know, or at least have a good bound for, the asymptotic behaviour on $x$ of amount of integers less o equal than $x$ that are square free and with exactly $k$ primes on its decomposition. That is the cardinal of the following set
$$
mathcalJ_T(x,k) = n in mathbbN : n le x, Omega(n)=k , n mbox is square free .$$
Other way to describe this cardinal, using the Möbius function, is
$$ |mathcalJ_T(x,k) | = sum_Omega(n) = k, n le x |mu(n)|.$$
I am looking for the asymptotic behaviour on $x$, but this will depend also on $k$ in some way. The bound given using
$$
sum_Omega(n) = k, n le x |mu(n)| le sum_n le x |mu(n)| le frac6xpi^2 + O(sqrtx) ll x,
$$
is not good enought for my purposes.
Thanks in advanced, any reference or idea is helpful
nt.number-theory analytic-number-theory prime-numbers asymptotics
nt.number-theory analytic-number-theory prime-numbers asymptotics
New contributor
New contributor
edited 8 hours ago
GH from MO
60.4k5152231
60.4k5152231
New contributor
asked 8 hours ago
Martin MansillaMartin Mansilla
161
161
New contributor
New contributor
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_ n mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
add a comment |
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_ n mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_ n mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_ n mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "504"
;
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
);
);
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
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%2fmathoverflow.net%2fquestions%2f332797%2festimate-related-to-the-m%25c3%25b6bius-function%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
add a comment |
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
add a comment |
$begingroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
$endgroup$
You can derive a very precise asymptotic expansion of your quantity by the Selberg-Delange method.
I recommend that you adapt, to your situation, the arguments of Section II.6.1 of Tenenbaum: Introduction to analytic and probabilistic number theory. The starting point of your analysis should be the formula
$$sum_text$n$ square-freez^omega(n)n^-s=prod_pleft(1+fraczp^sright).$$
Then you will need to "factor out" $zeta(s)^z$ and proceed as in the mentioned chapter, where the analysis is carried out without the restriction that $n$ is square-free.
edited 7 hours ago
answered 8 hours ago
GH from MOGH from MO
60.4k5152231
60.4k5152231
add a comment |
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
add a comment |
$begingroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
$endgroup$
It hasn't been pointed out yet that you can derive the answer simply directly from the statement of the Selberg–Sathe theorem, which gives (for fixed $k$) the asymptotic formulas
beginalign*
# nle xcolon omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)! \
# nle xcolon Omega(n) = k &sim frac xlog x frac(loglog x)^k-1(k-1)!.
endalign*
(Here $omega$ and $Omega$ count the prime factors of $n$ without and with multiplicity, respectively.)
Note that any nonsquarefree integer $n$ with $Omega(n) = k$ satisfies $omega(n) = j$ for some $1le jle k-1$. Therefore the bounds
beginalign*
# nle xcolon Omega(n) = k,, ntext is squarefree &le # nle xcolon Omega(n) = k \
# nle xcolon Omega(n) = k,, ntext is squarefree &ge # nle xcolon Omega(n) = k - sum_j=1^k-1 # nle xcolon omega(n) = j ,
endalign*
together with the Selberg–Sathe asymptotic formulas above, immediately imply that
$$
# nle xcolon Omega(n) = k,, ntext is squarefree sim frac xlog x frac(loglog x)^k-1(k-1)!.
$$
This argument can be made to work with some uniformity in $k$ as well (I think $k=o(loglog x)$ is what is needed).
answered 2 hours ago
Greg MartinGreg Martin
9,09213862
9,09213862
add a comment |
add a comment |
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Martin Mansilla is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f332797%2festimate-related-to-the-m%25c3%25b6bius-function%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
$begingroup$
Hi it seems you want k fairly small relative to x in which case you can achieve this quite classically. Prof. Harcos has given you exactly the technique you need to do it: Selberg-Delange. But if you prefer to just cite a result, then express via Möbius inversion (using sum_ n mu(d)) the number of such squarefree integers leq x with omega(n) = k as an alternating sum of, for each dleq x^1/2, the number of integers leq x/d^2 with exactly k - omega(d) prime factors, and then insert the classical asymptotic for the number of such integers due to, I think, Selberg. Link below:
$endgroup$
– alpoge
3 hours ago
$begingroup$
mathoverflow.net/questions/35927/…
$endgroup$
– alpoge
3 hours ago
$begingroup$
Let me take a moment to point out the following in an answer on that thread. It is noted that Gauss empirically observed that the number of integers leq x with two prime factors seems to grow like x loglogx / logx. Who among us, even with our computers, has seen loglogx grow?! (Canonical quote about loglogx going to infinity with dignity.) What intuition...
$endgroup$
– alpoge
3 hours ago