Number of matrices with bounded products of rows and columnsAsymptotic Formula for a Mertens Style SumLeast prime in an arithmetic progression and the Selberg sieveBest upper bound on the number of divisors of $n$ that are larger than $N$.An upper bound to sum of ratios of gcd's and productsIndependence between the number of prime factors of $n$ and $n+2$On an open problem of GelfondProof of a theorem about the size of the number of sign changes of Hecke eigenvaluesBound on the number of primitive divisors of the $n$th Fibonacci number
Number of matrices with bounded products of rows and columns
Asymptotic Formula for a Mertens Style SumLeast prime in an arithmetic progression and the Selberg sieveBest upper bound on the number of divisors of $n$ that are larger than $N$.An upper bound to sum of ratios of gcd's and productsIndependence between the number of prime factors of $n$ and $n+2$On an open problem of GelfondProof of a theorem about the size of the number of sign changes of Hecke eigenvaluesBound on the number of primitive divisors of the $n$th Fibonacci number
$begingroup$
Fix an integer $d geq 2$ and for every real number $x$ let $M_d(x)$ be number of $d times d$ matrices $(a_ij)$ satisfying: every $a_ij$ is a positive integer, the product of every row does not exceed $x$, and the product of every column does not exceed $x$.
I'm looking for a good upper bound for $M_d(x)$ as $x to +infty$.
If we forget about the condition on the columns, since it is well known that the number of $d$-tuples $(b_1, dots, b_d)$ of positive integers satisfying $b_1 cdots b_d leq x$ is $ll_d x (log x)^d - 1$ (a generalization of Dirichlet divisor problem), we get the upper bound
$$M_d(x) ll_d x^d (log x)^d(d-1).$$
In the special case $d = 2$, we have that $a_12, a_21 leq min(x / a_11, x / a_22)$ and consequently
$$M_2(x) leq sum_a_11, a_22 leq x minleft(fracxa_11, fracxa_22right)^2 ll sum_a_11 leq a_22 leq x left(fracxa_22right)^2 ll x^2 log x ,$$
which is a better upper bound than the general one given in the above paragraph. However, I have no idea of how to generalize this trick to $d geq 3$ (if possible).
Has this problem been studied before? Do you have any idea/suggestion about it?
nt.number-theory matrices analytic-number-theory divisors
New contributor
Kate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
|
show 1 more comment
$begingroup$
Fix an integer $d geq 2$ and for every real number $x$ let $M_d(x)$ be number of $d times d$ matrices $(a_ij)$ satisfying: every $a_ij$ is a positive integer, the product of every row does not exceed $x$, and the product of every column does not exceed $x$.
I'm looking for a good upper bound for $M_d(x)$ as $x to +infty$.
If we forget about the condition on the columns, since it is well known that the number of $d$-tuples $(b_1, dots, b_d)$ of positive integers satisfying $b_1 cdots b_d leq x$ is $ll_d x (log x)^d - 1$ (a generalization of Dirichlet divisor problem), we get the upper bound
$$M_d(x) ll_d x^d (log x)^d(d-1).$$
In the special case $d = 2$, we have that $a_12, a_21 leq min(x / a_11, x / a_22)$ and consequently
$$M_2(x) leq sum_a_11, a_22 leq x minleft(fracxa_11, fracxa_22right)^2 ll sum_a_11 leq a_22 leq x left(fracxa_22right)^2 ll x^2 log x ,$$
which is a better upper bound than the general one given in the above paragraph. However, I have no idea of how to generalize this trick to $d geq 3$ (if possible).
Has this problem been studied before? Do you have any idea/suggestion about it?
nt.number-theory matrices analytic-number-theory divisors
New contributor
Kate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Do you believe the $2times 2$ bound is sharp?
$endgroup$
– Igor Rivin
9 hours ago
$begingroup$
@IgorRivin I don't have any heuristic supporting such belief.
$endgroup$
– Kate
8 hours ago
$begingroup$
It is obviously not far from sharp, since diagonal matrices give you $x^2.$
$endgroup$
– Igor Rivin
8 hours ago
$begingroup$
What happens when you take logarithms of the aij? Gerhard "Maybe Row-Sum Literature Will Help" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
5 hours ago
$begingroup$
@GerhardPaseman I think logarithms do not help very much. They turn the problem from multiplicative to additive, but the variables are not integers anymore, making things much more difficult...
$endgroup$
– Kate
5 hours ago
|
show 1 more comment
$begingroup$
Fix an integer $d geq 2$ and for every real number $x$ let $M_d(x)$ be number of $d times d$ matrices $(a_ij)$ satisfying: every $a_ij$ is a positive integer, the product of every row does not exceed $x$, and the product of every column does not exceed $x$.
I'm looking for a good upper bound for $M_d(x)$ as $x to +infty$.
If we forget about the condition on the columns, since it is well known that the number of $d$-tuples $(b_1, dots, b_d)$ of positive integers satisfying $b_1 cdots b_d leq x$ is $ll_d x (log x)^d - 1$ (a generalization of Dirichlet divisor problem), we get the upper bound
$$M_d(x) ll_d x^d (log x)^d(d-1).$$
In the special case $d = 2$, we have that $a_12, a_21 leq min(x / a_11, x / a_22)$ and consequently
$$M_2(x) leq sum_a_11, a_22 leq x minleft(fracxa_11, fracxa_22right)^2 ll sum_a_11 leq a_22 leq x left(fracxa_22right)^2 ll x^2 log x ,$$
which is a better upper bound than the general one given in the above paragraph. However, I have no idea of how to generalize this trick to $d geq 3$ (if possible).
Has this problem been studied before? Do you have any idea/suggestion about it?
nt.number-theory matrices analytic-number-theory divisors
New contributor
Kate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Fix an integer $d geq 2$ and for every real number $x$ let $M_d(x)$ be number of $d times d$ matrices $(a_ij)$ satisfying: every $a_ij$ is a positive integer, the product of every row does not exceed $x$, and the product of every column does not exceed $x$.
I'm looking for a good upper bound for $M_d(x)$ as $x to +infty$.
If we forget about the condition on the columns, since it is well known that the number of $d$-tuples $(b_1, dots, b_d)$ of positive integers satisfying $b_1 cdots b_d leq x$ is $ll_d x (log x)^d - 1$ (a generalization of Dirichlet divisor problem), we get the upper bound
$$M_d(x) ll_d x^d (log x)^d(d-1).$$
In the special case $d = 2$, we have that $a_12, a_21 leq min(x / a_11, x / a_22)$ and consequently
$$M_2(x) leq sum_a_11, a_22 leq x minleft(fracxa_11, fracxa_22right)^2 ll sum_a_11 leq a_22 leq x left(fracxa_22right)^2 ll x^2 log x ,$$
which is a better upper bound than the general one given in the above paragraph. However, I have no idea of how to generalize this trick to $d geq 3$ (if possible).
Has this problem been studied before? Do you have any idea/suggestion about it?
nt.number-theory matrices analytic-number-theory divisors
nt.number-theory matrices analytic-number-theory divisors
New contributor
Kate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kate is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
GH from MO
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asked 9 hours ago
KateKate
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$begingroup$
Do you believe the $2times 2$ bound is sharp?
$endgroup$
– Igor Rivin
9 hours ago
$begingroup$
@IgorRivin I don't have any heuristic supporting such belief.
$endgroup$
– Kate
8 hours ago
$begingroup$
It is obviously not far from sharp, since diagonal matrices give you $x^2.$
$endgroup$
– Igor Rivin
8 hours ago
$begingroup$
What happens when you take logarithms of the aij? Gerhard "Maybe Row-Sum Literature Will Help" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
5 hours ago
$begingroup$
@GerhardPaseman I think logarithms do not help very much. They turn the problem from multiplicative to additive, but the variables are not integers anymore, making things much more difficult...
$endgroup$
– Kate
5 hours ago
|
show 1 more comment
$begingroup$
Do you believe the $2times 2$ bound is sharp?
$endgroup$
– Igor Rivin
9 hours ago
$begingroup$
@IgorRivin I don't have any heuristic supporting such belief.
$endgroup$
– Kate
8 hours ago
$begingroup$
It is obviously not far from sharp, since diagonal matrices give you $x^2.$
$endgroup$
– Igor Rivin
8 hours ago
$begingroup$
What happens when you take logarithms of the aij? Gerhard "Maybe Row-Sum Literature Will Help" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
5 hours ago
$begingroup$
@GerhardPaseman I think logarithms do not help very much. They turn the problem from multiplicative to additive, but the variables are not integers anymore, making things much more difficult...
$endgroup$
– Kate
5 hours ago
$begingroup$
Do you believe the $2times 2$ bound is sharp?
$endgroup$
– Igor Rivin
9 hours ago
$begingroup$
Do you believe the $2times 2$ bound is sharp?
$endgroup$
– Igor Rivin
9 hours ago
$begingroup$
@IgorRivin I don't have any heuristic supporting such belief.
$endgroup$
– Kate
8 hours ago
$begingroup$
@IgorRivin I don't have any heuristic supporting such belief.
$endgroup$
– Kate
8 hours ago
$begingroup$
It is obviously not far from sharp, since diagonal matrices give you $x^2.$
$endgroup$
– Igor Rivin
8 hours ago
$begingroup$
It is obviously not far from sharp, since diagonal matrices give you $x^2.$
$endgroup$
– Igor Rivin
8 hours ago
$begingroup$
What happens when you take logarithms of the aij? Gerhard "Maybe Row-Sum Literature Will Help" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
5 hours ago
$begingroup$
What happens when you take logarithms of the aij? Gerhard "Maybe Row-Sum Literature Will Help" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
5 hours ago
$begingroup$
@GerhardPaseman I think logarithms do not help very much. They turn the problem from multiplicative to additive, but the variables are not integers anymore, making things much more difficult...
$endgroup$
– Kate
5 hours ago
$begingroup$
@GerhardPaseman I think logarithms do not help very much. They turn the problem from multiplicative to additive, but the variables are not integers anymore, making things much more difficult...
$endgroup$
– Kate
5 hours ago
|
show 1 more comment
2 Answers
2
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$begingroup$
This problem was considered in passing in the proof of Theorem 4.1 in Granville and Soundararajan, see the argument starting at the bottom of page 17. They show (in your notation) that $M_d(x)$ is of order $x^d (log x)^(d-1)^2$. You should also look at work of Harper, Nikeghbali and Radziwill which shows an asymptotic formula for closely related objects (see Theorem 3 there).
$endgroup$
1
$begingroup$
Thank you very much!
$endgroup$
– Kate
5 hours ago
add a comment |
$begingroup$
Nice question. Some thoughts on lower bounds: for $d=2$, the order of magnitude $x^2log x$ is correct—here is an argument giving such a lower bound.
For any real numbers $U,V$ such that $UV=x$ (and say $U,Vge2$), any $2times 2$ matrix such that $a_11,a_22inbig(frac U2,Ubig]$ and $a_21,a_12inbig(frac V2,Vbig]$ is counted by $M_2(x)$, and the number of such matrices is $gg U^2V^2 = x^2$. We can sum this dyadic-interval lower bound over $U=2^k$ for a suitable range of $k$ to obtain the lower bound $M_2(x) gg x^2log x$.
A similar argument gives a lower bound for general $dge3$. If $U_1U_2cdots U_d=x$, with the convention that $U_d+j=U_j$, then the matrices such that $a_ij in big( fracU_i+j2, U_i+jbig]$ for all $1le i,jle d$ contribute $gg U_1^d cdots U_d^d = x^d$ to $M_d(x)$. We then sum over all $U_1=2^k_1,dots,U_d=2^k_d$ such that $k_1+cdots+k_dle(log x)/log 2$; the number of such $d$-tuples of positive integers is $gg_d(log x)^d$, giving the lower bound $M_d(x) gg_d x^d(log x)^d$.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
This problem was considered in passing in the proof of Theorem 4.1 in Granville and Soundararajan, see the argument starting at the bottom of page 17. They show (in your notation) that $M_d(x)$ is of order $x^d (log x)^(d-1)^2$. You should also look at work of Harper, Nikeghbali and Radziwill which shows an asymptotic formula for closely related objects (see Theorem 3 there).
$endgroup$
1
$begingroup$
Thank you very much!
$endgroup$
– Kate
5 hours ago
add a comment |
$begingroup$
This problem was considered in passing in the proof of Theorem 4.1 in Granville and Soundararajan, see the argument starting at the bottom of page 17. They show (in your notation) that $M_d(x)$ is of order $x^d (log x)^(d-1)^2$. You should also look at work of Harper, Nikeghbali and Radziwill which shows an asymptotic formula for closely related objects (see Theorem 3 there).
$endgroup$
1
$begingroup$
Thank you very much!
$endgroup$
– Kate
5 hours ago
add a comment |
$begingroup$
This problem was considered in passing in the proof of Theorem 4.1 in Granville and Soundararajan, see the argument starting at the bottom of page 17. They show (in your notation) that $M_d(x)$ is of order $x^d (log x)^(d-1)^2$. You should also look at work of Harper, Nikeghbali and Radziwill which shows an asymptotic formula for closely related objects (see Theorem 3 there).
$endgroup$
This problem was considered in passing in the proof of Theorem 4.1 in Granville and Soundararajan, see the argument starting at the bottom of page 17. They show (in your notation) that $M_d(x)$ is of order $x^d (log x)^(d-1)^2$. You should also look at work of Harper, Nikeghbali and Radziwill which shows an asymptotic formula for closely related objects (see Theorem 3 there).
answered 5 hours ago
LuciaLucia
36.5k5 gold badges157 silver badges185 bronze badges
36.5k5 gold badges157 silver badges185 bronze badges
1
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Thank you very much!
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– Kate
5 hours ago
add a comment |
1
$begingroup$
Thank you very much!
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– Kate
5 hours ago
1
1
$begingroup$
Thank you very much!
$endgroup$
– Kate
5 hours ago
$begingroup$
Thank you very much!
$endgroup$
– Kate
5 hours ago
add a comment |
$begingroup$
Nice question. Some thoughts on lower bounds: for $d=2$, the order of magnitude $x^2log x$ is correct—here is an argument giving such a lower bound.
For any real numbers $U,V$ such that $UV=x$ (and say $U,Vge2$), any $2times 2$ matrix such that $a_11,a_22inbig(frac U2,Ubig]$ and $a_21,a_12inbig(frac V2,Vbig]$ is counted by $M_2(x)$, and the number of such matrices is $gg U^2V^2 = x^2$. We can sum this dyadic-interval lower bound over $U=2^k$ for a suitable range of $k$ to obtain the lower bound $M_2(x) gg x^2log x$.
A similar argument gives a lower bound for general $dge3$. If $U_1U_2cdots U_d=x$, with the convention that $U_d+j=U_j$, then the matrices such that $a_ij in big( fracU_i+j2, U_i+jbig]$ for all $1le i,jle d$ contribute $gg U_1^d cdots U_d^d = x^d$ to $M_d(x)$. We then sum over all $U_1=2^k_1,dots,U_d=2^k_d$ such that $k_1+cdots+k_dle(log x)/log 2$; the number of such $d$-tuples of positive integers is $gg_d(log x)^d$, giving the lower bound $M_d(x) gg_d x^d(log x)^d$.
$endgroup$
add a comment |
$begingroup$
Nice question. Some thoughts on lower bounds: for $d=2$, the order of magnitude $x^2log x$ is correct—here is an argument giving such a lower bound.
For any real numbers $U,V$ such that $UV=x$ (and say $U,Vge2$), any $2times 2$ matrix such that $a_11,a_22inbig(frac U2,Ubig]$ and $a_21,a_12inbig(frac V2,Vbig]$ is counted by $M_2(x)$, and the number of such matrices is $gg U^2V^2 = x^2$. We can sum this dyadic-interval lower bound over $U=2^k$ for a suitable range of $k$ to obtain the lower bound $M_2(x) gg x^2log x$.
A similar argument gives a lower bound for general $dge3$. If $U_1U_2cdots U_d=x$, with the convention that $U_d+j=U_j$, then the matrices such that $a_ij in big( fracU_i+j2, U_i+jbig]$ for all $1le i,jle d$ contribute $gg U_1^d cdots U_d^d = x^d$ to $M_d(x)$. We then sum over all $U_1=2^k_1,dots,U_d=2^k_d$ such that $k_1+cdots+k_dle(log x)/log 2$; the number of such $d$-tuples of positive integers is $gg_d(log x)^d$, giving the lower bound $M_d(x) gg_d x^d(log x)^d$.
$endgroup$
add a comment |
$begingroup$
Nice question. Some thoughts on lower bounds: for $d=2$, the order of magnitude $x^2log x$ is correct—here is an argument giving such a lower bound.
For any real numbers $U,V$ such that $UV=x$ (and say $U,Vge2$), any $2times 2$ matrix such that $a_11,a_22inbig(frac U2,Ubig]$ and $a_21,a_12inbig(frac V2,Vbig]$ is counted by $M_2(x)$, and the number of such matrices is $gg U^2V^2 = x^2$. We can sum this dyadic-interval lower bound over $U=2^k$ for a suitable range of $k$ to obtain the lower bound $M_2(x) gg x^2log x$.
A similar argument gives a lower bound for general $dge3$. If $U_1U_2cdots U_d=x$, with the convention that $U_d+j=U_j$, then the matrices such that $a_ij in big( fracU_i+j2, U_i+jbig]$ for all $1le i,jle d$ contribute $gg U_1^d cdots U_d^d = x^d$ to $M_d(x)$. We then sum over all $U_1=2^k_1,dots,U_d=2^k_d$ such that $k_1+cdots+k_dle(log x)/log 2$; the number of such $d$-tuples of positive integers is $gg_d(log x)^d$, giving the lower bound $M_d(x) gg_d x^d(log x)^d$.
$endgroup$
Nice question. Some thoughts on lower bounds: for $d=2$, the order of magnitude $x^2log x$ is correct—here is an argument giving such a lower bound.
For any real numbers $U,V$ such that $UV=x$ (and say $U,Vge2$), any $2times 2$ matrix such that $a_11,a_22inbig(frac U2,Ubig]$ and $a_21,a_12inbig(frac V2,Vbig]$ is counted by $M_2(x)$, and the number of such matrices is $gg U^2V^2 = x^2$. We can sum this dyadic-interval lower bound over $U=2^k$ for a suitable range of $k$ to obtain the lower bound $M_2(x) gg x^2log x$.
A similar argument gives a lower bound for general $dge3$. If $U_1U_2cdots U_d=x$, with the convention that $U_d+j=U_j$, then the matrices such that $a_ij in big( fracU_i+j2, U_i+jbig]$ for all $1le i,jle d$ contribute $gg U_1^d cdots U_d^d = x^d$ to $M_d(x)$. We then sum over all $U_1=2^k_1,dots,U_d=2^k_d$ such that $k_1+cdots+k_dle(log x)/log 2$; the number of such $d$-tuples of positive integers is $gg_d(log x)^d$, giving the lower bound $M_d(x) gg_d x^d(log x)^d$.
edited 5 hours ago
answered 6 hours ago
Greg MartinGreg Martin
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9,7851 gold badge39 silver badges63 bronze badges
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$begingroup$
Do you believe the $2times 2$ bound is sharp?
$endgroup$
– Igor Rivin
9 hours ago
$begingroup$
@IgorRivin I don't have any heuristic supporting such belief.
$endgroup$
– Kate
8 hours ago
$begingroup$
It is obviously not far from sharp, since diagonal matrices give you $x^2.$
$endgroup$
– Igor Rivin
8 hours ago
$begingroup$
What happens when you take logarithms of the aij? Gerhard "Maybe Row-Sum Literature Will Help" Paseman, 2019.08.17.
$endgroup$
– Gerhard Paseman
5 hours ago
$begingroup$
@GerhardPaseman I think logarithms do not help very much. They turn the problem from multiplicative to additive, but the variables are not integers anymore, making things much more difficult...
$endgroup$
– Kate
5 hours ago