generalized multiplication tables - williams collegedimitris koukoulopoulos university of illinois...
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![Page 1: Generalized multiplication tables - Williams CollegeDimitris Koukoulopoulos University of Illinois at Urbana - Champaign June 4, 2009 1 The multiplication table and higher dimensional](https://reader035.vdocuments.site/reader035/viewer/2022071603/613e6ed069193359046d1e1a/html5/thumbnails/1.jpg)
The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Generalized multiplication tables
Dimitris Koukoulopoulos
University of Illinois at Urbana - Champaign
June 4, 2009
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Consider the multiplication table
1 2 3 4 · · · N1 1 2 3 4 · · · N2 2 4 6 8 · · · 2N3 3 6 9 12 · · · 3N4 4 8 12 16 · · · 4N...
......
......
...N N 2N 3N 4N · · · N2
Question (Erdos, 1955)How many distinct integers does this table contain?
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
We may ask the same question for products of 3 integers, inwhich case we have a multiplication ‘box’, or 4 integers, and soon. To this end, define
Ak+1(N) := |{n1 · · · nk+1 : ni ≤ N (1 ≤ i ≤ k + 1)}|.
Then the problem is equivalent to estimating Ak+1(N).
The key to understanding the combinatorics of themultiplication table is the function
Hk+1(x ,y , z) := |{n ≤ x :∃d1 · · · dk |n such thatyi < di ≤ zi (1 ≤ i ≤ k)}|,
where y = (y1, . . . , yk ) and z = (z1, . . . , zk ).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
We may ask the same question for products of 3 integers, inwhich case we have a multiplication ‘box’, or 4 integers, and soon. To this end, define
Ak+1(N) := |{n1 · · · nk+1 : ni ≤ N (1 ≤ i ≤ k + 1)}|.
Then the problem is equivalent to estimating Ak+1(N).
The key to understanding the combinatorics of themultiplication table is the function
Hk+1(x ,y , z) := |{n ≤ x :∃d1 · · · dk |n such thatyi < di ≤ zi (1 ≤ i ≤ k)}|,
where y = (y1, . . . , yk ) and z = (z1, . . . , zk ).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
The transition from Hk+1(x ,y , z) to Ak+1(N) is achieved via theelementary inequalities
Hk+1
(Nk+1
2k ,(N
2, . . . ,
N2
), (N, . . . ,N)
)≤ Ak+1(N)
≤∑
2mi≤√
N1≤i≤k
Hk+1
(Nk+1
2m1+···+mk,
(N
2m1+1 , . . . ,N
2mk+1
),
(N
2m1, . . . ,
N2mk
))
+ Nk+1/2.
Note that it suffices to study Hk+1(x ,y , z) whenz = 2y ;the numbers log y1, . . . , log yk all have the same order ofmagnitude.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
The transition from Hk+1(x ,y , z) to Ak+1(N) is achieved via theelementary inequalities
Hk+1
(Nk+1
2k ,(N
2, . . . ,
N2
), (N, . . . ,N)
)≤ Ak+1(N)
≤∑
2mi≤√
N1≤i≤k
Hk+1
(Nk+1
2m1+···+mk,
(N
2m1+1 , . . . ,N
2mk+1
),
(N
2m1, . . . ,
N2mk
))
+ Nk+1/2.
Note that it suffices to study Hk+1(x ,y , z) whenz = 2y ;the numbers log y1, . . . , log yk all have the same order ofmagnitude.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Set Q(u) := u log u − u + 1.
Then the following estimate holds.
Theorem (Ford (2004), K. (2008))Let k ≥ 1, 0 < δ ≤ 1 and c ≥ 1. Assume that x ≥ 1 and3 ≤ y1 ≤ y2 ≤ · · · ≤ yk ≤ yc
1 with xy1···yk
≥ max{2k , yδ1}. Then
Hk+1(x ,y ,2y) �k ,δ,cx
(log y1)Q( k
log(k+1))(log log y1)3/2
.
Consequently
Corollary
Ak+1(N) �kNk+1
(log N)Q( k
log(k+1))(log log N)3/2
(N ≥ 3).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Set Q(u) := u log u − u + 1. Then the following estimate holds.
Theorem (Ford (2004), K. (2008))Let k ≥ 1, 0 < δ ≤ 1 and c ≥ 1. Assume that x ≥ 1 and3 ≤ y1 ≤ y2 ≤ · · · ≤ yk ≤ yc
1 with xy1···yk
≥ max{2k , yδ1}. Then
Hk+1(x ,y ,2y) �k ,δ,cx
(log y1)Q( k
log(k+1))(log log y1)3/2
.
Consequently
Corollary
Ak+1(N) �kNk+1
(log N)Q( k
log(k+1))(log log N)3/2
(N ≥ 3).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Set Q(u) := u log u − u + 1. Then the following estimate holds.
Theorem (Ford (2004), K. (2008))Let k ≥ 1, 0 < δ ≤ 1 and c ≥ 1. Assume that x ≥ 1 and3 ≤ y1 ≤ y2 ≤ · · · ≤ yk ≤ yc
1 with xy1···yk
≥ max{2k , yδ1}. Then
Hk+1(x ,y ,2y) �k ,δ,cx
(log y1)Q( k
log(k+1))(log log y1)3/2
.
Consequently
Corollary
Ak+1(N) �kNk+1
(log N)Q( k
log(k+1))(log log N)3/2
(N ≥ 3).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
The following heuristic was first given by Ford when k = 1.
Assume that log y1, . . . , log yk have the same order ofmagnitude. For n ∈ N write n = ab, where
a =∏
pe‖n,p≤2y1
pe.
Assume that µ2(a) = 1 and a ≤ yC1 .
Consider the set
Dk+1(a) = {(log d1, . . . , log dk ) : d1 · · · dk |a}.
Main assumption: Dk+1(a) is well-distributed in [0, log a]k .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
The following heuristic was first given by Ford when k = 1.
Assume that log y1, . . . , log yk have the same order ofmagnitude. For n ∈ N write n = ab, where
a =∏
pe‖n,p≤2y1
pe.
Assume that µ2(a) = 1 and a ≤ yC1 .
Consider the set
Dk+1(a) = {(log d1, . . . , log dk ) : d1 · · · dk |a}.
Main assumption: Dk+1(a) is well-distributed in [0, log a]k .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
The following heuristic was first given by Ford when k = 1.
Assume that log y1, . . . , log yk have the same order ofmagnitude. For n ∈ N write n = ab, where
a =∏
pe‖n,p≤2y1
pe.
Assume that µ2(a) = 1 and a ≤ yC1 .
Consider the set
Dk+1(a) = {(log d1, . . . , log dk ) : d1 · · · dk |a}.
Main assumption: Dk+1(a) is well-distributed in [0, log a]k .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
The following heuristic was first given by Ford when k = 1.
Assume that log y1, . . . , log yk have the same order ofmagnitude. For n ∈ N write n = ab, where
a =∏
pe‖n,p≤2y1
pe.
Assume that µ2(a) = 1 and a ≤ yC1 .
Consider the set
Dk+1(a) = {(log d1, . . . , log dk ) : d1 · · · dk |a}.
Main assumption: Dk+1(a) is well-distributed in [0, log a]k .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Then we would expect that
τk+1(a,y ,2y) := |{d1 · · · dk |a : yi < di ≤ 2yi (1 ≤ i ≤ k)}|
=∣∣∣Dk+1(a) ∩
k∏i=1
(log yi , log yi + log 2]∣∣∣
≈ |Dk+1(a)|(log 2)k
(log a)k ≈(k + 1)ω(a)
(log y1)k .
This expression is ≥ 1 when
ω(a) ≥ m :=⌊ k
log(k + 1)log log y1
⌋+ O(1).
We have
|{n ≤ x : ω(a) = r}| ≈ xlog y1
(log log y1)r
r !.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Then we would expect that
τk+1(a,y ,2y) := |{d1 · · · dk |a : yi < di ≤ 2yi (1 ≤ i ≤ k)}|
=∣∣∣Dk+1(a) ∩
k∏i=1
(log yi , log yi + log 2]∣∣∣
≈ |Dk+1(a)|(log 2)k
(log a)k ≈(k + 1)ω(a)
(log y1)k .
This expression is ≥ 1 when
ω(a) ≥ m :=⌊ k
log(k + 1)log log y1
⌋+ O(1).
We have
|{n ≤ x : ω(a) = r}| ≈ xlog y1
(log log y1)r
r !.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Then we would expect that
τk+1(a,y ,2y) := |{d1 · · · dk |a : yi < di ≤ 2yi (1 ≤ i ≤ k)}|
=∣∣∣Dk+1(a) ∩
k∏i=1
(log yi , log yi + log 2]∣∣∣
≈ |Dk+1(a)|(log 2)k
(log a)k ≈(k + 1)ω(a)
(log y1)k .
This expression is ≥ 1 when
ω(a) ≥ m :=⌊ k
log(k + 1)log log y1
⌋+ O(1).
We have
|{n ≤ x : ω(a) = r}| ≈ xlog y1
(log log y1)r
r !.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Therefore
Hk+1(x ,y ,2y) ≈ xlog y1
∑r≥m
(log log y1)r
r !
� x
(log y1)Q( k
log(k+1))(log log y1)1/2
.
This is slightly too big. The problem arises from the fact thatDk+1(a) is usually not well-distributed, but instead it has manyclumps.
To see this define
Lk+1(a) := Vol( ⋃
d1···dk |a
[log(d1/2), log d1)×· · ·×[log(dk/2), log dk )),
which is a quantitative measure of how well-distributed Dk+1(a)is.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Therefore
Hk+1(x ,y ,2y) ≈ xlog y1
∑r≥m
(log log y1)r
r !
� x
(log y1)Q( k
log(k+1))(log log y1)1/2
.
This is slightly too big. The problem arises from the fact thatDk+1(a) is usually not well-distributed, but instead it has manyclumps.
To see this define
Lk+1(a) := Vol( ⋃
d1···dk |a
[log(d1/2), log d1)×· · ·×[log(dk/2), log dk )),
which is a quantitative measure of how well-distributed Dk+1(a)is.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
Therefore
Hk+1(x ,y ,2y) ≈ xlog y1
∑r≥m
(log log y1)r
r !
� x
(log y1)Q( k
log(k+1))(log log y1)1/2
.
This is slightly too big. The problem arises from the fact thatDk+1(a) is usually not well-distributed, but instead it has manyclumps.
To see this define
Lk+1(a) := Vol( ⋃
d1···dk |a
[log(d1/2), log d1)×· · ·×[log(dk/2), log dk )),
which is a quantitative measure of how well-distributed Dk+1(a)is.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
If a = p1 · · · pm, where p1 < · · · < pm ≤ 2y1, then we expect that
log log pj ∼jm
log log y1 = jlog(k + 1)
k+ O(1).
But with probability tending to 1 there is a j so that
log log pj ≤ jlog(k + 1)
k− (log log y1)
1/3,
which implies that
Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)
. (k + 1)m exp{−(log log y1)1/3}.
This is much less than τk+1(a) = (k + 1)m and so most of thetime Dk+1(a) contains large clusters of points.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
If a = p1 · · · pm, where p1 < · · · < pm ≤ 2y1, then we expect that
log log pj ∼jm
log log y1 = jlog(k + 1)
k+ O(1).
But with probability tending to 1 there is a j so that
log log pj ≤ jlog(k + 1)
k− (log log y1)
1/3,
which implies that
Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)
. (k + 1)m exp{−(log log y1)1/3}.
This is much less than τk+1(a) = (k + 1)m and so most of thetime Dk+1(a) contains large clusters of points.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
If a = p1 · · · pm, where p1 < · · · < pm ≤ 2y1, then we expect that
log log pj ∼jm
log log y1 = jlog(k + 1)
k+ O(1).
But with probability tending to 1 there is a j so that
log log pj ≤ jlog(k + 1)
k− (log log y1)
1/3,
which implies that
Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)
. (k + 1)m exp{−(log log y1)1/3}.
This is much less than τk+1(a) = (k + 1)m and so most of thetime Dk+1(a) contains large clusters of points.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
If a = p1 · · · pm, where p1 < · · · < pm ≤ 2y1, then we expect that
log log pj ∼jm
log log y1 = jlog(k + 1)
k+ O(1).
But with probability tending to 1 there is a j so that
log log pj ≤ jlog(k + 1)
k− (log log y1)
1/3,
which implies that
Lk+1(a) ≤ τk+1(pj+1 · · · pm)Lk+1(p1 · · · pj)
. (k + 1)m exp{−(log log y1)1/3}.
This is much less than τk+1(a) = (k + 1)m and so most of thetime Dk+1(a) contains large clusters of points.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
We must focus on abnormal a’s for which
log log pj ≥ jlog(k + 1)
k−O(1) (1 ≤ j ≤ m).
The probability that a has this property is about
1m� 1
log log y1(Ford).
This leads to the refined heuristic estimate
Hk+1(x ,y ,2y) ≈ x
(log y1)Q( 1
log ρ)(log log y1)3/2
,
which is the correct one.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
We must focus on abnormal a’s for which
log log pj ≥ jlog(k + 1)
k−O(1) (1 ≤ j ≤ m).
The probability that a has this property is about
1m� 1
log log y1(Ford).
This leads to the refined heuristic estimate
Hk+1(x ,y ,2y) ≈ x
(log y1)Q( 1
log ρ)(log log y1)3/2
,
which is the correct one.
10
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
The multiplication table problemStatement of resultsA heuristic argument
We must focus on abnormal a’s for which
log log pj ≥ jlog(k + 1)
k−O(1) (1 ≤ j ≤ m).
The probability that a has this property is about
1m� 1
log log y1(Ford).
This leads to the refined heuristic estimate
Hk+1(x ,y ,2y) ≈ x
(log y1)Q( 1
log ρ)(log log y1)3/2
,
which is the correct one.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
The natural generalization of the multiplication table problem isthe estimation of
Ak+1(N1, . . . ,Nk+1) := |{n1 · · · nk+1 : ni ≤ Ni (1 ≤ i ≤ k + 1)}|.
As before, to understand this question we study Hk+1(x ,y ,2y).
The difference is that we now drop the assumption that thenumbers log y1, . . . , log yk are of the same order of magnitude.
When k = 1, Ford’s result immediately implies that
CorollaryLet 3 ≤ N1 ≤ N2. Then
A2(N1,N2) �N1N2
(log N1)Q( 1
log 2 )(log log N1)3/2
.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
The natural generalization of the multiplication table problem isthe estimation of
Ak+1(N1, . . . ,Nk+1) := |{n1 · · · nk+1 : ni ≤ Ni (1 ≤ i ≤ k + 1)}|.
As before, to understand this question we study Hk+1(x ,y ,2y).
The difference is that we now drop the assumption that thenumbers log y1, . . . , log yk are of the same order of magnitude.
When k = 1, Ford’s result immediately implies that
CorollaryLet 3 ≤ N1 ≤ N2. Then
A2(N1,N2) �N1N2
(log N1)Q( 1
log 2 )(log log N1)3/2
.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
The natural generalization of the multiplication table problem isthe estimation of
Ak+1(N1, . . . ,Nk+1) := |{n1 · · · nk+1 : ni ≤ Ni (1 ≤ i ≤ k + 1)}|.
As before, to understand this question we study Hk+1(x ,y ,2y).
The difference is that we now drop the assumption that thenumbers log y1, . . . , log yk are of the same order of magnitude.
When k = 1, Ford’s result immediately implies that
CorollaryLet 3 ≤ N1 ≤ N2. Then
A2(N1,N2) �N1N2
(log N1)Q( 1
log 2 )(log log N1)3/2
.
11
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
For n ∈ N write n = a1 · · · akb, where
ai =∏
pe‖n,2yi−1<p≤2yi
pe.
Assume that µ2(ai) = 1 and ai ≤ yCi for 1 ≤ i ≤ k .
Set
Dk+1(a) := {(log d1, . . . , log dk ) : d1 · · · di |a1 · · · ai (1 ≤ i ≤ k)}
and assume that Dk+1(a) is well-distributed in[0, log y1]× · · · × [0, log yk ].
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
For n ∈ N write n = a1 · · · akb, where
ai =∏
pe‖n,2yi−1<p≤2yi
pe.
Assume that µ2(ai) = 1 and ai ≤ yCi for 1 ≤ i ≤ k .
Set
Dk+1(a) := {(log d1, . . . , log dk ) : d1 · · · di |a1 · · · ai (1 ≤ i ≤ k)}
and assume that Dk+1(a) is well-distributed in[0, log y1]× · · · × [0, log yk ].
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
Then
τ(n,y ,2y) = {(d1, . . . ,dk ) : d1 · · · di |a1 · · · ai ,
yi < di ≤ 2yi (1 ≤ i ≤ k)}
=∣∣∣Dk+1(a) ∩
k∏i=1
(log yi , log yi + log 2]∣∣∣
≈∏k
i=1(k − i + 2)ω(ai )∏ki=1 log yi
.
Set `i := log 3 log yilog yi−1
and
H :={
(r1, . . . , rk ) ∈ (N∪{0})k :k∑
i=1
ri log(k−i+2) ≥k∑
i=1
`i(k−i+1)}.
Then, heuristically, τ(n,y ,2y) ≥ 1 if-f (ω(a1), . . . , ω(ak )) ∈H .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
Then
τ(n,y ,2y) = {(d1, . . . ,dk ) : d1 · · · di |a1 · · · ai ,
yi < di ≤ 2yi (1 ≤ i ≤ k)}
=∣∣∣Dk+1(a) ∩
k∏i=1
(log yi , log yi + log 2]∣∣∣
≈∏k
i=1(k − i + 2)ω(ai )∏ki=1 log yi
.
Set `i := log 3 log yilog yi−1
and
H :={
(r1, . . . , rk ) ∈ (N∪{0})k :k∑
i=1
ri log(k−i+2) ≥k∑
i=1
`i(k−i+1)}.
Then, heuristically, τ(n,y ,2y) ≥ 1 if-f (ω(a1), . . . , ω(ak )) ∈H .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
Then
τ(n,y ,2y) = {(d1, . . . ,dk ) : d1 · · · di |a1 · · · ai ,
yi < di ≤ 2yi (1 ≤ i ≤ k)}
=∣∣∣Dk+1(a) ∩
k∏i=1
(log yi , log yi + log 2]∣∣∣
≈∏k
i=1(k − i + 2)ω(ai )∏ki=1 log yi
.
Set `i := log 3 log yilog yi−1
and
H :={
(r1, . . . , rk ) ∈ (N∪{0})k :k∑
i=1
ri log(k−i+2) ≥k∑
i=1
`i(k−i+1)}.
Then, heuristically, τ(n,y ,2y) ≥ 1 if-f (ω(a1), . . . , ω(ak )) ∈H .13
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
We have that
|{n ≤ x : ω(ai) = ri (1 ≤ i ≤ k)}| ≈ xlog yk
k∏i=1
`rii
ri !.
So using Taylor’s theorem and Lagrange multipliers we find that
Hk+1(x ,y ,2y) ≈ xlog yk
∑r∈H
k∏i=1
`rii
ri !
� x√log log yk
∏ki=1
(log yi
log yi−1
)Q((k−i+2)α),
where α = α(k ,y) satisfiesk∑
i=1
(k − i + 2)α log(k − i + 2)`i =k∑
i=1
(k − i + 1)`i .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
We have that
|{n ≤ x : ω(ai) = ri (1 ≤ i ≤ k)}| ≈ xlog yk
k∏i=1
`rii
ri !.
So using Taylor’s theorem and Lagrange multipliers we find that
Hk+1(x ,y ,2y) ≈ xlog yk
∑r∈H
k∏i=1
`rii
ri !
� x√log log yk
∏ki=1
(log yi
log yi−1
)Q((k−i+2)α),
where α = α(k ,y) satisfiesk∑
i=1
(k − i + 2)α log(k − i + 2)`i =k∑
i=1
(k − i + 1)`i .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
Indeed, we may prove the following estimate.
Theorem (K, 2008)
Let 3 = y0 ≤ y1 ≤ · · · ≤ yk with xy1···yk
≥ max{2k , yk}. Then
Hk+1(x ,y ,2y)
x�k
min{
1,(log log 3yi0−1)(log 3 log yk
log yi0)
`i0
}√
log log yk
k∏i=1
( log yi
log yi−1
)Q((k−i+2)α),
where i0 is such that `i0 = max1≤i≤k `i .
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
When k = 2, this upper bound is the correct order ofH3(x ,y ,2y).
Theorem (K, 2008)
Let 3 ≤ y1 ≤ y2 so that xy1y2≥ max{4, y2}. Then
H3(x ,y ,2y)
x�
(log log 3y1)(log 3 log y2log y1
)
(log log y2)5/2(log y1)Q(3α)(
log y2log y1
)Q(2α),
where y = (y1, y2).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
As a direct corollary we obtain the order of magnitude of thenumber of integers in a 3-dimensional multiplication table.
CorollaryFor 3 ≤ N1 ≤ N2 ≤ N3 we have that
A3(N1,N2,N3)
N1N2N3�
(log log N1)(log 3 log N2log N1
)
(log log N2)5/2(log N1)Q(3α)(
log N2log N1
)Q(2α).
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
In general, we may reduce the counting in Hk+1(x ,y ,2y) (localdistribution of factorizations) to estimating averages of
Lk+1(a) := Vol( ⋃
d1···di |a1···ai1≤i≤k
[log(d1/2), log d1)×· · · [log(dk/2), log dk ))
(global distribution of factorizations).
Proposition (Ford (2004), K. (2008))Let k ≥ 1, x ≥ 1 and 3 ≤ y1 ≤ · · · ≤ yk with
xy1···yk
≥ max{2k , yk}. Then
Hk+1(x ,y ,2y)
x�k
k∏i=1
( log yi
log yi−1
)−(k−i+2) ∑ai∈P(yi−1,yi )
1≤i≤k
Lk+1(a)
a1 · · · ak.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
In general, we may reduce the counting in Hk+1(x ,y ,2y) (localdistribution of factorizations) to estimating averages of
Lk+1(a) := Vol( ⋃
d1···di |a1···ai1≤i≤k
[log(d1/2), log d1)×· · · [log(dk/2), log dk ))
(global distribution of factorizations).
Proposition (Ford (2004), K. (2008))Let k ≥ 1, x ≥ 1 and 3 ≤ y1 ≤ · · · ≤ yk with
xy1···yk
≥ max{2k , yk}. Then
Hk+1(x ,y ,2y)
x�k
k∏i=1
( log yi
log yi−1
)−(k−i+2) ∑ai∈P(yi−1,yi )
1≤i≤k
Lk+1(a)
a1 · · · ak.
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The multiplication table and higher dimensional analoguesGeneralized multiplication tables
Basic set-upAdjusting the heuristic argumentGeneral upper boundsThe case k = 2Local to global estimates
Thank you for your attention!
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