Distribution of Missing Sums in Sumsets...Distribution of Missing Sums in Sumsets Oleg Lazarev,...

Introduction Bounds on Distribution Variance Consecutive Missing Sums Future Research

Distribution of Missing Sums in Sumsets

Oleg Lazarev, Princeton UniversityAdvisor: Steven Miller, Williams College

SMALL Research Program 2011, Williams College

2011 Young Mathematicians ConferenceThe Ohio State University, August 21, 2011

1


Background

Let A ⊆ N ∪ {0}.

DefinitionSumset: A + A = {x + y : x , y ∈ A}Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}

Example: if A = {1,2,5}, then

A + A = {2,3,4,6,7,10}

Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.

Key Question: What is the structure of A + A?

2


Background

Let A ⊆ N ∪ {0}.

DefinitionSumset: A + A = {x + y : x , y ∈ A}

Interval: [a,b] = {x ∈ N : a ≤ x ≤ b}


A + A = {2,3,4,6,7,10}



3


Background

Let A ⊆ N ∪ {0}.



A + A = {2,3,4,6,7,10}



4


Background

Let A ⊆ N ∪ {0}.



A + A = {2,3,4,6,7,10}



5


Background

Let A ⊆ N ∪ {0}.



A + A = {2,3,4,6,7,10}

Why study sumsets?

Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.


6


Background

Let A ⊆ N ∪ {0}.



A + A = {2,3,4,6,7,10}

Why study sumsets?Goldbach’s conjecture: {4,6,8, · · · } ⊆ P + P.

Fermat’s last theorem: let An be the nth powers and thenask if (An + An) ∩ An = ∅ for all n > 2.


7


Background

Let A ⊆ N ∪ {0}.



A + A = {2,3,4,6,7,10}



8


Background

Let A ⊆ N ∪ {0}.



A + A = {2,3,4,6,7,10}


Key Question: What is the structure of A + A?9


Structure of Random Sets

Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].

Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?

Theorem: Martin, O’Bryant

E|A + A| = 2n − 1− 10 + O((3/4)n/2)

Theorem: ZhaoFor each fixed k , P(A ⊆ [0,n − 1] : |A + A| = 2n − 1− k) has alimit as n→∞

Note: Both theorem can be more naturally stated in terms ofmissing sums (independent of n).

10



Consider finite A ⊆ [0,n− 1] chosen randomly with uniformdistribution from all subsets of [0,n − 1].Question: What is the structure of A + A for such A? whatthe distribution of |A + A| for such A?


E|A + A| = 2n − 1− 10 + O((3/4)n/2)



11





E|A + A| = 2n − 1− 10 + O((3/4)n/2)



12





E|A + A| = 2n − 1− 10 + O((3/4)n/2)



13





E|A + A| = 2n − 1− 10 + O((3/4)n/2)



14


Structure of Random Sets, Continued

Why is the expectation so high? E |A + A| ∼ 2n − 11.

Main characteristic of typical A + A: middle is full.Many ways to write middle elements as sums

50 100 150 200

5

10

15

20

25

Figure: Comparison of predicted and observed number ofrepresentations of possible elements of the sumset, Miller

Key fact: if k < n, then P(k 6∈ A + A) ∼(3

4

)k/2

15



Why is the expectation so high? E |A + A| ∼ 2n − 11.Main characteristic of typical A + A: middle is full.

Many ways to write middle elements as sums

50 100 150 200

5

10

15

20

25



4

)k/2

16



Why is the expectation so high? E |A + A| ∼ 2n − 11.Main characteristic of typical A + A: middle is full.Many ways to write middle elements as sums

50 100 150 200

5

10

15

20

25



4

)k/2

17



Why is the expectation so high? E |A + A| ∼ 2n − 11.Main characteristic of typical A + A: middle is full.Many ways to write middle elements as sums

50 100 150 200

5

10

15

20

25



4

)k/2

18


New Results

Theorem: Bounds on the distribution (Lazarev, 2011)

0.70k � P(A + A has k missing sums)� 0.80k

From numerical data, conjecture that

P(A + A has k missing sums) ∼ 0.78k

and Log(P(A + A has k missing sums)) is eventually linear:

Figure: Log P(k missing sums) for 5 · 108 trial A of size 120

19


New Results







20


New Results







21


New Results







22


More Results

Theorem: Variance (Lazarev, 2011)Exact series for the variance (too long to fit on page...);also Var ∼ 35.98.

Theorem: Asymptotic formula for distribution ofconsecutive missing sums (Lazarev, 2011)

(12

)(k+i)/2

� P(A+A is missing [k , k+i])�(

12

)(k+i)/2

(1+εi)k

Key Obstacle: Dependent random variablesKey Approach: Graph theory! (and Fibonacci Numbers?)

23


More Results



(12

)(k+i)/2


12

)(k+i)/2

(1+εi)k


24


More Results



(12

)(k+i)/2


12

)(k+i)/2

(1+εi)k

Key Obstacle: Dependent random variables

Key Approach: Graph theory! (and Fibonacci Numbers?)

25


More Results



(12

)(k+i)/2


12

)(k+i)/2

(1+εi)k

Key Obstacle: Dependent random variablesKey Approach: Graph theory!

(and Fibonacci Numbers?)

26


More Results



(12

)(k+i)/2


12

)(k+i)/2

(1+εi)k


27


Bounds on the Distribution

28


Bound on Distribution: Upper Bound

Weaker Upper bound: P(A + A has k missing sums) < 0.93k

Proof sketch:

Recall P(k 6∈ A + A) =(3

4

)k/2

If k elements are missing, then missing one at least k/2from the edges

P(A + A has k missing sums) < P(k/2 6∈ A + A) <(34

)k/4 ∼ 0.93k

Note: Since the expectation is around 10 and expansion decay,all the higher moments are bounded as n→∞.

29




Proof sketch:Recall P(k 6∈ A + A) =

(34

)k/2



)k/4 ∼ 0.93k


30





(34

)k/2



)k/4 ∼ 0.93k


31





(34

)k/2



)k/4 ∼ 0.93k


32





(34

)k/2



)k/4 ∼ 0.93k


33





(34

)k/2



)k/4 ∼ 0.93k


34


Bound on Distribution: Lower Bound

Lower bound: P(A + A has k missing sums) > 0.01 · 0.70k

Proof sketch: Construction.

Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n

P(A + A has k missing sums) > 0.01(1

2

)k/2 ∼ 0.01 · 0.70k

35




Proof sketch: Construction.Let the first k/2 be missing from A

For the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n


2

)k/2 ∼ 0.01 · 0.70k

36




Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills in

Martin/O’Bryant: P( fills in) > 0.01 independent of n


2

)k/2 ∼ 0.01 · 0.70k

37




Proof sketch: Construction.Let the first k/2 be missing from AFor the rest of elements, pick any set that fills inMartin/O’Bryant: P( fills in) > 0.01 independent of n


2

)k/2 ∼ 0.01 · 0.70k

38






2

)k/2 ∼ 0.01 · 0.70k

39






2

)k/2 ∼ 0.01 · 0.70k

40






2

)k/2 ∼ 0.01 · 0.70k41


Variance

42


Variance

Can study Var|A + A| instead.Recall that

Var|A + A| = E(|A + A|2)− (E|A + A|)2

Therefore just need E(|A + A|2):

E(|A + A|2) =12n

∑A⊆[0,n−1]

|A + A|2

=12n

∑A⊆[0,n−1]

∑i,j∈A+A

1

=12n

∑0≤i,j≤2n−2

∑A:i,j∈A+A

1

=∑

0≤i,j≤2n−2

P(A : i and j ∈ A + A)

43


Variance

Can study Var|A + A| instead.Recall that

Var|A + A| = E(|A + A|2)− (E|A + A|)2

Therefore just need E(|A + A|2):

E(|A + A|2) =12n

∑A⊆[0,n−1]

|A + A|2

=12n

∑A⊆[0,n−1]

∑i,j∈A+A

1

=12n

∑0≤i,j≤2n−2

∑A:i,j∈A+A

1

=∑

0≤i,j≤2n−2

P(A : i and j ∈ A + A)

44


Problem: Dependent Random Variables

It is sufficient to study P(A : i and j 6∈ A + A):

P(A : i and j ∈ A + A) = 1− P(A : i 6∈ A + A)− P(A : j 6∈ A + A)+P(A : i and j 6∈ A + A).

Example: P(A : 3 and 7 6∈ A + A)Conditions:

i = 3 : 0 or 3 6∈ A j = 7 : 0 or 7 6∈ Aand 1 or 2 6∈ A and 1 or 6 6∈ A

and 2 or 5 6∈ Aand 3 or 4 6∈ A.

Since there are common integers in both lists, the events3 6∈ A + A and 7 6∈ A + A are dependent.

45





Example: P(A : 3 and 7 6∈ A + A)

Conditions:


and 2 or 5 6∈ Aand 3 or 4 6∈ A.


46







and 2 or 5 6∈ Aand 3 or 4 6∈ A.


47







and 2 or 5 6∈ Aand 3 or 4 6∈ A.


48


Solution: Use Graphs!

Transform the conditions into a graph!

For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.

Example i = 3, j = 7:

01

2

34

5

6

7

7 0 3 4

6 1 2 5

One-to-one correspondence between conditions/edges(and integers/vertices).

49



Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.

Then connect two vertices if add up to 3 or 7.


01

2

34

5

6

7

7 0 3 4

6 1 2 5


50



Transform the conditions into a graph!For each integers in [0,7], add a vertex with that integer.Then connect two vertices if add up to 3 or 7.


01

2

34

5

6

7

7 0 3 4

6 1 2 5


51





01

2

34

5

6

7

7 0 3 4

6 1 2 5


52





01

2

34

5

6

7

7 0 3 4

6 1 2 5


53





01

2

34

5

6

7

7 0 3 4

6 1 2 5


54





01

2

34

5

6

7

7 0 3 4

6 1 2 5


55


Interpretation of Graphs

Transformed into:

7 0 3 4 6 1 2 5

Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!

56



Transformed into:

7 0 3 4 6 1 2 5

Need to pick integers so that each condition is satisfied.

Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!

57



Transformed into:

7 0 3 4 6 1 2 5

Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.

So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!

58



Transformed into:

7 0 3 4 6 1 2 5

Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!

Each integer is equally likely so can ignore vertex labels!

59



Transformed into:

7 0 3 4 6 1 2 5

Need to pick integers so that each condition is satisfied.Therefore, need to pick vertices so that each edge has avertex chosen.So need to pick a vertex cover!Each integer is equally likely so can ignore vertex labels!

60


Vertex Covers

Have:

7 0 3 4 6 1 2 5

Example:7,0,4 and 6,2 form a vertex cover⇐⇒If 7,0,4,6,2 6∈ A, then 3,7 6∈ A + A

Lemma(Lazarev, 2011)

P(i , j 6∈ A + A) = P(pick a vertex cover for graph)

61


Vertex Covers

Have:

7 0 3 4 6 1 2 5

Example:7,0,4 and 6,2 form a vertex cover

⇐⇒If 7,0,4,6,2 6∈ A, then 3,7 6∈ A + A



62


Vertex Covers

Have:

7 0 3 4 6 1 2 5




63


Vertex Covers

Have:

7 0 3 4 6 1 2 5




64


Number of Vertex Covers

Condition graphs are always ‘segment’ graphs. So we just needg(n), the number of vertex covers for a ‘segment’ graph with nvertices.

Case 1: If the first vertex is chosen:x ? ? ? ?

Need an vertex cover for the rest of the graph: g(n − 1).Case 2: If the first vertex is not chosen:

o x ? ? ?

Need an vertex cover for the rest of the graph: g(n − 2).Fibonacci recursive relationship!

g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

65




Case 1: If the first vertex is chosen:

x ? ? ? ?


o x ? ? ?


g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

66






o x ? ? ?


g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

67





Need an vertex cover for the rest of the graph: g(n − 1).

Case 2: If the first vertex is not chosen:

o x ? ? ?


g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

68






o x ? ? ?


g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

69






o x ? ? ?

Need an vertex cover for the rest of the graph: g(n − 2).

Fibonacci recursive relationship!

g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

70






o x ? ? ?


g(n) = g(n − 1) + g(n − 2)

=⇒ g(n) = Fn+2

71






o x ? ? ?


g(n) = g(n − 1) + g(n − 2)=⇒ g(n) = Fn+2

72


General i , j

In particular

P(3 and 7 6∈ A + A) =128 F4+2F4+2 =

14

since there were two graphs each of length 4.

In general, graphs will have many components withdifferent lengths.For odd i < j < n:

P(A : i and j 6∈ A + A)

=1

2j+1 F12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

× F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4

Need to find the formulas for when i , j are even....And to get variance, need to sum up over all i < j < 2n....

73


General i , j

In particular

P(3 and 7 6∈ A + A) =128 F4+2F4+2 =

14


In general, graphs will have many components withdifferent lengths.

For odd i < j < n:

P(A : i and j 6∈ A + A)

=1

2j+1 F12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

× F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4


74


General i , j

In particular

P(3 and 7 6∈ A + A) =128 F4+2F4+2 =

14



P(A : i and j 6∈ A + A)

=1

2j+1 F12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

× F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4


75


General i , j

In particular

P(3 and 7 6∈ A + A) =128 F4+2F4+2 =

14



P(A : i and j 6∈ A + A)

=1

2j+1 F12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

× F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4

Need to find the formulas for when i , j are even....

And to get variance, need to sum up over all i < j < 2n....

76


General i , j

In particular

P(3 and 7 6∈ A + A) =128 F4+2F4+2 =

14



P(A : i and j 6∈ A + A)

=1

2j+1 F12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

× F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4


77


Variance Formula

Var|A + A| = −40 + 4∑

i<j<n

P(i, j 6∈ A + A)

= −40 + O(cn)

+4∑

i,j odd

1

2j+1F

12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4

+4∑

i even,j odd

1

2j+1F

2⌈

i/2+1j−i

⌉+1

F12

((j−i−1)

⌈i+1j−i

⌉−(i+1)+2

⌈i/2+1

j−i

⌉−1)

2⌈

i+1j−i

⌉+2

F12

(j+1−(j−i−1)

⌈i+1j−i

⌉−2⌈

i/2+1j−i

⌉)2⌈

i+1j−i

⌉+4

+4∑

i odd,j even

1

2j+1F

2⌈

j/2+1j−i

⌉F12

((j−i−1)

⌈i+1j−i

⌉−(i+1)+2

⌈j/2+1

j−i

⌉−2)

2⌈

i+1j−i

⌉+2

F12

(j+2−(j−i−1)

⌈i+1j−i

⌉+2⌈

j/2+1j−i

⌉)2⌈

i+1j−i

⌉+4

+4∑

i,j even

1

2j+1F

2⌈

i/2+1j−i

⌉+1

F2⌈ j+2

j−i

⌉ ×

F12

((j−i−2)

⌈i+1j−i

⌉−(i+1)+2

⌈i/2+1

j−i

⌉+2⌈

j/2+1j−i

⌉−3)

2⌈

i+1j−i

⌉+2

F12

(j+2−(j−i−2)

⌈i+1j−i +2

⌈i/2+1

j−i

⌉+2⌈

j/2+1j−i

⌉⌉)2⌈

i+1j−i

⌉+4

So clearlyVar |A + A| ∼ 35.98

78


Variance Formula

Var|A + A| = −40 + 4∑

i<j<n

P(i, j 6∈ A + A)

= −40 + O(cn)

+4∑

i,j odd

1

2j+1F

12

((j−i)

⌈i+1j−i

⌉−(i+1)

)2⌈

i+1j−i

⌉+2

F12

(j+1−(j−i)

⌈i+1j−i

⌉)2⌈

i+1j−i

⌉+4

+4∑

i even,j odd

1

2j+1F

2⌈

i/2+1j−i

⌉+1

F12

((j−i−1)

⌈i+1j−i

⌉−(i+1)+2

⌈i/2+1

j−i

⌉−1)

2⌈

i+1j−i

⌉+2

F12

(j+1−(j−i−1)

⌈i+1j−i

⌉−2⌈

i/2+1j−i

⌉)2⌈

i+1j−i

⌉+4

+4∑

i odd,j even

1

2j+1F

2⌈

j/2+1j−i

⌉F12

((j−i−1)

⌈i+1j−i

⌉−(i+1)+2

⌈j/2+1

j−i

⌉−2)

2⌈

i+1j−i

⌉+2

F12

(j+2−(j−i−1)

⌈i+1j−i

⌉+2⌈

j/2+1j−i

⌉)2⌈

i+1j−i

⌉+4

+4∑

i,j even

1

2j+1F

2⌈

i/2+1j−i

⌉+1

F2⌈ j+2

j−i

⌉ ×

F12

((j−i−2)

⌈i+1j−i

⌉−(i+1)+2

⌈i/2+1

j−i

⌉+2⌈

j/2+1j−i

⌉−3)

2⌈

i+1j−i

⌉+2

F12

(j+2−(j−i−2)

⌈i+1j−i +2

⌈i/2+1

j−i

⌉+2⌈

j/2+1j−i

⌉⌉)2⌈

i+1j−i

⌉+4

So clearlyVar |A + A| ∼ 35.98

79


Consecutive Missing Sums

80


Improved Upper Bound

Use P(i , j 6∈ A + A) to improve the upper bound forP(A + A has k missing sums )

From previous section:

P(i , j 6∈ A + A) ∼(

φ2

2 · 51/4

)j (51/4

φ

)i

where φ is the golden ratioIf A + A is missing at k elements, then missing at least twoelements greater than k :

P(A+A has k missing sums ) <

(φ2

2 · 51/4

)k (51/4

φ

)k

∼ 0.80k

General idea: The better we know P(a1, · · · ,aj 6∈ A + A),the better we know the P(A + A has k missing sums )

81





P(i , j 6∈ A + A) ∼(

φ2

2 · 51/4

)j (51/4

φ

)i

where φ is the golden ratio

If A + A is missing at k elements, then missing at least twoelements greater than k :


(φ2

2 · 51/4

)k (51/4

φ

)k

∼ 0.80k


82





P(i , j 6∈ A + A) ∼(

φ2

2 · 51/4

)j (51/4

φ

)i



(φ2

2 · 51/4

)k (51/4

φ

)k

∼ 0.80k


83





P(i , j 6∈ A + A) ∼(

φ2

2 · 51/4

)j (51/4

φ

)i



(φ2

2 · 51/4

)k (51/4

φ

)k

∼ 0.80k


84


Consecutive Missing Sums in A+A

Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)

Example: P(16,17,18,19,20 6∈ A + A)

Start with original graph and remove some conditions (edges):

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

=⇒ Transforms to:

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

85



Will study the particular case of P(a1, · · · ,aj 6∈ A + A) ofconsecutive missing sums: P(k , k + 1, · · · , k + i 6∈ A + A)Example: P(16,17,18,19,20 6∈ A + A)


0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

=⇒ Transforms to:

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

86





0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

=⇒ Transforms to:

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

87





0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

=⇒ Transforms to:

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

88





0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

=⇒ Transforms to:

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

89





0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

=⇒ Transforms to:

0 1 2 3 4 5 6 7 8

20 19 18 17 16 15 14 13 12 11 10 9

90



So get around 20/6 complete bipartite graphs like:

0 1 2

18 17 16

For a vertex cover, need one side to have all verticeschosen; occurs with probability ≤ 1

8 + 18 = 1

4

By independence, P(16,17,18,19,20) ≤(1

4

)20/6

In generalP(k , k + 1, k + 2, k + 3, k + 4) ≤

(14

)(k+4)/6 ∼ 0.79k+4

which is a slight improvement!

91




0 1 2

18 17 16


8 + 18 = 1

4


4

)20/6


(14

)(k+4)/6 ∼ 0.79k+4


92




0 1 2

18 17 16


8 + 18 = 1

4


4

)20/6


(14

)(k+4)/6 ∼ 0.79k+4


93




0 1 2

18 17 16


8 + 18 = 1

4


4

)20/6


(14

)(k+4)/6 ∼ 0.79k+4


94



But most general is:P(k , k + 1, · · · , k + i 6∈ A + A) ≤

(12

)(k+i)/2(1 + εi)

k

However the trivial lower bound is:(12

)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)Why interesting? Bounds almost match!Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.

95




(12

)(k+i)/2(1 + εi)

k


)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)

Why interesting? Bounds almost match!Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.

96




(12

)(k+i)/2(1 + εi)

k


)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)Why interesting? Bounds almost match!

Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.

97




(12

)(k+i)/2(1 + εi)

k


)(k+i)/2 ≤ P(k , k + 1, · · · , k + i 6∈ A + A)Why interesting? Bounds almost match!Essentially the only way to miss a block of i consecutivesums is to miss all elements before the block as well.

98


Summary

We found

Exponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums

by developing graph theoretic framework for studyingdependent random variables.

99


Summary

We foundExponential bounds for distribution of number of missingsums

Series for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums


100


Summary

We foundExponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbers

Asymptotic formula for consecutive missing sumsby developing graph theoretic framework for studyingdependent random variables.

101


Summary

We foundExponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums


102


Summary

We foundExponential bounds for distribution of number of missingsumsSeries for the variance in terms of Fibonacci numbersAsymptotic formula for consecutive missing sums


103


Future Research

And currently investigating/other research ideas:

Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}

104


Future Research

And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?

Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}

105


Future Research

And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?

Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}

106


Future Research

And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphs

Distribution of missing differences inA− A = {x − y : x , y ∈ A}

107


Future Research

And currently investigating/other research ideas:Is distribution of missing sums approximately exponential?Is P(k + a1, · · · , k + am) ∼ λk for any fixed a1, · · · ,am?Higher moments: third moment involves P(i , j , k) whichhas more complicated graphsDistribution of missing differences inA− A = {x − y : x , y ∈ A}

108


Thank you!

109

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