the complementary bell numbers - lsu math

IntroductionConstruction of the R-Matrix

ResultsConclusion

The Complementary Bell NumbersExplored via a Matrix Constructed with Rising Factorials

Jonathan Broom, Stefan Hannie, Sarah Seger

Ole Miss,ULL,LSU

July 6, 2012

Jonathan Broom, Stefan Hannie, Sarah Seger The Complementary Bell Numbers

ResultsConclusion

1 IntroductionFactorialsStirling NumbersBell Numbers

2 Construction of the R-Matrixλj(x)BasisCoefficientsMatrices

3 ResultsInfinite MatricesFinite Matrices

4 ConclusionConclusionAcknowledgementsWorks Cited

ResultsConclusion

FactorialsStirling NumbersBell Numbers

Factorials

The falling factorial is denoted (x)r

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

The rising factorial is denoted x (r)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

Rising factorial example: Let x = 7 and r = 4

7(4) = 7(8)(9)(10) = 5040

Note that both (x)r and x (r) are polynomials of degree r .

ResultsConclusion

Factorials

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

7(4) = 7(8)(9)(10) = 5040

ResultsConclusion

Factorials

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

7(4) = 7(8)(9)(10) = 5040

ResultsConclusion

Factorials

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

7(4) = 7(8)(9)(10) = 5040

ResultsConclusion

Factorials

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

7(4) = 7(8)(9)(10) = 5040

ResultsConclusion

Factorials

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

7(4) = 7(8)(9)(10) = 5040

ResultsConclusion

Factorials

(x)r = x(x − 1)(x − 2) · · · (x − r + 1)

x (r) = x(x + 1)(x + 2) · · · (x + r − 1)

7(4) = 7(8)(9)(10) = 5040

ResultsConclusion

Stirling Numbers of the Second Kind

The Stirling Numbers of the Second Kind are denoted S(n, k).

They are the number of ways you can partition n elementsinto k non-empty blocks.

For example, take a set containing 3 items {a, b, c}

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

Another example for S(3, k):

Figure: S(3, 1) = 1 Figure: S(3, 2) = 3 Figure: S(3, 3) = 1

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3

{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}

{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}

{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1

{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

Figure: S(3, 1) = 1

Figure: S(3, 2) = 3 Figure: S(3, 3) = 1

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

Figure: S(3, 1) = 1 Figure: S(3, 2) = 3

Figure: S(3, 3) = 1

ResultsConclusion

S(3, 1) = 1{{a, b, c}}

S(3, 2) = 3{{a}, {b, c}}{{b}, {a, c}}{{c}, {a, b}}

S(3, 3) = 1{{a}, {b}, {c}}

ResultsConclusion

Similarly S(4, k):

Figure: S(4, 1) = 1 Figure: S(4, 2) = 7

Figure: S(4, 3) = 6 Figure: S(4, 4) = 1

Note: From the examples, it is clear that S(n, 1) = S(n, n) = 1.

ResultsConclusion

Similarly S(4, k):

Figure: S(4, 1) = 1

Figure: S(4, 2) = 7

Figure: S(4, 3) = 6 Figure: S(4, 4) = 1

ResultsConclusion

Similarly S(4, k):

Figure: S(4, 1) = 1 Figure: S(4, 2) = 7

Figure: S(4, 3) = 6 Figure: S(4, 4) = 1

ResultsConclusion

Similarly S(4, k):

Figure: S(4, 1) = 1 Figure: S(4, 2) = 7

Figure: S(4, 3) = 6

Figure: S(4, 4) = 1

ResultsConclusion

Similarly S(4, k):

Figure: S(4, 1) = 1 Figure: S(4, 2) = 7

Figure: S(4, 3) = 6 Figure: S(4, 4) = 1

ResultsConclusion

Similarly S(4, k):

Figure: S(4, 1) = 1 Figure: S(4, 2) = 7

Figure: S(4, 3) = 6 Figure: S(4, 4) = 1

ResultsConclusion

Growth

SH7, 4L = 350

SH6, 3L = 90

SH8, 4L = 1701

2 4 6 8k

The points labeled are the k values that yield the maximum S(n, k) for a given n.

ResultsConclusion

Bell Numbers and Complementary Bell Numbers

The Bell Numbers are denoted B(n)

B(n) =n∑

S(n, k)

The Complementary Bell Numbers are denoted B̃(n)

B̃(n) =n∑

(−1)kS(n, k)

ResultsConclusion

B(n) =n∑

S(n, k)

B̃(n) =n∑

(−1)kS(n, k)

ResultsConclusion

B(n) =n∑

S(n, k)

B̃(n) =n∑

(−1)kS(n, k)

ResultsConclusion

B(n) =n∑

S(n, k)

B̃(n) =n∑

(−1)kS(n, k)

ResultsConclusion

B̃(n) Examples

Examples:

B̃(2) = 0

Odd Even{a, b} {a}, {b}

B̃(4) = 1

Odd Even

B̃(3) = 1

Odd Even

B̃(5) = −2

Odd Even

ResultsConclusion

B̃(n) Examples

Examples:

B̃(2) = 0

B̃(4) = 1

Odd Even

B̃(3) = 1

Odd Even

B̃(5) = −2

Odd Even

ResultsConclusion

B̃(n) Examples

Examples:

B̃(2) = 0

B̃(4) = 1

Odd Even

B̃(3) = 1

Odd Even

B̃(5) = −2

Odd Even

ResultsConclusion

B̃(n) Examples

Examples:

B̃(2) = 0

B̃(4) = 1

Odd Even

B̃(3) = 1

Odd Even

B̃(5) = −2

Odd Even

ResultsConclusion

B̃(n) Examples

Examples:

B̃(2) = 0

B̃(4) = 1

Odd Even

B̃(3) = 1

Odd Even

B̃(5) = −2

Odd Even

ResultsConclusion

Complementary Bell Numbers

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267

10 41311 −218012 −1773113 −5053314 11017615 196679716 993866917 8638718...

2 4 6 8n

BnH-1L

Figure: |B̃(n)| for n ≤ 8Jonathan Broom, Stefan Hannie, Sarah Seger The Complementary Bell Numbers

ResultsConclusion

Wilf’s Conjecture

H.S. Wilf’s Conjecture:

B̃(n) 6= 0 for all n > 2

ResultsConclusion

λj (x)BasisCoefficientsMatrices

The λj(x) Polynomials

There exist polynomials λj , for all n, j ≥ 0, that satisfy

B̃(n + j) =n∑

(−1)kλj(k)S(n, k)

λj(x) can be defined recursively as follows:

λ0(x) = 1

λn+1(x) = xλn(x)− λn (x + 1)

Alternate Form:

λ0(x) = 1

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

Note that λn(x) is a monic polynomial of degree n.

ResultsConclusion

B̃(n + j) =n∑

λ0(x) = 1

λn+1(x) = xλn(x)− λn (x + 1)

Alternate Form:

λ0(x) = 1

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

ResultsConclusion

B̃(n + j) =n∑

λ0(x) = 1

λn+1(x) = xλn(x)− λn (x + 1)

Alternate Form:

λ0(x) = 1

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

ResultsConclusion

B̃(n + j) =n∑

λ0(x) = 1

λn+1(x) = xλn(x)− λn (x + 1)

Alternate Form:

λ0(x) = 1

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

ResultsConclusion

Rising Factorials as a Basis for Pn

Theorem

For each n ≥ 0, the set of rising factorials{

x (k) : 0 ≤ k ≤ n}

is abasis for Pn, the vector space of polynomials of degree less than orequal to n.

xn =n∑

(−1)n+kS(n, k)x (k) for all n ≥ 0

ResultsConclusion

Rising Factorials as a Basis for Pn

Theorem

For each n ≥ 0, the set of rising factorials{

x (k) : 0 ≤ k ≤ n}

is abasis for Pn, the vector space of polynomials of degree less than orequal to n.

xn =n∑

(−1)n+kS(n, k)x (k) for all n ≥ 0

ResultsConclusion

The Coefficients of the R-Matrix

By the previous theorem:

λn(x) =n∑

an(k)x(k)

By the recurrence relation of λn(x):

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

Therefore:

n+1∑k=0

an+1(k) (x − 1)(k) =n∑

an(k) (x − 1) (x − 1)(k) −n∑

an(k)x(k)

For all n ≥ 0 and for all 0 ≤ k ≤ n + 1,

an+1(k)− (k + 1) an+1(k +1) = an(k − 1)− 2 (k + 1) an(k)+ (k + 1)2 an(k +1)

ResultsConclusion

λn(x) =n∑

an(k)x(k)

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

Therefore:

n+1∑k=0

an+1(k) (x − 1)(k) =n∑

an(k) (x − 1) (x − 1)(k) −n∑

an(k)x(k)

ResultsConclusion

λn(x) =n∑

an(k)x(k)

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

Therefore:

n+1∑k=0

an+1(k) (x − 1)(k) =n∑

an(k) (x − 1) (x − 1)(k) −n∑

an(k)x(k)

ResultsConclusion

λn(x) =n∑

an(k)x(k)

λn+1(x − 1) = (x − 1)λn(x − 1)− λn(x)

Therefore:

n+1∑k=0

an+1(k) (x − 1)(k) =n∑

an(k) (x − 1) (x − 1)(k) −n∑

an(k)x(k)

ResultsConclusion

The A-Matrix and B-Matrix

Let Aan+1 = Ban, then A,B are infinite matrices whose entries aredefined by

A(i , j) =

1 if j = i

− (i + 1) if j = i + 1

0 if j i + 1

B(i , j) =

1 if j = i − 1

−2 (i + 1) if j = i

(i + 1)2 if j = i + 1

0 if |i − j | > 1

1 −1 0 0 . . .0 1 −2 0 . . .0 0 1 −3 . . .0 0 0 1 . . ....

......

.... . .

−2 1 0 0 . . .1 −4 4 0 . . .0 1 −6 9 . . .0 0 1 −8 . . ....

......

.... . .

ResultsConclusion

A(i , j) =

1 if j = i

− (i + 1) if j = i + 1

0 if j i + 1

B(i , j) =

1 if j = i − 1

−2 (i + 1) if j = i

(i + 1)2 if j = i + 1

0 if |i − j | > 1

1 −1 0 0 . . .0 1 −2 0 . . .0 0 1 −3 . . .0 0 0 1 . . ....

......

.... . .

−2 1 0 0 . . .1 −4 4 0 . . .0 1 −6 9 . . .0 0 1 −8 . . ....

......

.... . .

ResultsConclusion

A(i , j) =

1 if j = i

− (i + 1) if j = i + 1

0 if j i + 1

B(i , j) =

1 if j = i − 1

−2 (i + 1) if j = i

(i + 1)2 if j = i + 1

0 if |i − j | > 1

1 −1 0 0 . . .0 1 −2 0 . . .0 0 1 −3 . . .0 0 0 1 . . ....

......

.... . .

−2 1 0 0 . . .1 −4 4 0 . . .0 1 −6 9 . . .0 0 1 −8 . . ....

......

.... . .

ResultsConclusion

A(i , j) =

1 if j = i

− (i + 1) if j = i + 1

0 if j i + 1

B(i , j) =

1 if j = i − 1

−2 (i + 1) if j = i

(i + 1)2 if j = i + 1

0 if |i − j | > 1

1 −1 0 0 . . .0 1 −2 0 . . .0 0 1 −3 . . .0 0 0 1 . . ....

......

.... . .

−2 1 0 0 . . .1 −4 4 0 . . .0 1 −6 9 . . .0 0 1 −8 . . ....

......

.... . .

ResultsConclusion

A(i , j) =

1 if j = i

− (i + 1) if j = i + 1

0 if j i + 1

B(i , j) =

1 if j = i − 1

−2 (i + 1) if j = i

(i + 1)2 if j = i + 1

0 if |i − j | > 1

1 −1 0 0 . . .0 1 −2 0 . . .0 0 1 −3 . . .0 0 0 1 . . ....

......

.... . .

−2 1 0 0 . . .1 −4 4 0 . . .0 1 −6 9 . . .0 0 1 −8 . . ....

......

.... . .

ResultsConclusion

The A−1-Matrix

Taking Aan+1 = Ban, we solve for an+1. Therefore:

an+1 = A−1Ban

A−1(i , j) =

{j!i! if j ≥ i

0 if j < i A−1 =

1 1 2 6 24 . . .0 1 2 6 24 . . .0 0 1 3 12 . . .0 0 0 1 4 . . .0 0 0 0 1 . . ....

......

ResultsConclusion

The A−1-Matrix

an+1 = A−1Ban

A−1(i , j) =

{j!i! if j ≥ i

0 if j < i

A−1 =

1 1 2 6 24 . . .0 1 2 6 24 . . .0 0 1 3 12 . . .0 0 0 1 4 . . .0 0 0 0 1 . . ....

......

ResultsConclusion

The A−1-Matrix

an+1 = A−1Ban

A−1(i , j) =

{j!i! if j ≥ i

0 if j < i A−1 =

1 1 2 6 24 . . .0 1 2 6 24 . . .0 0 1 3 12 . . .0 0 0 1 4 . . .0 0 0 0 1 . . ....

......

ResultsConclusion

The R-Matrix

Taking an+1 = A−1Ban, we call R = A−1B. Therefore:

an+1 = Ran

R(i , j) =

− j!

i! if j > i

−(i + 1) if j = i

1 if j = i − 1

0 if j < i − 1

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . ....

......

ResultsConclusion

The R-Matrix

an+1 = Ran

R(i , j) =

− j!

i! if j > i

−(i + 1) if j = i

1 if j = i − 1

0 if j < i − 1

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . ....

......

ResultsConclusion

The R-Matrix

an+1 = Ran

R(i , j) =

− j!

i! if j > i

−(i + 1) if j = i

1 if j = i − 1

0 if j < i − 1

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . ....

......

ResultsConclusion

Infinite MatricesFinite Matrices

The Lower Section

For each n ∈ N, the nth power of R is defined and Rn(i , j) = 0 ifj < i − n.

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . .

.. . .

0 1 4 18 96 . . .−3 1 2 12 72 . . .

1 −5 4 3 24 . . .0 1 −7 9 4 . . .0 0 1 −9 16 . . .

.. . .

1 2 4 6 −24 . . .4 3 10 30 96 . . .

−6 13 −1 24 96 . . .1 −9 28 −17 44 . . .0 1 −12 49 −51 . . .

.. . .

1 −1 −12 −78 −504 . . .−1 0 −14 −96 −648 . . .19 −21 13 −39 −312 . . .

−10 45 −85 76 −76 . . .1 −14 83 −217 249 . . .

.. . .

ResultsConclusion

The Lower Section

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . .

.. . .

0 1 4 18 96 . . .−3 1 2 12 72 . . .

1 −5 4 3 24 . . .0 1 −7 9 4 . . .0 0 1 −9 16 . . .

.. . .

1 2 4 6 −24 . . .4 3 10 30 96 . . .

−6 13 −1 24 96 . . .1 −9 28 −17 44 . . .0 1 −12 49 −51 . . .

.. . .

1 −1 −12 −78 −504 . . .−1 0 −14 −96 −648 . . .19 −21 13 −39 −312 . . .

−10 45 −85 76 −76 . . .1 −14 83 −217 249 . . .

.. . .

ResultsConclusion

The Lower Section

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . .

.. . .

0 1 4 18 96 . . .−3 1 2 12 72 . . .

1 −5 4 3 24 . . .0 1 −7 9 4 . . .0 0 1 −9 16 . . .

.. . .

1 2 4 6 −24 . . .4 3 10 30 96 . . .

−6 13 −1 24 96 . . .1 −9 28 −17 44 . . .0 1 −12 49 −51 . . .

.. . .

1 −1 −12 −78 −504 . . .−1 0 −14 −96 −648 . . .19 −21 13 −39 −312 . . .

−10 45 −85 76 −76 . . .1 −14 83 −217 249 . . .

.. . .

ResultsConclusion

The Lower Section

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . .

.. . .

0 1 4 18 96 . . .−3 1 2 12 72 . . .

1 −5 4 3 24 . . .0 1 −7 9 4 . . .0 0 1 −9 16 . . .

.. . .

1 2 4 6 −24 . . .4 3 10 30 96 . . .

−6 13 −1 24 96 . . .1 −9 28 −17 44 . . .0 1 −12 49 −51 . . .

.. . .

1 −1 −12 −78 −504 . . .−1 0 −14 −96 −648 . . .19 −21 13 −39 −312 . . .

−10 45 −85 76 −76 . . .1 −14 83 −217 249 . . .

.. . .

ResultsConclusion

The Lower Section

−1 −1 −2 −6 −24 . . .1 −2 −2 −6 −24 . . .0 1 −3 −3 −12 . . .0 0 1 −4 −4 . . .0 0 0 1 −5 . . .

.. . .

0 1 4 18 96 . . .−3 1 2 12 72 . . .

1 −5 4 3 24 . . .0 1 −7 9 4 . . .0 0 1 −9 16 . . .

.. . .

1 2 4 6 −24 . . .4 3 10 30 96 . . .

−6 13 −1 24 96 . . .1 −9 28 −17 44 . . .0 1 −12 49 −51 . . .

.. . .

1 −1 −12 −78 −504 . . .−1 0 −14 −96 −648 . . .19 −21 13 −39 −312 . . .

−10 45 −85 76 −76 . . .1 −14 83 −217 249 . . .

.. . .

ResultsConclusion

The Top Row

For all n ≥ 1 and j ≥ 0, the (0, j)th entry of Rn is divisible by j!.

j j!0 11 12 23 64 245 1206 720...

−1 −1 −2 −6 −24 −120 −720 . . .1 −2 −2 −6 −24 −120 −720 . . .0 1 −3 −3 −12 −60 −360 . . .0 0 1 −4 −4 −20 −120 . . .

.. . .

1 −1 −12 −78 −504 36840 −953280 . . .−1 0 −14 −96 −648 35640 −923760 . . .19 −21 13 −39 −312 17700 −443880 . . .

−10 45 −85 76 −76 6000 −141000 . . .

.. . .

For R4: −783!

= −13, −5044!

= −21, 368405!

= 307, −9532806!

= −1324

ResultsConclusion

The Top Row

j j!0 11 12 23 64 245 1206 720...

−1 −1 −2 −6 −24 −120 −720 . . .1 −2 −2 −6 −24 −120 −720 . . .0 1 −3 −3 −12 −60 −360 . . .0 0 1 −4 −4 −20 −120 . . .

.. . .

1 −1 −12 −78 −504 36840 −953280 . . .−1 0 −14 −96 −648 35640 −923760 . . .19 −21 13 −39 −312 17700 −443880 . . .

−10 45 −85 76 −76 6000 −141000 . . .

.. . .

For R4: −783!

= −13, −5044!

= −21, 368405!

= 307, −9532806!

= −1324

ResultsConclusion

The Top Row

j j!0 11 12 23 64 245 1206 720...

−1 −1 −2 −6 −24 −120 −720 . . .1 −2 −2 −6 −24 −120 −720 . . .0 1 −3 −3 −12 −60 −360 . . .0 0 1 −4 −4 −20 −120 . . .

.. . .

1 −1 −12 −78 −504 36840 −953280 . . .−1 0 −14 −96 −648 35640 −923760 . . .19 −21 13 −39 −312 17700 −443880 . . .

−10 45 −85 76 −76 6000 −141000 . . .

.. . .

For R4: −783!

= −13, −5044!

= −21, 368405!

= 307, −9532806!

= −1324

ResultsConclusion

The Top Row

j j!0 11 12 23 64 245 1206 720...

−1 −1 −2 −6 −24 −120 −720 . . .1 −2 −2 −6 −24 −120 −720 . . .0 1 −3 −3 −12 −60 −360 . . .0 0 1 −4 −4 −20 −120 . . .

.. . .

1 −1 −12 −78 −504 36840 −953280 . . .−1 0 −14 −96 −648 35640 −923760 . . .19 −21 13 −39 −312 17700 −443880 . . .

−10 45 −85 76 −76 6000 −141000 . . .

.. . .

For R4: −783!

= −13, −5044!

= −21, 368405!

= 307, −9532806!

= −1324

ResultsConclusion

The Top Row

j j!0 11 12 23 64 245 1206 720...

−1 −1 −2 −6 −24 −120 −720 . . .1 −2 −2 −6 −24 −120 −720 . . .0 1 −3 −3 −12 −60 −360 . . .0 0 1 −4 −4 −20 −120 . . .

.. . .

1 −1 −12 −78 −504 36840 −953280 . . .−1 0 −14 −96 −648 35640 −923760 . . .19 −21 13 −39 −312 17700 −443880 . . .

−10 45 −85 76 −76 6000 −141000 . . .

.. . .

For R4: −783!

= −13, −5044!

= −21, 368405!

= 307, −9532806!

= −1324

ResultsConclusion

The Top Left Entry

Theorem

For all n ∈ N, B̃(n) = Rn(0, 0).

n B̃(n)1 −12 03 14 15 −26 −9...

−1 −1 −2 −6 . . .1 −2 −2 −6 . . .0 1 −3 −3 . . .0 0 1 −4 . . .

.. . .

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

0 1 4 18 . . .−3 1 2 12 . . .

1 −5 4 3 . . .0 1 −7 9 . . .

.. . .

−9 −18 −4 40644 . . .−14 −27 −36 40548 . . .

76 −106 47 20286 . . .−220 536 −898 7473 . . .

.. . .

ResultsConclusion

The Top Left Entry

Theorem

For all n ∈ N, B̃(n) = Rn(0, 0).

n B̃(n)1 −12 03 14 15 −26 −9...

−1 −1 −2 −6 . . .1 −2 −2 −6 . . .0 1 −3 −3 . . .0 0 1 −4 . . .

.. . .

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

0 1 4 18 . . .−3 1 2 12 . . .

1 −5 4 3 . . .0 1 −7 9 . . .

.. . .

−9 −18 −4 40644 . . .−14 −27 −36 40548 . . .

76 −106 47 20286 . . .−220 536 −898 7473 . . .

.. . .

ResultsConclusion

The Top Left Entry

Theorem

For all n ∈ N, B̃(n) = Rn(0, 0).

n B̃(n)1 −12 03 14 15 −26 −9...

−1 −1 −2 −6 . . .1 −2 −2 −6 . . .0 1 −3 −3 . . .0 0 1 −4 . . .

.. . .

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

0 1 4 18 . . .−3 1 2 12 . . .

1 −5 4 3 . . .0 1 −7 9 . . .

.. . .

−9 −18 −4 40644 . . .−14 −27 −36 40548 . . .

76 −106 47 20286 . . .−220 536 −898 7473 . . .

.. . .

ResultsConclusion

The Top Left Entry

Theorem

For all n ∈ N, B̃(n) = Rn(0, 0).

n B̃(n)1 −12 03 14 15 −26 −9...

−1 −1 −2 −6 . . .1 −2 −2 −6 . . .0 1 −3 −3 . . .0 0 1 −4 . . .

.. . .

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

0 1 4 18 . . .−3 1 2 12 . . .

1 −5 4 3 . . .0 1 −7 9 . . .

.. . .

−9 −18 −4 40644 . . .−14 −27 −36 40548 . . .

76 −106 47 20286 . . .−220 536 −898 7473 . . .

.. . .

ResultsConclusion

The Top Left Entry

Theorem

For all n ∈ N, B̃(n) = Rn(0, 0).

n B̃(n)1 −12 03 14 15 −26 −9...

−1 −1 −2 −6 . . .1 −2 −2 −6 . . .0 1 −3 −3 . . .0 0 1 −4 . . .

.. . .

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

0 1 4 18 . . .−3 1 2 12 . . .

1 −5 4 3 . . .0 1 −7 9 . . .

.. . .

−9 −18 −4 40644 . . .−14 −27 −36 40548 . . .

76 −106 47 20286 . . .−220 536 −898 7473 . . .

.. . .

ResultsConclusion

The Top Left Entry

Theorem

For all n ∈ N, B̃(n) = Rn(0, 0).

n B̃(n)1 −12 03 14 15 −26 −9...

−1 −1 −2 −6 . . .1 −2 −2 −6 . . .0 1 −3 −3 . . .0 0 1 −4 . . .

.. . .

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

0 1 4 18 . . .−3 1 2 12 . . .

1 −5 4 3 . . .0 1 −7 9 . . .

.. . .

−9 −18 −4 40644 . . .−14 −27 −36 40548 . . .

76 −106 47 20286 . . .−220 536 −898 7473 . . .

.. . .

ResultsConclusion

The Top Row

For all m, n ≥ 1 and for each 0 ≤ j ≤ 2m − 1,

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Row

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Row

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Row

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Row

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Row

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Row

Rnm(0, j) ≡ Rn(0, j) mod 22m−1.

1 −1 −12 −78 . . .−1 0 −14 −96 . . .19 −21 13 −39 . . .

−10 45 −85 76 . . .

.. . .

(1 11 0

1 7 4 27 0 2 03 7 1 56 1 3 4

−2 −11 −42 −156 . . .1 −13 −52 −216 . . .

−40 36 −74 −183 . . .55 −165 261 −335 . . .

.. . .

(0 11 1

6 5 6 41 3 4 04 0 6 53 3 5 5

ResultsConclusion

The Top Left Entry

Theorem

For all n,m ∈ N,

B̃(n) ≡ Rnm(0, 0)(mod 22m−1)

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267...

22 −323 1422 −188425 −301 1124 −1008

−28 −96 382 −24359 −205 373 −283

−2 ≡ 22 ≡ 6 mod 8

46203 −112360 161308 −13968631762 −66157 80710 −76050

9756 −18293 24253 −3675010181 −20787 33462 −30421

267 ≡ 46203 ≡ 3 mod 8

ResultsConclusion

The Top Left Entry

Theorem

For all n,m ∈ N,

B̃(n) ≡ Rnm(0, 0)(mod 22m−1)

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267...

22 −323 1422 −188425 −301 1124 −1008

−28 −96 382 −24359 −205 373 −283

−2 ≡ 22 ≡ 6 mod 8

46203 −112360 161308 −13968631762 −66157 80710 −76050

9756 −18293 24253 −3675010181 −20787 33462 −30421

267 ≡ 46203 ≡ 3 mod 8

ResultsConclusion

The Top Left Entry

Theorem

For all n,m ∈ N,

B̃(n) ≡ Rnm(0, 0)(mod 22m−1)

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267...

22 −323 1422 −188425 −301 1124 −1008

−28 −96 382 −24359 −205 373 −283

−2 ≡ 22 ≡ 6 mod 8

46203 −112360 161308 −13968631762 −66157 80710 −76050

9756 −18293 24253 −3675010181 −20787 33462 −30421

267 ≡ 46203 ≡ 3 mod 8

ResultsConclusion

The Top Left Entry

Theorem

For all n,m ∈ N,

B̃(n) ≡ Rnm(0, 0)(mod 22m−1)

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267...

22 −323 1422 −188425 −301 1124 −1008

−28 −96 382 −24359 −205 373 −283

−2 ≡ 22 ≡ 6 mod 8

46203 −112360 161308 −13968631762 −66157 80710 −76050

9756 −18293 24253 −3675010181 −20787 33462 −30421

267 ≡ 46203 ≡ 3 mod 8

ResultsConclusion

The Top Left Entry

Theorem

For all n,m ∈ N,

B̃(n) ≡ Rnm(0, 0)(mod 22m−1)

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267...

22 −323 1422 −188425 −301 1124 −1008

−28 −96 382 −24359 −205 373 −283

−2 ≡ 22 ≡ 6 mod 8

46203 −112360 161308 −13968631762 −66157 80710 −76050

9756 −18293 24253 −3675010181 −20787 33462 −30421

267 ≡ 46203 ≡ 3 mod 8

ResultsConclusion

The Top Left Entry

Theorem

For all n,m ∈ N,

B̃(n) ≡ Rnm(0, 0)(mod 22m−1)

n B̃(n)0 11 −12 03 14 15 −26 −97 −98 509 267...

22 −323 1422 −188425 −301 1124 −1008

−28 −96 382 −24359 −205 373 −283

−2 ≡ 22 ≡ 6 mod 8

46203 −112360 161308 −13968631762 −66157 80710 −76050

9756 −18293 24253 −3675010181 −20787 33462 −30421

267 ≡ 46203 ≡ 3 mod 8

ResultsConclusion

ConclusionAcknowledgementsWorks Cited

Conclusion

In Conclusion:

Additional Results

Alternate Bases

ResultsConclusion

Conclusion

In Conclusion:

Additional Results

Alternate Bases

ResultsConclusion

Conclusion

In Conclusion:

Additional Results

Alternate Bases

ResultsConclusion

Acknowledgements

We would like to thank LSU for hosting the SMILE Program.Thank you NSF for funding the VIGRE program. Thank you to Dr.De Angelis for spending his summer with us. Thank you to SimonPfeil for mentoring us.

ResultsConclusion

Works Cited

T. Amdeberhan, V. De Angelis, and V.H. Moll.Complementary Bell Numbers: Arithmetical Properties andWilf’s Conjecture. 2011.

http://www-history.mcs.st-and.ac.uk/Miscellaneous/StirlingBell/stirling2.html

the complementary bell numbers - lsu math

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