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Experimental Approach to the Hankel Transform of Catalan Number - - PowerPoint PPT Presentation

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Experimental Approach to the Hankel Transform of Catalan Number Combinations Wenyang Qian qianweny@grinnell.edu Department of Mathematics and Statistics, Grinnell College Oct 4, 2010 Wenyang Qian (Grinnell College) Hankel Transform of

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Experimental Approach to the Hankel Transform of Catalan Number Combinations

Wenyang Qian qianweny@grinnell.edu

Department of Mathematics and Statistics, Grinnell College

Oct 4, 2010

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 1 / 40

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Brief

1 Introductions 2 Experiments 3 Conjectures Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 2 / 40

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Review: Fibonacci Numbers

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 3 / 40

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Review: Fibonacci Numbers

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 3 / 40

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Review: Fibonacci Numbers

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...

More precisely, F0 = 0, F1 = 1; Fn = Fn−1 + Fn−2, n 2.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 3 / 40

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Review: Recurrence Relation

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 4 / 40

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Review: Recurrence Relation

By definition, a recurrence relation is an equation that recursively defines a sequence: each term of the sequence is defined as a function of the preceding terms.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 4 / 40

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Review: Recurrence Relation

By definition, a recurrence relation is an equation that recursively defines a sequence: each term of the sequence is defined as a function of the preceding terms. For instance, a recurrence relation for the sequence X = {x2, x4, x6, x8, ...} can be defined as xn = x2xn−1.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 4 / 40

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Catalan Numbers

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 5 / 40

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Catalan Numbers

1, 1, 2, 5, 14, 42, 132, 429, ...

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 5 / 40

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Catalan Numbers

1, 1, 2, 5, 14, 42, 132, 429, ...

By definition, C0 = 1; Cn+1 =

n

  • i=0

CiCn−i = C0Cn + C1Cn−1 + · · · + CnC0, n ≥ 1.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 5 / 40

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Examples in Combinatorics

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 6 / 40

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Examples in Combinatorics

Monotonic Path:

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 6 / 40

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Examples in Combinatorics

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 7 / 40

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Examples in Combinatorics

Convex Polygon:

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 7 / 40

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Review: Determinants

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 8 / 40

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Review: Determinants

The determinant is a number associated with a square matrix, we can use co-factor expansion to find the determinant of a matrix.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 8 / 40

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Review: Determinants

The determinant is a number associated with a square matrix, we can use co-factor expansion to find the determinant of a matrix.

  • a

b c d e f g h i

  • =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 8 / 40

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Review: Determinants

The determinant is a number associated with a square matrix, we can use co-factor expansion to find the determinant of a matrix.

  • a

b c d e f g h i

  • = a.
  • e

f h i

  • Wenyang Qian (Grinnell College)

Hankel Transform of Catalan Numbers Oct 4, 2010 8 / 40

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Review: Determinants

The determinant is a number associated with a square matrix, we can use co-factor expansion to find the determinant of a matrix.

  • a

b c d e f g h i

  • = a.
  • e

f h i

  • − b.
  • d

f g i

  • Wenyang Qian (Grinnell College)

Hankel Transform of Catalan Numbers Oct 4, 2010 8 / 40

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Review: Determinants

The determinant is a number associated with a square matrix, we can use co-factor expansion to find the determinant of a matrix.

  • a

b c d e f g h i

  • = a.
  • e

f h i

  • − b.
  • d

f g i

  • + c.
  • d

e g h

  • Wenyang Qian (Grinnell College)

Hankel Transform of Catalan Numbers Oct 4, 2010 8 / 40

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Hankel Matrix

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 9 / 40

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Hankel Matrix

Given an integer sequence A = {a0, a1, a2, a3, a4, ...}, the Hankel matrix H can be defined as an infinite square matrix,

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 9 / 40

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Hankel Matrix

Given an integer sequence A = {a0, a1, a2, a3, a4, ...}, the Hankel matrix H can be defined as an infinite square matrix, H =        a0 a1 a2 a3 · · · a1 a2 a3 a4 · · · a2 a3 a4 a5 · · · a3 a4 a5 a6 · · · . . . . . . . . . . . . ...        .

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 9 / 40

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The Hankel Determinant of Order n

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 10 / 40

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The Hankel Determinant of Order n

The Hankel determinant of order n is the the determinant of the upper-left n by n submatrix of H.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 10 / 40

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The Hankel Determinant of Order n

The Hankel determinant of order n is the the determinant of the upper-left n by n submatrix of H. hn =

  • a0

a1 a2 a3 · · · an−1 a1 a2 a3 a4 · · · an a2 a3 a4 a5 · · · an+1 a3 a4 a5 a6 · · · an+2 . . . . . . . . . . . . ... . . . an−1 an an+1 an+2 · · · a2n−1

  • Wenyang Qian (Grinnell College)

Hankel Transform of Catalan Numbers Oct 4, 2010 10 / 40

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Hankel Transform

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 11 / 40

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Hankel Transform

The Hankel transform is a sequence transform that returns successive Hankel determinants formed by the original sequence.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 11 / 40

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Hankel Transform

The Hankel transform is a sequence transform that returns successive Hankel determinants formed by the original sequence. H{A} = {h1, h2, h3, h4, ...}where,

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 11 / 40

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Hankel Transform

The Hankel transform is a sequence transform that returns successive Hankel determinants formed by the original sequence. H{A} = {h1, h2, h3, h4, ...}where, h1 =

  • a0
  • = a0;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 11 / 40

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Hankel Transform

The Hankel transform is a sequence transform that returns successive Hankel determinants formed by the original sequence. H{A} = {h1, h2, h3, h4, ...}where, h1 =

  • a0
  • = a0;

h2 =

  • a0

a1 a1 a2

  • = a0a2 − a2

1;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 11 / 40

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Hankel Transform

The Hankel transform is a sequence transform that returns successive Hankel determinants formed by the original sequence. H{A} = {h1, h2, h3, h4, ...}where, h1 =

  • a0
  • = a0;

h2 =

  • a0

a1 a1 a2

  • = a0a2 − a2

1;

h3 =

  • a0

a1 a2 a1 a2 a3 a2 a3 a4

  • = a0a2a4 + 2a1a2a3 − a0a2

3 − a2 1a4 − a3 2;

· · · · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 11 / 40

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Early Work

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002. For example, the Hankel transform of the original Catalan numbers Cn:

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002. For example, the Hankel transform of the original Catalan numbers Cn:

  • 1
  • = 1;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002. For example, the Hankel transform of the original Catalan numbers Cn:

  • 1
  • = 1;
  • 1

1 1 2

  • = 1;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002. For example, the Hankel transform of the original Catalan numbers Cn:

  • 1
  • = 1;
  • 1

1 1 2

  • = 1;
  • 1

1 2 1 2 5 2 5 14

  • = 1;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002. For example, the Hankel transform of the original Catalan numbers Cn:

  • 1
  • = 1;
  • 1

1 1 2

  • = 1;
  • 1

1 2 1 2 5 2 5 14

  • = 1;
  • 1

1 2 5 1 2 5 14 2 5 14 42 5 14 42 132

  • = 1; · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Layman discovered intriguing properties of Catalan numbers when blended with Hankel transform in 2001. Afterwards, Cvetkovic proved them in 2002. For example, the Hankel transform of the original Catalan numbers Cn:

  • 1
  • = 1;
  • 1

1 1 2

  • = 1;
  • 1

1 2 1 2 5 2 5 14

  • = 1;
  • 1

1 2 5 1 2 5 14 2 5 14 42 5 14 42 132

  • = 1; · · ·

H{Cn} = {1, 1, 1, · · · }

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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Early Work

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;
  • 2

3 7 3 7 19 7 19 56

  • =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;
  • 2

3 7 3 7 19 7 19 56

  • = 13;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;
  • 2

3 7 3 7 19 7 19 56

  • = 13;
  • 2

3 7 19 3 7 19 56 7 19 56 174 19 56 174 561

  • =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;
  • 2

3 7 3 7 19 7 19 56

  • = 13;
  • 2

3 7 19 3 7 19 56 7 19 56 174 19 56 174 561

  • = 34; · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;
  • 2

3 7 3 7 19 7 19 56

  • = 13;
  • 2

3 7 19 3 7 19 56 7 19 56 174 19 56 174 561

  • = 34; · · ·

H{Cn + Cn+1} = {2, 5, 13, 34, · · · }

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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Early Work

Another example, the Hankel transform of the sum of adjacent Catalan numbers, Cn + Cn+1 = {1, 1, 2, 5, 14, 42, 132, ...} + {1, 2, 5, 14, 42, 132, 429, ...} = {2, 3, 7, 19, 56, 174, 561, ...}:

  • 2
  • = 2;
  • 2

3 3 7

  • = 5;
  • 2

3 7 3 7 19 7 19 56

  • = 13;
  • 2

3 7 19 3 7 19 56 7 19 56 174 19 56 174 561

  • = 34; · · ·

H{Cn + Cn+1} = {2, 5, 13, 34, · · · } H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

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H{Cn + Cn+1}, a recurrence relation?

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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H{Cn + Cn+1}, a recurrence relation?

H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3 Pick any element, say 13, observe that

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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H{Cn + Cn+1}, a recurrence relation?

H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3 Pick any element, say 13, observe that 13 = 5 + (8)

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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H{Cn + Cn+1}, a recurrence relation?

H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3 Pick any element, say 13, observe that 13 = 5 + (8) 13 = 5 + 5 + (3)

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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H{Cn + Cn+1}, a recurrence relation?

H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3 Pick any element, say 13, observe that 13 = 5 + (8) 13 = 5 + 5 + (3) 13 = 5 ∗ 2 + (5 − 2)

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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H{Cn + Cn+1}, a recurrence relation?

H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3 Pick any element, say 13, observe that 13 = 5 + (8) 13 = 5 + 5 + (3) 13 = 5 ∗ 2 + (5 − 2) 13 = 5 ∗ 3 − 2

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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H{Cn + Cn+1}, a recurrence relation?

H{Cn + Cn+1} = {(0), (1), (1), 2, (3), 5, (8), 13, (21), 34, ...} = F2n+3 Pick any element, say 13, observe that 13 = 5 + (8) 13 = 5 + 5 + (3) 13 = 5 ∗ 2 + (5 − 2) 13 = 5 ∗ 3 − 2

Conjecture

The recurrence relation for H{Cn + Cn+1} is, hn+1 = 3hn − hn−1.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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Goals

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 15 / 40

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Goals

We want to classify the properties of all Hankel transforms of linear combinations of the adjacent Catalan numbers.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 15 / 40

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Goals

We want to classify the properties of all Hankel transforms of linear combinations of the adjacent Catalan numbers. Today, we focus on the conjectures/experimental part of the project.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 15 / 40

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SLIDE 63

A simplest case: aCn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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A simplest case: aCn

aCn = {1a, 1a, 2a, 5a, 14a, 42a, 132a, ...}, such that

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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SLIDE 65

A simplest case: aCn

aCn = {1a, 1a, 2a, 5a, 14a, 42a, 132a, ...}, such that

  • 1a
  • = 1a;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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SLIDE 66

A simplest case: aCn

aCn = {1a, 1a, 2a, 5a, 14a, 42a, 132a, ...}, such that

  • 1a
  • = 1a;
  • 1a

1a 1a 2a

  • = 1a2;

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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SLIDE 67

A simplest case: aCn

aCn = {1a, 1a, 2a, 5a, 14a, 42a, 132a, ...}, such that

  • 1a
  • = 1a;
  • 1a

1a 1a 2a

  • = 1a2;
  • 1a

1a 2a 1a 2a 5a 2a 5a 14a

  • = 1a3; · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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SLIDE 68

A simplest case: aCn

aCn = {1a, 1a, 2a, 5a, 14a, 42a, 132a, ...}, such that

  • 1a
  • = 1a;
  • 1a

1a 1a 2a

  • = 1a2;
  • 1a

1a 2a 1a 2a 5a 2a 5a 14a

  • = 1a3; · · ·

H{aCn} = {a, a2, a3, a4, · · · }

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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SLIDE 69

A simplest case: aCn

aCn = {1a, 1a, 2a, 5a, 14a, 42a, 132a, ...}, such that

  • 1a
  • = 1a;
  • 1a

1a 1a 2a

  • = 1a2;
  • 1a

1a 2a 1a 2a 5a 2a 5a 14a

  • = 1a3; · · ·

H{aCn} = {a, a2, a3, a4, · · · }

Conjecture

The recurrence relation for H{aCn} is hn+1 = ahn.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

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SLIDE 70

aCn + bCn+1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

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aCn + bCn+1

Basic Methods?

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

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aCn + bCn+1

Basic Methods? List and Compare!

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

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SLIDE 73

aCn + bCn+1

Basic Methods? List and Compare! Tool: Wolfram Mathematica 7.0

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

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SLIDE 74

aCn + bCn+1 Continue[a]

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

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aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

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aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

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SLIDE 77

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-78
SLIDE 78

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-79
SLIDE 79

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-80
SLIDE 80

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-81
SLIDE 81

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-82
SLIDE 82

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-83
SLIDE 83

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-84
SLIDE 84

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-85
SLIDE 85

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-86
SLIDE 86

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1 H{3Cn + Cn+1} : hn+1 =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-87
SLIDE 87

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1 H{3Cn + Cn+1} : hn+1 = 5hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-88
SLIDE 88

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1 H{3Cn + Cn+1} : hn+1 = 5hn − hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-89
SLIDE 89

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1 H{3Cn + Cn+1} : hn+1 = 5hn − hn−1 H{4Cn + Cn+1} : hn+1 =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-90
SLIDE 90

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1 H{3Cn + Cn+1} : hn+1 = 5hn − hn−1 H{4Cn + Cn+1} : hn+1 = 6hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-91
SLIDE 91

aCn + bCn+1 Continue[a]

H{0Cn + Cn+1} = {1, 1, 1, 1, ...} H{1Cn + Cn+1} = {2, 5, 13, 34, ...} H{2Cn + Cn+1} = {3, 11, 41, 153, ...} H{3Cn + Cn+1} = {4, 19, 91, 436, ...} H{4Cn + Cn+1} = {5, 29, 169, 985, ...} · · · · · · H{0Cn + Cn+1} : hn+1 = hn H{1Cn + Cn+1} : hn+1 = 3hn − hn−1 H{2Cn + Cn+1} : hn+1 = 4hn − hn−1 H{3Cn + Cn+1} : hn+1 = 5hn − hn−1 H{4Cn + Cn+1} : hn+1 = 6hn − hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

slide-92
SLIDE 92

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-93
SLIDE 93

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-94
SLIDE 94

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-95
SLIDE 95

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-96
SLIDE 96

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-97
SLIDE 97

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-98
SLIDE 98

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-99
SLIDE 99

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-100
SLIDE 100

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-101
SLIDE 101

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-102
SLIDE 102

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-103
SLIDE 103

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1 H{Cn + 3Cn+1} : hn+1 =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-104
SLIDE 104

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1 H{Cn + 3Cn+1} : hn+1 = 7hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-105
SLIDE 105

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1 H{Cn + 3Cn+1} : hn+1 = 7hn − 9hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-106
SLIDE 106

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1 H{Cn + 3Cn+1} : hn+1 = 7hn − 9hn−1 H{Cn + 4Cn+1} : hn+1 =

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-107
SLIDE 107

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1 H{Cn + 3Cn+1} : hn+1 = 7hn − 9hn−1 H{Cn + 4Cn+1} : hn+1 = 9hn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-108
SLIDE 108

aCn + bCn+1 Continue[b]

H{Cn + 0Cn+1} = {1, 1, 1, 1, ...} H{Cn + 1Cn+1} = {2, 5, 13, 34, ...} H{Cn + 2Cn+1} = {3, 11, 43, 171, ...} H{Cn + 3Cn+1} = {4, 19, 97, 508, ...} H{Cn + 4Cn+1} = {5, 29, 181, 1165, ...} · · · · · · H{Cn + 0Cn+1} : hn+1 = hn H{Cn + 1Cn+1} : hn+1 = 3hn − hn−1 H{Cn + 2Cn+1} : hn+1 = 5hn − 4hn−1 H{Cn + 3Cn+1} : hn+1 = 7hn − 9hn−1 H{Cn + 4Cn+1} : hn+1 = 9hn − 16hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

slide-109
SLIDE 109

H{aCn + bCn+1} Comparison Table

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 20 / 40

slide-110
SLIDE 110

H{aCn + bCn+1} Comparison Table

Simply notation: Rewrite the recurrence relation as a coefficient column vector and a variable row vector whose product is 0.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 20 / 40

slide-111
SLIDE 111

H{aCn + bCn+1} Comparison Table

Simply notation: Rewrite the recurrence relation as a coefficient column vector and a variable row vector whose product is 0. x0hn+1 + x1hn + x2hn−1 = 0

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 20 / 40

slide-112
SLIDE 112

H{aCn + bCn+1} Comparison Table

Simply notation: Rewrite the recurrence relation as a coefficient column vector and a variable row vector whose product is 0. x0hn+1 + x1hn + x2hn−1 = 0 ⇓

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 20 / 40

slide-113
SLIDE 113

H{aCn + bCn+1} Comparison Table

Simply notation: Rewrite the recurrence relation as a coefficient column vector and a variable row vector whose product is 0. x0hn+1 + x1hn + x2hn−1 = 0 ⇓

  • hn+1

hn hn−1

  • .

  x0 x1 x2   = 0

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 20 / 40

slide-114
SLIDE 114

H{aCn + bCn+1} Comparison Table

Simply notation: Rewrite the recurrence relation as a coefficient column vector and a variable row vector whose product is 0. x0hn+1 + x1hn + x2hn−1 = 0 ⇓

  • hn+1

hn hn−1

  • .

  x0 x1 x2   = 0 ±

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 20 / 40

slide-115
SLIDE 115

H{aCn + bCn+1} Compare the coefficient vectors

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 21 / 40

slide-116
SLIDE 116

H{aCn + bCn+1} Compare the coefficient vectors

a,b b=1 b=2 b=3 a=1   1 −3 1     1 −5 4     1 −7 9  

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 21 / 40

slide-117
SLIDE 117

H{aCn + bCn+1} Compare the coefficient vectors

a,b b=1 b=2 b=3 a=1   1 −3 1     1 −5 4     1 −7 9   a=2   1 −4 1     1 −6 4     1 −8 9  

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 21 / 40

slide-118
SLIDE 118

H{aCn + bCn+1} Compare the coefficient vectors

a,b b=1 b=2 b=3 a=1   1 −3 1     1 −5 4     1 −7 9   a=2   1 −4 1     1 −6 4     1 −8 9   a=3   1 −5 1     1 −7 4     1 −9 9  

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 21 / 40

slide-119
SLIDE 119

H{aCn + bCn+1} Compare the coefficient vectors

a,b b=1 b=2 b=3 a=1   1 −3 1     1 −5 4     1 −7 9   a=2   1 −4 1     1 −6 4     1 −8 9   a=3   1 −5 1     1 −7 4     1 −9 9   Conjecture of the Coefficient Column Vector:

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 21 / 40

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SLIDE 120

H{aCn + bCn+1} Compare the coefficient vectors

a,b b=1 b=2 b=3 a=1   1 −3 1     1 −5 4     1 −7 9   a=2   1 −4 1     1 −6 4     1 −8 9   a=3   1 −5 1     1 −7 4     1 −9 9   Conjecture of the Coefficient Column Vector:   1 −a − 2b b2  

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 21 / 40

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SLIDE 121

H{aCn + bCn+1} Generalization

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 22 / 40

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SLIDE 122

H{aCn + bCn+1} Generalization

Conjecture

The general recurrence relation for H{aCn + bCn+1} is, hn+1 = (a + 2b)hn − b2hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 22 / 40

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SLIDE 123

H{aCn + bCn+1} Generalization

Conjecture

The general recurrence relation for H{aCn + bCn+1} is, hn+1 = (a + 2b)hn − b2hn−1 But, is this method efficient?

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 22 / 40

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SLIDE 124

Review: Null Space

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 23 / 40

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SLIDE 125

Review: Null Space

By definition, the null space or the kernel of a n by n square matrix M is the set of all the n dimensional vectors x for which M x = 0.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 23 / 40

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SLIDE 126

Review: Null Space

By definition, the null space or the kernel of a n by n square matrix M is the set of all the n dimensional vectors x for which M x = 0. Think: What happens if we replace M for the Hankel matrix of our desired sequence?

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 23 / 40

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SLIDE 127

General Methods

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 24 / 40

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SLIDE 128

General Methods

Suppose we want to know the recurrence relations of the sequence {h1, h2, h3, ..., hn}, by the definition of null-space,

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 24 / 40

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SLIDE 129

General Methods

Suppose we want to know the recurrence relations of the sequence {h1, h2, h3, ..., hn}, by the definition of null-space,        h1 h2 h3 · · · hn h2 h3 h4 · · · hn+1 h3 h4 h5 · · · hn+2 . . . . . . . . . ... . . . hn hn+1 hn+2 · · · h2n             x0 x1 . . . xn−1      =      . . .      .

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 24 / 40

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SLIDE 130

General Methods

Suppose we want to know the recurrence relations of the sequence {h1, h2, h3, ..., hn}, by the definition of null-space,        h1 h2 h3 · · · hn h2 h3 h4 · · · hn+1 h3 h4 h5 · · · hn+2 . . . . . . . . . ... . . . hn hn+1 hn+2 · · · h2n             x0 x1 . . . xn−1      =      . . .      . ↓

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 24 / 40

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SLIDE 131

General Methods

Suppose we want to know the recurrence relations of the sequence {h1, h2, h3, ..., hn}, by the definition of null-space,        h1 h2 h3 · · · hn h2 h3 h4 · · · hn+1 h3 h4 h5 · · · hn+2 . . . . . . . . . ... . . . hn hn+1 hn+2 · · · h2n             x0 x1 . . . xn−1      =      . . .      . ↓

  • h1

h2 h3 · · · hn

    x0 x1 . . . xn−1      = 0.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 24 / 40

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SLIDE 132

General Methods

Suppose we want to know the recurrence relations of the sequence {h1, h2, h3, ..., hn}, by the definition of null-space,        h1 h2 h3 · · · hn h2 h3 h4 · · · hn+1 h3 h4 h5 · · · hn+2 . . . . . . . . . ... . . . hn hn+1 hn+2 · · · h2n             x0 x1 . . . xn−1      =      . . .      . ↓

  • h1

h2 h3 · · · hn

    x0 x1 . . . xn−1      = 0. The null-space of the Hankel matrix is the coefficient column vector.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 24 / 40

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SLIDE 133

General Methods

Suppose we want to know the recurrence relations of the sequence {h1, h2, h3, ..., hn}, by the definition of null-space,        h1 h2 h3 · · · hn h2 h3 h4 · · · hn+1 h3 h4 h5 · · · hn+2 . . . . . . . . . ... . . . hn hn+1 hn+2 · · · h2n             x0 x1 . . . xn−1      =      . . .      . ↓

  • h1

h2 h3 · · · hn

    x0 x1 . . . xn−1      = 0. The null-space of the Hankel matrix is the Recurrence Relations.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 25 / 40

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SLIDE 134

Flow Chart

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 26 / 40

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SLIDE 135

Flow Chart

1 Construct the sequence of linear combinations of Catalan numbers,

aCn + bCn+1 + cCn+2...

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 26 / 40

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SLIDE 136

Flow Chart

1 Construct the sequence of linear combinations of Catalan numbers,

aCn + bCn+1 + cCn+2...

2 Compute the Hankel transform of the sequence, associated with

values of coefficients.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 26 / 40

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SLIDE 137

Flow Chart

1 Construct the sequence of linear combinations of Catalan numbers,

aCn + bCn+1 + cCn+2...

2 Compute the Hankel transform of the sequence, associated with

values of coefficients.

3 Find* the recurrence relations (coefficient column vector) by

calculating the null-space of the Hankel matrix of the Hankel transform.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 26 / 40

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SLIDE 138

Flow Chart

1 Construct the sequence of linear combinations of Catalan numbers,

aCn + bCn+1 + cCn+2...

2 Compute the Hankel transform of the sequence, associated with

values of coefficients.

3 Find* the recurrence relations (coefficient column vector) by

calculating the null-space of the Hankel matrix of the Hankel transform.

4 Compare* coefficient column vectors associated to each linear

combinations and conjecture the general form of the recurrence relations.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 26 / 40

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SLIDE 139

Flow Chart

1 Construct the sequence of linear combinations of Catalan numbers,

aCn + bCn+1 + cCn+2...

2 Compute the Hankel transform of the sequence, associated with

values of coefficients.

3 Find* the recurrence relations (coefficient column vector) by

calculating the null-space of the Hankel matrix of the Hankel transform.

4 Compare* coefficient column vectors associated to each linear

combinations and conjecture the general form of the recurrence relations. Advantage/Disadvantage...

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 26 / 40

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SLIDE 140

Conjectures: H{aCn}, H{aCn + bCn+1}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 27 / 40

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SLIDE 141

Conjectures: H{aCn}, H{aCn + bCn+1}

Conjecture

H{aCn} satisfies the recurrence relation, A0 ∗ hn+1 + A1 ∗ hn = 0, where A0 = 1 A1 = −a

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 27 / 40

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SLIDE 142

Conjectures: H{aCn}, H{aCn + bCn+1}

Conjecture

H{aCn} satisfies the recurrence relation, A0 ∗ hn+1 + A1 ∗ hn = 0, where A0 = 1 A1 = −a

Conjecture

H{aCn + bCn+1} satisfies the recurrence relation, B0 ∗ hn+1 + B1 ∗ hn + B2 ∗ hn−1 = 0, where B0 = 1 B1 = −a − 2b B2 = b2

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 27 / 40

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SLIDE 143

Conjectures: H{aCn + bCn+1 + cCn+2}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 28 / 40

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SLIDE 144

Conjectures: H{aCn + bCn+1 + cCn+2}

Conjecture

H{aCn + bCn+1 + cCn+2} satisfies the recurrence relation, C0 ∗ hn+1 + C1 ∗ hn + C2 ∗ hn−1 + C3 ∗ hn−2 + C4 ∗ hn−3 = 0, where C0 = 1 C1 = −a − 2b − 4c C2 = b2 + c(−2a + 4b + 6c) C3 = c2(−a − 2b − 4c) C4 = c4

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 28 / 40

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SLIDE 145

H{aCn + bCn+1 + cCn+2 + dCn+3}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 29 / 40

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SLIDE 146

H{aCn + bCn+1 + cCn+2 + dCn+3}

Conjecture

H{aCn + bCn+1 + cCn+2 + dCn+3} satisfies the recurrence relation, D0 ∗ hn+1 + D1 ∗ hn + D2 ∗ hn−1 + · · · + D7 ∗ hn−6 + D8 ∗ hn−7 = 0,where

D0 = 1 D1 = −a − 2b − 4c − 8d D2 = b2 + c(−2a + 4b + 6c) + d(−12a + 4b + 24c + 28d) D3 = c2(−a − 2b − 4c) + d(2ab + 4b2 − 4ac − 24c2 − 7ad + 2bd − 60cd − 56d2) D4 = c4 + d(−4bc2 + 8c3) + d2(a2 + 4ab + 6b2 + 12ac − 8bc + 36c2) + d3(40a − 8b + 80c) + 70d4 D5 = d2(c2(−a − 2b − 4c) + d(2ab + 4b2 − 4ac − 24c2 − 7ad + 2bd − 60cd − 56d D6 = d4(b2 + c(−2a + 4b + 6c) + d(−12a + 4b + 24c + 28d)) D7 = d6(−a − 2b − 4c − 8d) D8 = d8

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 29 / 40

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SLIDE 147

H{aCn + bCn+1 + cCn+2 + dCn+3 + eCn+4}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 30 / 40

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SLIDE 148

H{aCn + bCn+1 + cCn+2 + dCn+3 + eCn+4}

Conjecture

H{aCn + bCn+1 + cCn+2 + dCn+3 + eCn+4} satisfies the recurrence relation, E0 ∗ hn+1 + E1 ∗ hn + E2 ∗ hn−1 + · · · + E15 ∗ hn−14 + E16 ∗ hn−15 = 0, E0 = 1 E1 = −a − 2b − 4c − 8d − 16e E2 = b2 + c(−2a + 4b + 6c) + d(−12a + 4b + 24c + 28d)+ e(−52a − 8b + 40c + 112d + 120e) ... = ... E14 = e12 ∗ (b2 + c(−2a + 4b + 6c) + d(−12a + 4b + 24c + 28d)+ e(−52a − 8b + 40c + 112d + 120e)) E15 = e14 ∗ (−a − 2b − 4c − 8d − 16e) E16 = e16

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 30 / 40

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SLIDE 149

More Results

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 31 / 40

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SLIDE 150

More Results

H{aCn + bCn+1 + cCn+2 + dCn+3 + eCn+4 + fCn+5}

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 31 / 40

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SLIDE 151

More Results

H{aCn + bCn+1 + cCn+2 + dCn+3 + eCn+4 + fCn+5} H{aCn + bCn+1 + cCn+2 + dCn+3 + eCn+4 + fCn+5 + gCn+6} · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 31 / 40

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SLIDE 152

More Conjectures: 1.Order

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 153

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 154

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 155

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 156

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1 2-term case: hn+1 = (a + 2b)hn − b2hn−1

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 157

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1 2-term case: hn+1 = (a + 2b)hn − b2hn−1 2

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 158

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1 2-term case: hn+1 = (a + 2b)hn − b2hn−1 2 3-term case: ... 4

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 159

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1 2-term case: hn+1 = (a + 2b)hn − b2hn−1 2 3-term case: ... 4 4-term case: ... 8

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 160

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1 2-term case: hn+1 = (a + 2b)hn − b2hn−1 2 3-term case: ... 4 4-term case: ... 8 5-term case: ... 16

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 161

More Conjectures: 1.Order

The order of the recurrence relations grows as a function of 2n−1. 1-term case: hn+1 = ahn 1 2-term case: hn+1 = (a + 2b)hn − b2hn−1 2 3-term case: ... 4 4-term case: ... 8 5-term case: ... 16 · · · · · ·

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 32 / 40

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SLIDE 162

More Conjectures: 2.Symmetry

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 33 / 40

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SLIDE 163

More Conjectures: 2.Symmetry

By observations, we realize that the formula of the coefficients for each term of the recurrence relations is symmetric about the center coefficient.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 33 / 40

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SLIDE 164

More Conjectures: 2.Symmetry

By observations, we realize that the formula of the coefficients for each term of the recurrence relations is symmetric about the center coefficient. For the particular case listed above, B2 = b2 ∗ B0, C3 = c2 ∗ C1, C4 = c4 ∗ C0, D6 = d4 ∗ D2, D7 = d6 ∗ D1 and D8 = d8 ∗ D0.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 33 / 40

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SLIDE 165

Symmetry of H{aCn + bCn+1 + cCn+2}

Conjecture

H{aCn + bCn+1 + cCn+2} satisfies the recurrence relation, C0 ∗ hn+1 + C1 ∗ hn + C2 ∗ hn−1 + C3 ∗ hn−2 + C4 ∗ hn−3 = 0, where C0 = 1 C1 = −a − 2b − 4c C2 = · · · C3 = c2(−a − 2b − 4c) C4 = c4

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 34 / 40

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SLIDE 166

Symmetry of H{aCn + bCn+1 + cCn+2 + dCn+3}

Conjecture

H{aCn + bCn+1 + cCn+2 + dCn+3} satisfies the recurrence relation, D0 ∗ hn+1 + D1 ∗ hn + D2 ∗ hn−1 + · · · + D7 ∗ hn−6 + D8 ∗ hn−7 = 0,where

D0 = 1 D1 = −a − 2b − 4c − 8d D2 = b2 + c(−2a + 4b + 6c) + d(−12a + 4b + 24c + 28d) D3 = c2(−a − 2b − 4c) + d(2ab + 4b2 − 4ac − 24c2 − 7ad + 2bd − 60cd − 56d2) D4 = · · · D5 = d2(c2(−a − 2b − 4c) + d(2ab + 4b2 − 4ac − 24c2 − 7ad + 2bd − 60cd − 56d D6 = d4(b2 + c(−2a + 4b + 6c) + d(−12a + 4b + 24c + 28d)) D7 = d6(−a − 2b − 4c − 8d) D8 = d8

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 35 / 40

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SLIDE 167

More Conjectures: 2.Symmetry

By observations, we realize that the formula of the coefficients for each term of the recurrence relations is symmetric about the center coefficient. For the particular case listed above, B2 = b2 ∗ B0, C3 = c2 ∗ C1, C4 = c4 ∗ C0, D6 = d4 ∗ D2, D7 = d6 ∗ D1 and D8 = d8 ∗ D0.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 36 / 40

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SLIDE 168

More Conjectures: 2.Symmetry

By observations, we realize that the formula of the coefficients for each term of the recurrence relations is symmetric about the center coefficient. For the particular case listed above, B2 = b2 ∗ B0, C3 = c2 ∗ C1, C4 = c4 ∗ C0, D6 = d4 ∗ D2, D7 = d6 ∗ D1 and D8 = d8 ∗ D0.

Conjecture

Tthe recurrence relations for q terms of adjacent Catalan numbers must satisfy the relation, Qn+2m = q2m ∗ Qn.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 36 / 40

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SLIDE 169

Next Step: Proof

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 170

Next Step: Proof

Conjectures ⇒ Theorems

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 171

Next Step: Proof

Conjectures ⇒ Theorems ⇑

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 172

Next Step: Proof

Conjectures ⇒ Theorems ⇑

1 Cofactor Expansion Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 173

Next Step: Proof

Conjectures ⇒ Theorems ⇑

1 Cofactor Expansion 2 Basic properties of Determinants Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 174

Next Step: Proof

Conjectures ⇒ Theorems ⇑

1 Cofactor Expansion 2 Basic properties of Determinants 3 Special property Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 175

Next Step: Proof

Conjectures ⇒ Theorems ⇑

1 Cofactor Expansion 2 Basic properties of Determinants 3 Special property 4 Matrix Characteristic Polynomials* Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 37 / 40

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SLIDE 176

Reference

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 38 / 40

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SLIDE 177

Reference

  • 1. J. W. Layman, “The Hankel Transform and Some of its Properties”, Journal of

Integer Sequences, Article 01.1.5, Volume 4, 2001.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 38 / 40

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SLIDE 178

Reference

  • 1. J. W. Layman, “The Hankel Transform and Some of its Properties”, Journal of

Integer Sequences, Article 01.1.5, Volume 4, 2001.

  • 2. Marc Chamberland and Christopher French, “Generalized Catalan Numbers

and Generalized Hankel Transformations”, Journal of Integer Sequences, Article 07.1.1, Volume 10, 2007.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 38 / 40

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SLIDE 179

Reference

  • 1. J. W. Layman, “The Hankel Transform and Some of its Properties”, Journal of

Integer Sequences, Article 01.1.5, Volume 4, 2001.

  • 2. Marc Chamberland and Christopher French, “Generalized Catalan Numbers

and Generalized Hankel Transformations”, Journal of Integer Sequences, Article 07.1.1, Volume 10, 2007.

  • 3. A. Cvetkovic, P. Rajkovic, and M. Ivkovic, “Catalan Numbers, the Hankel

Transform, and Fibonacci Numbers”, Journal of Integer Sequences, Article 02.1.3, Volume 5, 2002.

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 38 / 40

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SLIDE 180

Acknowledgements

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 39 / 40

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SLIDE 181

Acknowledgements

Grinnell College

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 39 / 40

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SLIDE 182

Acknowledgements

Grinnell College Michael Dougherty, Wenyang Qian and Ben Saderholm

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 39 / 40

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SLIDE 183

Acknowledgements

Grinnell College Michael Dougherty, Wenyang Qian and Ben Saderholm Professor Christopher French

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 39 / 40

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SLIDE 184

Acknowledgements

Grinnell College Michael Dougherty, Wenyang Qian and Ben Saderholm Professor Christopher French Mathematics Department People: Ben, Colin, Cyrus, Klevi, Fatemeh, Michael, Solomon...

Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 39 / 40

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SLIDE 185

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Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 40 / 40