experimental approach to the hankel transform of catalan
play

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

Nov 17, 2022 •253 likes •2.12k views

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

  1. 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 C n : � = 1; � � � 1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

  2. 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 C n : � � 1 1 � = 1; � � � � � 1 � = 1; � � 1 2 � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

  3. 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 C n : � � 1 1 2 � � � � 1 1 � = 1; � � � � � � � 1 � = 1; 1 2 5 = 1; � � � � 1 2 � � � 2 5 14 � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

  4. 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 C n : � � 1 1 2 5 � � � � 1 1 2 � � � � � � 1 1 1 2 5 14 � = 1; � � � � � � � � � 1 � = 1; 1 2 5 = 1; = 1; · · · � � � � � � 1 2 2 5 14 42 � � � � � 2 5 14 � � � � 5 14 42 132 � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

  5. 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 C n : � � 1 1 2 5 � � � � 1 1 2 � � � � � � 1 1 1 2 5 14 � = 1; � � � � � � � � � 1 � = 1; 1 2 5 = 1; = 1; · · · � � � � � � 1 2 2 5 14 42 � � � � � 2 5 14 � � � � 5 14 42 132 � � H { C n } = { 1 , 1 , 1 , · · · } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 12 / 40

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

  7. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +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

  8. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +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

  9. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 � = 2; � � � � � 2 � = � � 3 7 � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  10. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 � = 2; � � � � � 2 � = 5; � � 3 7 � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  11. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 7 � � � � 2 3 � = 2; � � � � � � � 2 � = 5; 3 7 19 = � � � � 3 7 � � � 7 19 56 � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  12. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 7 � � � � 2 3 � = 2; � � � � � � � 2 � = 5; 3 7 19 = 13; � � � � 3 7 � � � 7 19 56 � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  13. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 7 19 � � � � 2 3 7 � � � � � � 2 3 3 7 19 56 � = 2; � � � � � � � � � 2 � = 5; 3 7 19 = 13; = � � � � � � 3 7 7 19 56 174 � � � � � 7 19 56 � � � � 19 56 174 561 � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  14. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 7 19 � � � � 2 3 7 � � � � � � 2 3 3 7 19 56 � = 2; � � � � � � � � � 2 � = 5; 3 7 19 = 13; = 34; · · · � � � � � � 3 7 7 19 56 174 � � � � � 7 19 56 � � � � 19 56 174 561 � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  15. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 7 19 � � � � 2 3 7 � � � � � � 2 3 3 7 19 56 � = 2; � � � � � � � � � 2 � = 5; 3 7 19 = 13; = 34; · · · � � � � � � 3 7 7 19 56 174 � � � � � 7 19 56 � � � � 19 56 174 561 � � H { C n + C n +1 } = { 2 , 5 , 13 , 34 , · · · } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  16. Early Work Another example, the Hankel transform of the sum of adjacent Catalan numbers, C n + C n +1 = { 1 , 1 , 2 , 5 , 14 , 42 , 132 , ... } + { 1 , 2 , 5 , 14 , 42 , 132 , 429 , ... } = { 2 , 3 , 7 , 19 , 56 , 174 , 561 , ... } : � � 2 3 7 19 � � � � 2 3 7 � � � � � � 2 3 3 7 19 56 � = 2; � � � � � � � � � 2 � = 5; 3 7 19 = 13; = 34; · · · � � � � � � 3 7 7 19 56 174 � � � � � 7 19 56 � � � � 19 56 174 561 � � H { C n + C n +1 } = { 2 , 5 , 13 , 34 , · · · } H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +3 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 13 / 40

  17. H { C n + C n +1 } , a recurrence relation? Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

  18. H { C n + C n +1 } , a recurrence relation? H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +3 Pick any element, say 13, observe that Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

  19. H { C n + C n +1 } , a recurrence relation? H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +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

  20. H { C n + C n +1 } , a recurrence relation? H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +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

  21. H { C n + C n +1 } , a recurrence relation? H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +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

  22. H { C n + C n +1 } , a recurrence relation? H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +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

  23. H { C n + C n +1 } , a recurrence relation? H { C n + C n +1 } = { (0) , (1) , (1) , 2 , (3) , 5 , (8) , 13 , (21) , 34 , ... } = F 2 n +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 { C n + C n +1 } is, h n +1 = 3 h n − h n − 1 . Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 14 / 40

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

  25. 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

  26. 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

  27. A simplest case: aC n Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  28. A simplest case: aC n aC n = { 1 a , 1 a , 2 a , 5 a , 14 a , 42 a , 132 a , ... } , such that Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  29. A simplest case: aC n aC n = { 1 a , 1 a , 2 a , 5 a , 14 a , 42 a , 132 a , ... } , such that � = 1 a ; � � � 1 a Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  30. A simplest case: aC n aC n = { 1 a , 1 a , 2 a , 5 a , 14 a , 42 a , 132 a , ... } , such that � � 1 a 1 a � = 1 a ; � � � � � = 1 a 2 ; � 1 a � � 1 a 2 a � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  31. A simplest case: aC n aC n = { 1 a , 1 a , 2 a , 5 a , 14 a , 42 a , 132 a , ... } , such that � � 1 a 1 a 2 a � � � � 1 a 1 a � = 1 a ; � � � � = 1 a 2 ; � � � = 1 a 3 ; · · · � 1 a 1 a 2 a 5 a � � � � 1 a 2 a � � � 2 a 5 a 14 a � � Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  32. A simplest case: aC n aC n = { 1 a , 1 a , 2 a , 5 a , 14 a , 42 a , 132 a , ... } , such that � � 1 a 1 a 2 a � � � � 1 a 1 a � = 1 a ; � � � � = 1 a 2 ; � � � = 1 a 3 ; · · · � 1 a 1 a 2 a 5 a � � � � 1 a 2 a � � � 2 a 5 a 14 a � � H { aC n } = { a , a 2 , a 3 , a 4 , · · · } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  33. A simplest case: aC n aC n = { 1 a , 1 a , 2 a , 5 a , 14 a , 42 a , 132 a , ... } , such that � � 1 a 1 a 2 a � � � � 1 a 1 a � = 1 a ; � � � � = 1 a 2 ; � � � = 1 a 3 ; · · · � 1 a 1 a 2 a 5 a � � � � 1 a 2 a � � � 2 a 5 a 14 a � � H { aC n } = { a , a 2 , a 3 , a 4 , · · · } Conjecture The recurrence relation for H { aC n } is h n +1 = ah n . Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 16 / 40

  34. aC n + bC n +1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

  35. aC n + bC n +1 Basic Methods? Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

  36. aC n + bC n +1 Basic Methods? List and Compare! Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 17 / 40

  37. aC n + bC n +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

  38. aC n + bC n +1 Continue[a] Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  39. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  40. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  41. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  42. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  43. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  44. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  45. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  46. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  47. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  48. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  49. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  50. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 H { 3 C n + C n +1 } : h n +1 = Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  51. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 H { 3 C n + C n +1 } : h n +1 = 5 h n Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  52. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 H { 3 C n + C n +1 } : h n +1 = 5 h n − h n − 1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  53. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 H { 3 C n + C n +1 } : h n +1 = 5 h n − h n − 1 H { 4 C n + C n +1 } : h n +1 = Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  54. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 H { 3 C n + C n +1 } : h n +1 = 5 h n − h n − 1 H { 4 C n + C n +1 } : h n +1 = 6 h n Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  55. aC n + bC n +1 Continue[a] H { 0 C n + C n +1 } = { 1 , 1 , 1 , 1 , ... } H { 1 C n + C n +1 } = { 2 , 5 , 13 , 34 , ... } H { 2 C n + C n +1 } = { 3 , 11 , 41 , 153 , ... } H { 3 C n + C n +1 } = { 4 , 19 , 91 , 436 , ... } H { 4 C n + C n +1 } = { 5 , 29 , 169 , 985 , ... } · · · · · · H { 0 C n + C n +1 } : h n +1 = h n H { 1 C n + C n +1 } : h n +1 = 3 h n − h n − 1 H { 2 C n + C n +1 } : h n +1 = 4 h n − h n − 1 H { 3 C n + C n +1 } : h n +1 = 5 h n − h n − 1 H { 4 C n + C n +1 } : h n +1 = 6 h n − h n − 1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 18 / 40

  56. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  57. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  58. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  59. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } H { C n + 3 C n +1 } = { 4 , 19 , 97 , 508 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  60. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } H { C n + 3 C n +1 } = { 4 , 19 , 97 , 508 , ... } H { C n + 4 C n +1 } = { 5 , 29 , 181 , 1165 , ... } Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  61. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } H { C n + 3 C n +1 } = { 4 , 19 , 97 , 508 , ... } H { C n + 4 C n +1 } = { 5 , 29 , 181 , 1165 , ... } · · · · · · Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  62. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } H { C n + 3 C n +1 } = { 4 , 19 , 97 , 508 , ... } H { C n + 4 C n +1 } = { 5 , 29 , 181 , 1165 , ... } · · · · · · H { C n + 0 C n +1 } : h n +1 = h n Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  63. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } H { C n + 3 C n +1 } = { 4 , 19 , 97 , 508 , ... } H { C n + 4 C n +1 } = { 5 , 29 , 181 , 1165 , ... } · · · · · · H { C n + 0 C n +1 } : h n +1 = h n H { C n + 1 C n +1 } : h n +1 = 3 h n − h n − 1 Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

  64. aC n + bC n +1 Continue[b] H { C n + 0 C n +1 } = { 1 , 1 , 1 , 1 , ... } H { C n + 1 C n +1 } = { 2 , 5 , 13 , 34 , ... } H { C n + 2 C n +1 } = { 3 , 11 , 43 , 171 , ... } H { C n + 3 C n +1 } = { 4 , 19 , 97 , 508 , ... } H { C n + 4 C n +1 } = { 5 , 29 , 181 , 1165 , ... } · · · · · · H { C n + 0 C n +1 } : h n +1 = h n H { C n + 1 C n +1 } : h n +1 = 3 h n − h n − 1 H { C n + 2 C n +1 } : h n +1 = Wenyang Qian (Grinnell College) Hankel Transform of Catalan Numbers Oct 4, 2010 19 / 40

Download Presentation
Download Policy: The content available on the website is offered to you 'AS IS' for your personal information and use only. It cannot be commercialized, licensed, or distributed on other websites without prior consent from the author. To download a presentation, simply click this link. If you encounter any difficulties during the download process, it's possible that the publisher has removed the file from their server.

Recommend

Rational Catalan Numbers and Music Jedrzejewski Catalan Numbers Franck Jedrzejewski Rational

Rational Catalan Numbers and Music Jedrzejewski Catalan Numbers Franck Jedrzejewski Rational

Rational Catalan Numbers and Music Franck Rational Catalan Numbers and Music Jedrzejewski Catalan Numbers Franck Jedrzejewski Rational Catalan Numbers Dyck Path Paris-Saclay University - CEA - France Christoffel words Well-formed

655 views • 47 slides
Notes on the Catalan problem Daniele Paolo Scarpazza Politecnico di Milano

Notes on the Catalan problem Daniele Paolo Scarpazza Politecnico di Milano

Notes on the Catalan problem Daniele Paolo Scarpazza Politecnico di Milano <scarpazz@elet.polimi.it> March 16th, 2004 Daniele Paolo Scarpazza Notes on the Catalan problem [1] An overview of Catalan problems Catalan numbers appear

557 views • 51 slides
Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series Ira

Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series Ira

Hankel determinants, continued fractions, orthgonal polynomials, and hypergeometric series Ira M. Gessel with Jiang Zeng and Guoce Xin LaBRI June 8, 2007 Continued fractions and Hankel determinants There is a close relationship between

1.1k views • 52 slides
Hankel Matrices: From Words to Graphs Nadia Labai and Johann A. Makowsky Faculty of Computer

Hankel Matrices: From Words to Graphs Nadia Labai and Johann A. Makowsky Faculty of Computer

LATA invited lecture, March 2015 Hankel matrices Hankel Matrices: From Words to Graphs Nadia Labai and Johann A. Makowsky Faculty of Computer Science, Technion - Israel Institute of Technology, Haifa, Israel http://www.cs.technion.ac.il/

1.44k views • 55 slides
WHAT WOULD TREX DO? From Experimental Design to Analysis, the TREX Approach EXPERIMENTAL DESIGN

WHAT WOULD TREX DO? From Experimental Design to Analysis, the TREX Approach EXPERIMENTAL DESIGN

Your Research Question WHAT WOULD TREX DO? From Experimental Design to Analysis, the TREX Approach EXPERIMENTAL DESIGN What are my research goals? FFPE Clinical Where are my samples coming from? Fresh Tissue/Cells How much RNA

863 views • 39 slides
Topic 10: The Z Transform o Introduction to Z Transform o Relationship to the Fourier transform o

Topic 10: The Z Transform o Introduction to Z Transform o Relationship to the Fourier transform o

ELEC361: Signals And Systems Topic 10: The Z Transform o Introduction to Z Transform o Relationship to the Fourier transform o Z Transform and Examples o Region of Convergence of the Z Transform o Inverse Z Transform and Examples o Properties of Z

3.02k views • 69 slides
The Fourier Transform I Image Analysis The Fourier transform Those who combine a theoretical

The Fourier Transform I Image Analysis The Fourier transform Those who combine a theoretical

The Fourier Transform I Image Analysis The Fourier transform Those who combine a theoretical knowledge of Fourier transform properties with a knowledge of their physical interpretation are well prepared to approach most Niclas Brlin

366 views • 8 slides
Hecke algebras on homogeneous trees and relation with Hankel and Toeplitz matrices Janusz

Hecke algebras on homogeneous trees and relation with Hankel and Toeplitz matrices Janusz

Homogeneous trees Algebraic MASA: deg ( X ) > 2 Hecke MASA: general case. Integers: Toeplitz and Hankel Hecke algebras on homogeneous trees and relation with Hankel and Toeplitz matrices Janusz Wysocza nski

1.46k views • 133 slides
Coefficientwise total positivity (via continued fractions) for some Hankel matrices of

Coefficientwise total positivity (via continued fractions) for some Hankel matrices of

Coefficientwise total positivity (via continued fractions) for some Hankel matrices of combinatorial polynomials Alan Sokal New York University / University College London S eminaire de combinatoire Philippe Flajolet 5 June 2014 Key

478 views • 36 slides
The Catalan crisis from an economists point of view Clara Ponsat VIVES, KU Leuven December

The Catalan crisis from an economists point of view Clara Ponsat VIVES, KU Leuven December

The Catalan crisis from an economists point of view Clara Ponsat VIVES, KU Leuven December 15th, 2017 Preamble For the last 5 years Catalan governments have been asking the Spanish government to reach an agreement to hold a referendum

632 views • 27 slides
A Universal Bijection for Catalan Families R. Brak School of Mathematics and Statistics

A Universal Bijection for Catalan Families R. Brak School of Mathematics and Statistics

A Universal Bijection for Catalan Families R. Brak School of Mathematics and Statistics University of Melbourne October 25, 2018 Catalan Numbers n + 1 ( 2 n 1 C n = n ) 1 , 2 , 5 , 14 , 42 , 132 , 429 , n C n + 1 = C i C n i i =

304 views • 29 slides
A novel integral transform approach to solving partial differential equations in the curved

A novel integral transform approach to solving partial differential equations in the curved

A novel integral transform approach to solving partial differential equations in the curved space-times Karen Yagdjian University of Texas Rio Grande Valley Microlocal and Global Analysis, Interactions with Geometry Colloquium in honor of

692 views • 42 slides
Catalan intransitive verbs and argument realization Alex Alsina & Fengrong Yang

Catalan intransitive verbs and argument realization Alex Alsina & Fengrong Yang

Catalan intransitive verbs and argument realization Alex Alsina & Fengrong Yang Universitat Pompeu Fabra Goal To describe and analyze the behavior of the single core argument of intransitive verbs in Catalan: The mapping between

978 views • 50 slides
Video Segmentation for Video Segmentation for Surveillance Surveillance -- A Transform Domain

Video Segmentation for Video Segmentation for Surveillance Surveillance -- A Transform Domain

Video Segmentation for Video Segmentation for Surveillance Surveillance -- A Transform Domain Approach A Transform Domain Approach -- Juhua Zhu Dept. of Electrical Engineering April 23, 2005 1 1 Problem Formulation Problem Formulation

209 views • 9 slides
Binomials, Derangements, and Catalan Numbers Dr. Mattox Beckman University of Illinois at

Binomials, Derangements, and Catalan Numbers Dr. Mattox Beckman University of Illinois at

Introduction Fibonacci Numbers Binomial Coeffjcients Derangements Catalan Numbers Binomials, Derangements, and Catalan Numbers Dr. Mattox Beckman University of Illinois at Urbana-Champaign Department of Computer Science Introduction

175 views • 6 slides
Log-convexity of q -Catalan numbers Lynne Butler and undergraduate Pat Flanigan at Haverford

Log-convexity of q -Catalan numbers Lynne Butler and undergraduate Pat Flanigan at Haverford

Log-convexity of q -Catalan numbers Lynne Butler and undergraduate Pat Flanigan at Haverford College The sequence of Catalan numbers 1 , 1 , 2 , 5 , 14 , . . . is defined by C 0 = 1 and the recursion n C n +1 = C k C n k . k =0 1 + C

309 views • 8 slides
On certain properties of a perturbed Freud-type weight Abey Kelil University of Pretoria

On certain properties of a perturbed Freud-type weight Abey Kelil University of Pretoria

On certain properties of a perturbed Freud-type weight Abey Kelil University of Pretoria AIMS-Volkswagen Stiftung Workshop on Introduction to Orthogonal Polynomials and Applications Douala, Cameroon October 09, 2018 1 Plan of the talk 1

1.03k views • 25 slides
Induction and Recursion CMPS/MATH 2170: Discrete Mathematics Outline Mathematical induction

Induction and Recursion CMPS/MATH 2170: Discrete Mathematics Outline Mathematical induction

Induction and Recursion CMPS/MATH 2170: Discrete Mathematics Outline Mathematical induction (5.1) Strong induction (5.2) Recursive definitions (5.3) Recurrence Relations (8.1) Principle of Mathematical Induction Want to know

952 views • 21 slides
CSC 344 Algorithms and Complexity Lecture #10 Recurrences Recurrence Relations

CSC 344 Algorithms and Complexity Lecture #10 Recurrences Recurrence Relations

5/4/2016 CSC 344 Algorithms and Complexity Lecture #10 Recurrences Recurrence Relations Overview Connection to recursive algorithms Techniques for solving them Methods for generating a guess Induction proofs

498 views • 20 slides
The Nature of Epistemic Uncertainty in Linear Dynamical Systems S. Adhikari and A. Sarkar

The Nature of Epistemic Uncertainty in Linear Dynamical Systems S. Adhikari and A. Sarkar

The Nature of Epistemic Uncertainty in Linear Dynamical Systems S. Adhikari and A. Sarkar Department of Aerospace Engineering, University of Bristol, Bristol, U.K. Email: S.Adhikari@bristol.ac.uk Department of Civil and

624 views • 44 slides
The Open Bankart is Best for the Collision Athlete Theodore F. Schlegel, MD UC Health / Steadman

The Open Bankart is Best for the Collision Athlete Theodore F. Schlegel, MD UC Health / Steadman

Orthopaedic Summit Las Vegas, Nevada December 8 th , 2017 The Open Bankart is Best for the Collision Athlete Theodore F. Schlegel, MD UC Health / Steadman Hawkins Clinic Denver Disclosures Neither I, Theodore Schlegel, or a family member

385 views • 19 slides
State-of-the-Art Large Scale Language Modeling in 12 Hours With a Single GPU Nitish Shirish

State-of-the-Art Large Scale Language Modeling in 12 Hours With a Single GPU Nitish Shirish

State-of-the-Art Large Scale Language Modeling in 12 Hours With a Single GPU Nitish Shirish Keskar - @strongduality Stephen Merity - @smerity About Us Stephen Merity Nitish Shirish Keskar smerity.com keskarnitish.github.io M.S. from Harvard

695 views • 41 slides
A Recurrent BERT-based Model for Question Generation Ying-Hong Chan Yao-Chung Fan Department of

A Recurrent BERT-based Model for Question Generation Ying-Hong Chan Yao-Chung Fan Department of

A Recurrent BERT-based Model for Question Generation Ying-Hong Chan Yao-Chung Fan Department of Computer Science Department of Computer Science National Chung Hsing University National Chung Hsing University Taichung, Taiwan Taichung, Taiwan

1.13k views • 9 slides
Structural Primary Balance May 2017 The Economic Policy Secretariat (SPE) reports take into

Structural Primary Balance May 2017 The Economic Policy Secretariat (SPE) reports take into

MINISTRY OF FINANCE Structural Primary Balance May 2017 The Economic Policy Secretariat (SPE) reports take into account public data, whose primary sources are autonomous institutions, from public and private sectors. The aim is to organize these

287 views • 14 slides

More recommend