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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

  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

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