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Irregular triangle read by rows: T(n,k) is the size of the subpart that is adjacent to the k-th peak of the largest Dyck path of the symmetric representation of sigma(n), or T(n,k) = 0 if the mentioned subpart is already associated to a previous peak or if there is no subpart adjacent to the k-th peak, with n >= 1, k >= 1.
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%I #120 May 19 2022 16:43:49

%S 1,3,2,2,7,0,3,3,11,1,0,4,0,4,15,0,0,5,3,5,9,0,9,0,6,0,0,6,23,5,0,0,7,

%T 0,0,7,12,0,12,0,8,7,1,0,8,31,0,0,0,0,9,0,0,0,9,35,2,0,2,0,10,0,0,0,

%U 10,39,0,3,0,0,11,5,0,5,0,11,18,0,0,0,18,0,12,0,0,0,0,12,47,13,0,0,0,0,13,0,5,0,0,13

%N Irregular triangle read by rows: T(n,k) is the size of the subpart that is adjacent to the k-th peak of the largest Dyck path of the symmetric representation of sigma(n), or T(n,k) = 0 if the mentioned subpart is already associated to a previous peak or if there is no subpart adjacent to the k-th peak, with n >= 1, k >= 1.

%C Conjecture: row n is formed by the odd-indexed terms of the n-th row of triangle A280850 together with the even-indexed terms of the same row but listed in reverse order. Examples: the 15th row of A280850 is [8, 8, 7, 0, 1] so the 15th row of this triangle is [8, 7, 1, 0, 8]. The 75th row of A280850 is [38, 38, 21, 0, 3, 3, 0, 0, 0, 21, 0] so the 75h row of this triangle is [38, 21, 3, 0, 0, 0, 21, 0, 3, 0, 38].

%C For the definition of "subparts" see A279387.

%C For more information about the mentioned Dyck paths see A237593.

%C T(n,k) could be called the "charge" of the k-th peak of the largest Dyck path of the symmetric representation of sigma(n).

%C The number of zeros in row n is A238005(n). - _Omar E. Pol_, Sep 11 2021

%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>

%e Triangle begins (rows 1..28):

%e 1;

%e 3;

%e 2, 2;

%e 7, 0;

%e 3, 3;

%e 11, 1, 0;

%e 4, 0, 4;

%e 15, 0, 0;

%e 5, 3, 5;

%e 9, 0, 9, 0;

%e 6, 0, 0, 6;

%e 23, 5, 0, 0;

%e 7, 0, 0, 7;

%e 12, 0, 12, 0;

%e 8, 7, 1, 0, 8;

%e 31, 0, 0, 0, 0;

%e 9, 0, 0, 0, 9;

%e 35, 2, 0, 2, 0;

%e 10, 0, 0, 0, 10;

%e 39, 0, 3, 0, 0;

%e 11, 5, 0, 5, 0, 11;

%e 18, 0, 0, 0, 18, 0;

%e 12, 0, 0, 0, 0, 12;

%e 47, 13, 0, 0, 0, 0;

%e 13, 0, 5, 0, 0, 13;

%e 21, 0, 0, 0 21, 0;

%e 14, 6, 0, 6, 0, 14;

%e 55, 0, 0, 1, 0, 0, 0;

%e ...

%e For n = 15 we have that the 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8], so the diagram of the symmetric representation of sigma(15) is constructed in the third quadrant as shown below in Figure 1:

%e . _ _

%e . | | | |

%e . | | | |

%e . | | | |

%e . 8 | | | |

%e . | | | |

%e . | | | |

%e . | | | |

%e . |_|_ _ _ |_|_ _ _

%e . | |_ _ 8 | |_ _

%e . |_ | |_ _ |

%e . |_ |_ 7 |_| |_

%e . 8 |_ _| 1 |_ _|

%e . | 0 |

%e . |_ _ _ _ _ _ _ _ |_ _ _ _ _ _ _ _

%e . |_ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _|

%e . 8 8

%e .

%e . Figure 1. The symmetric Figure 2. After the dissection

%e . representation of sigma(15) of the symmetric representation

%e . has three parts of size 8 of sigma(15) into layers of

%e . because every part contains width 1 we can see four subparts,

%e . 8 cells, so the 15th row of so the 15th row of this triangle is

%e . triangle A237270 is [8, 8, 8]. [8, 7, 1, 0, 8]. See also below.

%e .

%e Illustration of first 50 terms (rows 1..16 of triangle) in an irregular spiral which can be find in the top view of the pyramid described in A244050:

%e .

%e . 12 _ _ _ _ _ _ _ _

%e . | _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7

%e . | | |_ _ _ _ _ _ _|

%e . 0 _| | |

%e . |_ _|9 _ _ _ _ _ _ |_ _ 0

%e . 12 _ _| | _ _ _ _ _|_ _ _ _ _ 5 |_ 0

%e . 0 _ _ _| | 0 _| | |_ _ _ _ _| |

%e . | _ _ _| 9 _|_ _| |_ _ 3 |_ _ _ 7

%e . | | 0 _ _| | 11 _ _ _ _ |_ | | |

%e . | | | _ _| 1 _| _ _ _|_ _ _ 3 |_|_ _ 5 | |

%e . | | | | 0 _|_| | |_ _ _| | | | |

%e . | | | | | _ _| |_ _ 3 | | | |

%e . | | | | | | 3 _ _ | | | | | |

%e . | | | | | | | _|_ 1 | | | | | |

%e . _|_| _|_| _|_| _|_| |_| _|_| _|_| _|_| _

%e . | | | | | | | | | | | | | | | |

%e . | | | | | | |_|_ _ _| | | | | | | |

%e . | | | | | | 2 |_ _|_ _| _| | | | | | |

%e . | | | | |_|_ 2 |_ _ _| 0 _ _| | | | | |

%e . | | | | 4 |_ 7 _| _ _|0 | | | |

%e . | | |_|_ _ 0 |_ _ _ _ | _| _ _ _| | | |

%e . | | 6 |_ |_ _ _ _|_ _ _ _| | 0 _| _ _ _|0 | |

%e . |_|_ _ _ 0 |_ 4 |_ _ _ _ _| _| _| | _ _ _| |

%e . 8 | |_ _ 0 | 15| _| _| | _ _ _|

%e . |_ _ | |_ _ _ _ _ _ | |_ _| 0 _| | 0

%e . 7 |_| |_ |_ _ _ _ _ _|_ _ _ _ _ _| | 5 _| _|

%e . 1 |_ _| 6 |_ _ _ _ _ _ _| _ _| _| 0

%e . 0 | 23| _ _| 0

%e . |_ _ _ _ _ _ _ _ | | 0

%e . |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |

%e . 8 |_ _ _ _ _ _ _ _ _|

%e . 31

%e .

%e The diagram contains 30 subparts equaling A060831(16), the total number of partitions of all positive integers <= 16 into consecutive parts.

%e For the construction of the spiral see A239660.

%e From _Omar E. Pol_, Nov 26 2020: (Start)

%e Also consider the infinite double-staircases diagram defined in A335616 (see the theorem). For n = 15 the diagram with first 15 levels looks like this:

%e .

%e Level "Double-staircases" diagram

%e . _

%e 1 _|1|_

%e 2 _|1 _ 1|_

%e 3 _|1 |1| 1|_

%e 4 _|1 _| |_ 1|_

%e 5 _|1 |1 _ 1| 1|_

%e 6 _|1 _| |1| |_ 1|_

%e 7 _|1 |1 | | 1| 1|_

%e 8 _|1 _| _| |_ |_ 1|_

%e 9 _|1 |1 |1 _ 1| 1| 1|_

%e 10 _|1 _| | |1| | |_ 1|_

%e 11 _|1 |1 _| | | |_ 1| 1|_

%e 12 _|1 _| |1 | | 1| |_ 1|_

%e 13 _|1 |1 | _| |_ | 1| 1|_

%e 14 _|1 _| _| |1 _ 1| |_ |_ 1|_

%e 15 |1 |1 |1 | |1| | 1| 1| 1|

%e .

%e Starting from A196020 and after the algorithm described n A280850 and the conjecture applied to the above diagram we have a new diagram as shown below:

%e .

%e Level "Ziggurat" diagram

%e . _

%e 6 |1|

%e 7 _ | | _

%e 8 _|1| _| |_ |1|_

%e 9 _|1 | |1 1| | 1|_

%e 10 _|1 | | | | 1|_

%e 11 _|1 | _| |_ | 1|_

%e 12 _|1 | |1 1| | 1|_

%e 13 _|1 | | | | 1|_

%e 14 _|1 | _| _ |_ | 1|_

%e 15 |1 | |1 |1| 1| | 1|

%e .

%e The 15th row

%e of A249351: [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]

%e The 15th row

%e of A237270: [ 8, 8, 8 ]

%e The 15th row

%e of this seq: [ 8, 7, 1, 0, 8 ]

%e The 15th row

%e of A280851: [ 8, 7, 1, 8 ]

%e .

%e (End)

%Y Row sums give A000203.

%Y Row n has length A003056(n).

%Y Column k starts in row A000217(k).

%Y Nonzero terms give A280851.

%Y The number of nonzero terms in row n is A001227(n).

%Y The triangle with n rows contain A060831(n) nonzero terms.

%Y Cf. A024916, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A238005, A239657, A239660, A239931-A239934, A240542, A244050, A245092, A250068, A250070, A261699, A262626, A279387, A279388, A279391, A280850.

%K nonn,tabf

%O 1,2

%A _Omar E. Pol_, Feb 10 2018