%I #32 Dec 15 2022 13:43:56

%S 1,1,1,1,2,1,1,2,1,2,2,1,1,1,3,1,1,3,2,1,1,1,3,2,1,1,1,3,3,1,1,1,1,4,

%T 2,1,1,1,4,2,2,1,1,1,1,1,3,4,1,1,1,1,3,4,2,1,1,1,1,2,4,2,2,1,1,1,1,1,

%U 3,5,2,1,1,1,1,1,1,3,5,2,1,1,1,1,1,3,5,2,2,1,1,1,1,1,1,1,5,4,2,1,1,1,1,1,1,5,4,2,2

%N Irregular triangle read by rows: T(n,k) is one half of the number of line segments of length 1 in the k-th antidiagonal of the Dyck path described in the n-th row of A237593.

%e Triangle begins (first 19 rows):

%e 1;

%e 1, 1;

%e 1, 2;

%e 1, 1, 2;

%e 1, 2, 2;

%e 1, 1, 1, 3;

%e 1, 1, 3, 2;

%e 1, 1, 1, 3, 2;

%e 1, 1, 1, 3, 3;

%e 1, 1, 1, 1, 4, 2;

%e 1, 1, 1, 4, 2, 2;

%e 1, 1, 1, 1, 1, 3, 4;

%e 1, 1, 1, 1, 3, 4, 2;

%e 1, 1, 1, 1, 2, 4, 2, 2;

%e 1, 1, 1, 1, 1, 3, 5, 2;

%e 1, 1, 1, 1, 1, 1, 3, 5, 2;

%e 1, 1, 1, 1, 1, 3, 5, 2, 2;

%e 1, 1, 1, 1, 1, 1, 1, 5, 4, 2;

%e 1, 1, 1, 1, 1, 1, 5, 4, 2, 2;

%e ...

%e For n = 10 the 10th row of A237593 is [6, 2, 1, 1, 1, 1, 2, 6]. When that row is interpreted as a symmetric Dyck path in the fourth quadrant using 20 line segments of length 1 the Dyck path looks like this:

%e .

%e |

%e |

%e |

%e |

%e |

%e _ _|

%e _|

%e _|

%e |

%e _ _ _ _ _ _|

%e .

%e The numbers of line segments of length 1 in the successive antidiagonals are respectively [2, 2, 2, 2, 8, 4] so the 10th row of triangle is [1, 1, 1, 1, 4, 2].

%Y Row sums give A000027.

%Y Row n has length A008619(n).

%Y Column 1 gives A000012.

%Y Cf. A196020, A235791, A236104, A237270, A237271, A237591, A237593, A245092, A262626, A339575.

%K nonn,tabf

%O 1,5

%A _Omar E. Pol_, Nov 19 2022