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Irregular triangle read by rows T(n,k) in which row n has length A000041(n-1) and every column k is A024916, n >= 1, k >= 1.
6

%I #72 Mar 09 2022 01:15:36

%S 1,4,8,1,15,4,1,21,8,4,1,1,33,15,8,4,4,1,1,41,21,15,8,8,4,4,1,1,1,1,

%T 56,33,21,15,15,8,8,4,4,4,4,1,1,1,1,69,41,33,21,21,15,15,8,8,8,8,4,4,

%U 4,4,1,1,1,1,1,1,1,87,56,41,33,33,21,21,15,15,15,15,8,8,8,8

%N Irregular triangle read by rows T(n,k) in which row n has length A000041(n-1) and every column k is A024916, n >= 1, k >= 1.

%C T(n,k) is the number of cubic cells (or cubes) in the k-th level starting from the base of the tower described in A221529 whose largest side of the base is equal to n (see example). - _Omar E. Pol_, Jan 08 2022

%F T(n,k) = A024916(A336811(n,k)).

%F T(n,k) = Sum_{j=1..n} A339278(j,k). - _Omar E. Pol_, Jan 08 2022

%e Triangle begins:

%e 1;

%e 4;

%e 8, 1;

%e 15, 4, 1;

%e 21, 8, 4, 1, 1;

%e 33, 15, 8, 4, 4, 1, 1;

%e 41, 21, 15, 8, 8, 4, 4, 1, 1, 1, 1;

%e 56, 33, 21, 15, 15, 8, 8, 4, 4, 4, 4, 1, 1, 1, 1;

%e 69, 41, 33, 21, 21, 15, 15, 8, 8, 8, 8, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1;

%e ...

%e For n = 9 the length of row 9 is A000041(9-1) = 22.

%e From _Omar E. Pol_, Jan 08 2022: (Start)

%e For n = 9 the lateral view and top view of the tower described in A221529 look like as shown below:

%e _

%e 22 1 | |

%e 21 1 | |

%e 20 1 | |

%e 19 1 | |

%e 18 1 | |

%e 17 1 | |

%e 16 1 |_|_

%e 15 4 | |

%e 14 4 | |

%e 13 4 | |

%e 12 4 |_ _|_

%e 11 8 | | |

%e 10 8 | | |

%e 9 8 | | |

%e 8 8 |_ _|_|_

%e 7 15 | | |

%e 6 15 |_ _ _| |_

%e 5 21 | | |

%e 4 21 |_ _ _|_ _|_

%e 3 33 |_ _ _ _| | |_

%e 2 41 |_ _ _ _|_|_ _|_ _

%e 1 69 |_ _ _ _ _|_ _|_ _|

%e .

%e Level Row 9 Lateral view

%e k T(9,k) of the tower

%e .

%e _ _ _ _ _ _ _ _ _

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

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

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

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

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

%e |_ _ _ _| |

%e |_ _ _ _| _ _|

%e | |

%e |_ _ _ _ _|

%e .

%e Top view

%e of the tower

%e .

%e For n = 9 and k = 1 there are 69 cubic cells in the level 1 starting from the base of the tower, so T(9,1) = 69.

%e For n = 9 and k = 22 there is only one cubic cell in the level 22 (the top) of the tower, so T(9,22) = 1.

%e The volume of the tower (also the total number of cubic cells) represents the 9th term of the convolution of A000203 and A000041 hence it's equal to A066186(9) = 270, equaling the sum of the 9th row of triangle. (End)

%o (PARI) f(n) = numbpart(n-1);

%o T(n, k) = {if (k > f(n), error("invalid k")); if (k==1, return (n)); my(s=0); while (k <= f(n-1), s++; n--; ); 1+s; } \\ A336811

%o g(n) = sum(k=1, n, n\k*k); \\ A024916

%o row(n) = vector(f(n), k, g(T(n,k))); \\ _Michel Marcus_, Jan 22 2022

%Y Row sums give A066186.

%Y Row lengths give A000041.

%Y The length of the m-th block in row n is A187219(m), m >= 1.

%Y Cf. A350637 (analog for the stepped pyramid described in A245092).

%Y Cf. A000203, A024916, A196020, A221529, A236104, A235791, A237270, A237271, A237593, A339278, A262626, A336811, A338156, A340035, A341149, A346533, A350333.

%K nonn,tabf

%O 1,2

%A _Omar E. Pol_, Jan 07 2021