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A280940 Irregular triangle read by rows: T(n,k) = number of subparts in the k-th part of the symmetric representation of sigma(n). 6

%I #27 Dec 31 2020 11:11:15

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

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

%N Irregular triangle read by rows: T(n,k) = number of subparts in the k-th part of the symmetric representation of sigma(n).

%C The "subparts" of the symmetric representation of sigma(n) are the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.

%C The number of subparts in the symmetric representation of sigma(n) equals the number of odd divisors of n.

%C For more information about "subparts" see A279387, A279388 and A279391.

%C Note that we can find the symmetric representation of sigma(n) as the terraces at the n-th level (starting from the top) of the stepped pyramid described in A245092.

%e Triangle begins (n = 1..21):

%e 1;

%e 1;

%e 1, 1;

%e 1;

%e 1, 1;

%e 2;

%e 1, 1;

%e 1;

%e 1, 1, 1;

%e 1, 1;

%e 1, 1;

%e 2;

%e 1, 1;

%e 1, 1;

%e 1, 2, 1;

%e 1;

%e 1, 1;

%e 3;

%e 1, 1;

%e 2;

%e 1, 1, 1, 1;

%e ...

%e For n = 12 we have that the 11th row of triangle A237593 is [6, 3, 1, 1, 1, 1, 3, 6] and the 12th row of the same triangle is [7, 2, 2, 1, 1, 2, 2, 7], so the diagram of the symmetric representation of sigma(12) = 28 is constructed as shown below in Figure 1:

%e . _ _

%e . | | | |

%e . | | | |

%e . | | | |

%e . | | | |

%e . | | | |

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

%e . 28 _| _ _| 23 _| _ _ _|

%e . _| | _| _| |

%e . | _| | _| _|

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

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

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

%e .

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

%e . representation of sigma(12) of the symmetric representation

%e . has only one part which of sigma(12) into layers of

%e . contains 28 cells, so width 1 we can see two "subparts"

%e . A237271(12) = 1, and that contain 23 and 5 cells

%e . A000203(12) = 28. respectively, so the 12th row of

%e . this triangle is [2], and the

%e . row sum is A001227(12) = 2, equaling

%e . the number of odd divisors of 12.

%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) = 24 is constructed as shown below in Figure 3:

%e . _ _

%e . | | | |

%e . | | | |

%e . | | | |

%e . | | | |

%e . 8 | | 8 | |

%e . | | | |

%e . | | | |

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

%e . 8 _ _| | 7 _ _| |

%e . | _| | _ _|

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

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

%e . 8 | 8 |

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

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

%e .

%e . Figure 3. The symmetric Figure 4. 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 A237271(15) = 3, The first and third part contains

%e . and A000203(15) = 8+8+8 = 24. one subpart each. The second part contains

%e . two subparts, so the 15th row of this

%e . triangle is [1, 2, 1], and the row

%e . sum is A001227(15) = 4, equaling the

%e . number of odd divisors of 15.

%e .

%Y Row sums give A001227 (number of odd divisors of n).

%Y Row lengths is A237271.

%Y Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A244050, A245092, A249351, A250068, A262626, A279387, A279388, A279391.

%K nonn,tabf,more

%O 1,8

%A _Omar E. Pol_, Jan 11 2017

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