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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).

%I A280940 #27 Dec 31 2020 11:11:15
%S A280940 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 A280940 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 A280940 Irregular triangle read by rows: T(n,k) = number of subparts in the k-th part of the symmetric representation of sigma(n).
%C A280940 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 A280940 The number of subparts in the symmetric representation of sigma(n) equals the number of odd divisors of n.
%C A280940 For more information about "subparts" see A279387, A279388 and A279391.
%C A280940 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 A280940 Triangle begins (n = 1..21):
%e A280940 1;
%e A280940 1;
%e A280940 1, 1;
%e A280940 1;
%e A280940 1, 1;
%e A280940 2;
%e A280940 1, 1;
%e A280940 1;
%e A280940 1, 1, 1;
%e A280940 1, 1;
%e A280940 1, 1;
%e A280940 2;
%e A280940 1, 1;
%e A280940 1, 1;
%e A280940 1, 2, 1;
%e A280940 1;
%e A280940 1, 1;
%e A280940 3;
%e A280940 1, 1;
%e A280940 2;
%e A280940 1, 1, 1, 1;
%e A280940 ...
%e A280940 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 A280940 .                          _                                    _
%e A280940 .                         | |                                  | |
%e A280940 .                         | |                                  | |
%e A280940 .                         | |                                  | |
%e A280940 .                         | |                                  | |
%e A280940 .                         | |                                  | |
%e A280940 .                    _ _ _| |                             _ _ _| |
%e A280940 .              28  _|    _ _|                       23  _|  _ _ _|
%e A280940 .                _|     |                             _|  _| |
%e A280940 .               |      _|                            |  _|  _|
%e A280940 .               |  _ _|                              | |_ _|
%e A280940 .    _ _ _ _ _ _| |                       _ _ _ _ _ _| |      5
%e A280940 .   |_ _ _ _ _ _ _|                      |_ _ _ _ _ _ _|
%e A280940 .
%e A280940 .   Figure 1. The symmetric            Figure 2. After the dissection
%e A280940 .   representation of sigma(12)        of the symmetric representation
%e A280940 .   has only one part which            of sigma(12) into layers of
%e A280940 .   contains 28 cells, so              width 1 we can see two "subparts"
%e A280940 .   A237271(12) = 1, and               that contain 23 and 5 cells
%e A280940 .   A000203(12) = 28.                  respectively, so the 12th row of
%e A280940 .                                      this triangle is [2], and the
%e A280940 .                                      row sum is A001227(12) = 2, equaling
%e A280940 .                                      the number of odd divisors of 12.
%e A280940 .
%e A280940 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 A280940 .                                _                                  _
%e A280940 .                               | |                                | |
%e A280940 .                               | |                                | |
%e A280940 .                               | |                                | |
%e A280940 .                               | |                                | |
%e A280940 .                           8   | |                            8   | |
%e A280940 .                               | |                                | |
%e A280940 .                               | |                                | |
%e A280940 .                          _ _ _|_|                           _ _ _|_|
%e A280940 .                   8  _ _| |                          7  _ _| |
%e A280940 .                     |    _|                            |  _ _|
%e A280940 .                    _|  _|                             _| |_|
%e A280940 .                   |_ _|                              |_ _|  1
%e A280940 .           8       |                          8       |
%e A280940 .    _ _ _ _ _ _ _ _|                   _ _ _ _ _ _ _ _|
%e A280940 .   |_ _ _ _ _ _ _ _|                  |_ _ _ _ _ _ _ _|
%e A280940 .
%e A280940 .   Figure 3. The symmetric            Figure 4. After the dissection
%e A280940 .   representation of sigma(15)        of the symmetric representation
%e A280940 .   has three parts of size 8          of sigma(15) into layers of
%e A280940 .   because every part contains        width 1 we can see four "subparts".
%e A280940 .   8 cells, so A237271(15) = 3,       The first and third part contains
%e A280940 .   and A000203(15) = 8+8+8 = 24.      one subpart each. The second part contains
%e A280940 .                                      two subparts, so the 15th row of this
%e A280940 .                                      triangle is [1, 2, 1], and the row
%e A280940 .                                      sum is A001227(15) = 4, equaling the
%e A280940 .                                      number of odd divisors of 15.
%e A280940 .
%Y A280940 Row sums give A001227 (number of odd divisors of n).
%Y A280940 Row lengths is A237271.
%Y A280940 Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A244050, A245092, A249351, A250068, A262626, A279387, A279388, A279391.
%K A280940 nonn,tabf,more
%O A280940 1,8
%A A280940 _Omar E. Pol_, Jan 11 2017