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%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