A279387 Irregular triangle read by rows: suppose the symmetric representation of sigma(n) consists of m = A250068(n) layers of width 1, arranged in increasing order; then T(n,k) (n >= 1, 1 <= k <= m) is the number of subparts in the k-th layer.
1, 1, 2, 1, 2, 1, 1, 2, 1, 3, 2, 2, 1, 1, 2, 2, 3, 1, 1, 2, 1, 2, 2, 1, 1, 4, 2, 2, 1, 1, 3, 2, 4, 1, 1, 2, 1, 3, 2, 1, 4, 2, 3, 1, 1, 2, 2, 2, 4, 1, 1, 2, 1, 3, 2, 2, 3, 3, 2, 2, 1, 1, 3, 3, 4, 2, 2, 1, 3, 4, 1, 1, 4, 2, 2, 1, 1, 2, 2, 2, 5, 1, 1, 4, 1, 3, 2, 2, 4, 3, 1, 2, 1, 1, 1, 2, 2
Offset: 1
Examples
Triangle begins (first 18 rows):
1;
1;
2;
1;
2;
1, 1;
2;
1;
3;
2;
2;
1, 1;
2;
2;
3, 1;
1;
2;
1, 2;
...
For n = 12, 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:
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. _ _ _ _ _ _| | 28 _ _ _ _ _ _| | 5
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. 23
.
. Figure 1. The symmetric Figure 2. After the dissection
. representation of sigma(12) of the symmetric representation
. has only one part which of sigma(12) into layers of
. contains 28 cells, so width 1 we can see two "subparts"
. A237271(12) = 1. that contain 23 and 5 cells
. respectively, so the 12th row of
. this triangle is [1, 1], and the
. row sum is A001227(12) = 2,
. equaling the number of odd divisors
. of 12.
.
For n = 15, 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:
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. _ _| | 8 _ _| | 8
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. |_ _| 8 |_ _| 1
. | | 7
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. 8 8
.
. Figure 3. The symmetric Figure 4. After the dissection
. representation of sigma(15) of the symmetric representation
. has three parts of size 8 of sigma(15) into layers of
. because every part contains width 1 we can see four "subparts".
. 8 cells, so A237271(15) = 3. The first layer has three subparts:
. [8, 7, 8]. The second layer has
. only one subpart of size 1, so
. the 15th row of this triangle is
. [3, 1], and the row sum is
. A001227(15) = 4, equaling the
. number of odd divisors of 15.
.
For n = 360, the 359th row of triangle A237593 is [180, 61, 30, 19, 12, 9, 7, 6, 4, 4, 3, 3, 2, 3, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1] and the 360th row of the same triangle is [181, 60, 31, 18, 13, 9, 7, 5, 5, 4, 3, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1], so have that the symmetric representation of sigma(360) = 1170 has only one part, five layers, and six subparts: [(719), (237), (139), (71), (2, 2)], so the 360th row of this triangle is [1, 1, 1, 1, 2], and the row sum is A001227(360) = 6, equaling the number of odd divisors of 360 (the diagram is too large to include).
From _Hartmut F. W. Hoft_, Dec 16 2016: (Start)
45 has 6 subparts of which 2 have symmetric duplicates and 2 span the center. Row length is 18 and "|" indicates the center marker for a row.
1 2 3 4 5 6 7 8 9|9 8 7 6 5 4 3 2 1 : position indices
1 0 1 1 2 1 1 1 2|2 1 1 1 2 1 1 0 1 : row 45 of A262045
1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 : layer 1
1 1|1 1 : layer 2
1 1 1 0 1 1 0 0 1| : row 45 of A237048 (odd divisors)
+ - + . + - . . +| : change in level ("." no change)
90 has 6 subparts and 3 layers (row length is 24).
1 2 3 4 5 6 7 8..10..12|.14..16..18..20..22..24 : position indices
1 1 2 1 2 2 2 2 3 3 3 2|2 3 3 3 2 2 2 2 1 2 1 1 : row 90 of A262045
1 1 1 1 1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 1 1 1 1 : layer 1
1 1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 1 : layer 2
1 1 1 | 1 1 1 : layer 3
1 0 1 1 1 0 0 0 1 0 0 1| : row 90 of A237048
+ . + - + . . . + . . -| : change in level ("." no change)
The process of successive levels provides two "default" dissections of the symmetric representation into subparts from the boundary at n towards the boundary at n-1 or in the reverse direction. (End)
From _Omar E. Pol_, Nov 24 2020: (Start)
For n = 18 we have that the 17th row of triangle A237593 is [9, 4, 2, 1, 1, 1, 1, 2, 4, 9] and the 18th row of the same triangle is [10, 3, 2, 2, 1, 1, 2, 2, 3, 10], so the diagram of the symmetric representation of sigma(18) = 39 is constructed as shown below in Figure 5:
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. | | 39 | _ _|
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. | | | | 2
. _ _ _ _ _ _ _ _ _| | _ _ _ _ _ _ _ _ _| |
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. 35
.
. Figure 5. The symmetric Figure 6. After the dissection
. representation of sigma(18) of the symmetric representation
. has one part of size 39, so of sigma(18) into layers of
. A237271(18) = 1. width 1 we can see three "subparts".
. The first layer has one subpart of
. size 35. The second layer has
. two subparts of size 2, so
. the 18th row of this triangle is
. [1, 2], and the row sum is
. A001227(18) = 3.
(End)
Crossrefs
The sum of row n equals A001227(n).
Hence, if n is odd, the sum of row n equals A000005(n).
Row n has length A250068(n).
Column 1 gives A237271.
Programs
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Mathematica
(* function a341969[ ] is defined in A341969 *) a279387[n_] := Module[{widthL=a341969[n], partL, cL, top, ft, sL}, partL=Select[SplitBy[widthL, #==0&], #!={0}&]; cL=Table[0, Max[widthL]]; While[partL!={}, top=Last[partL]; ft=First[top]; sL=Select[SplitBy[top, #==ft&], #!={ft}&]; cL[[ft]]++; partL=Join[Most[partL], sL]]; cL] Flatten[a279387[74]] (* the first 74 rows of the table; Hartmut F. W. Hoft, Feb 24 2021 *)
Extensions
Definition edited by Omar E. Pol and N. J. A. Sloane, Nov 25 2020
Comments