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%I A279387 #105 Mar 11 2021 21:26:54
%S A279387 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,
%T A279387 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,
%U A279387 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
%N 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.
%C A279387 The "subparts" of the symmetric representation of sigma(n) are defined to be the regions that arise after the dissection of the symmetric representation of sigma(n) into successive layers of width 1.
%C A279387 The number of layers of width 1 in the symmetric representation of sigma(n) is given in A250068.
%C A279387 The number of subparts in the first layer of the symmetric representation of sigma(n) is equal to A237271(n).
%C A279387 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.
%C A279387 (All above comments are essentially the same as the comments dated Nov 05 2016 at the old version of A275601, which was the same as A001227).
%C A279387 The sum of row n equals the number of subparts in the symmetric representation of sigma(n).
%C A279387 Conjecture:
%C A279387 The number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n.
%C A279387 From _Hartmut F. W. Hoft_, Dec 16 2016: (Start)
%C A279387 Proof:
%C A279387 Each row of the irregular triangle of A262045 can be interpreted as a step function of step sizes 1, 0, and -1. The numbers in row n are the widths of the segments in the parts of the symmetric representation of sigma(n). Each new subpart in a segment (in the left half) of row n starts at the same odd index that represents an odd divisor d of n in the irregular triangle of A237048. Either a subpart ends at an even index e, representing a second odd divisor, which satisfies d * e = oddpart(n), and thus the entire subpart is duplicated in the symmetric portion of the representation, or a subpart runs through the center and continues contiguously into the right half of the symmetric portion of the representation. In other words, the number of subparts in row n equals the number of odd divisors of n, i.e., the conjecture is true. (End)
%e A279387 Triangle begins (first 18 rows):
%e A279387 1;
%e A279387 1;
%e A279387 2;
%e A279387 1;
%e A279387 2;
%e A279387 1, 1;
%e A279387 2;
%e A279387 1;
%e A279387 3;
%e A279387 2;
%e A279387 2;
%e A279387 1, 1;
%e A279387 2;
%e A279387 2;
%e A279387 3, 1;
%e A279387 1;
%e A279387 2;
%e A279387 1, 2;
%e A279387 ...
%e A279387 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:
%e A279387 . _ _
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . _ _ _| | _ _ _| |
%e A279387 . _| _ _| _| _ _ _|
%e A279387 . _| | _| _| |
%e A279387 . | _| | _| _|
%e A279387 . | _ _| | |_ _|
%e A279387 . _ _ _ _ _ _| | 28 _ _ _ _ _ _| | 5
%e A279387 . |_ _ _ _ _ _ _| |_ _ _ _ _ _ _|
%e A279387 . 23
%e A279387 .
%e A279387 . Figure 1. The symmetric Figure 2. After the dissection
%e A279387 . representation of sigma(12) of the symmetric representation
%e A279387 . has only one part which of sigma(12) into layers of
%e A279387 . contains 28 cells, so width 1 we can see two "subparts"
%e A279387 . A237271(12) = 1. that contain 23 and 5 cells
%e A279387 . respectively, so the 12th row of
%e A279387 . this triangle is [1, 1], and the
%e A279387 . row sum is A001227(12) = 2,
%e A279387 . equaling the number of odd divisors
%e A279387 . of 12.
%e A279387 .
%e A279387 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:
%e A279387 . _ _
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . _ _ _|_| _ _ _|_|
%e A279387 . _ _| | 8 _ _| | 8
%e A279387 . | _| | _ _|
%e A279387 . _| _| _| |_|
%e A279387 . |_ _| 8 |_ _| 1
%e A279387 . | | 7
%e A279387 . _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _|
%e A279387 . |_ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _|
%e A279387 . 8 8
%e A279387 .
%e A279387 . Figure 3. The symmetric Figure 4. After the dissection
%e A279387 . representation of sigma(15) of the symmetric representation
%e A279387 . has three parts of size 8 of sigma(15) into layers of
%e A279387 . because every part contains width 1 we can see four "subparts".
%e A279387 . 8 cells, so A237271(15) = 3. The first layer has three subparts:
%e A279387 . [8, 7, 8]. The second layer has
%e A279387 . only one subpart of size 1, so
%e A279387 . the 15th row of this triangle is
%e A279387 . [3, 1], and the row sum is
%e A279387 . A001227(15) = 4, equaling the
%e A279387 . number of odd divisors of 15.
%e A279387 .
%e A279387 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).
%e A279387 From _Hartmut F. W. Hoft_, Dec 16 2016: (Start)
%e A279387 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.
%e A279387 1 2 3 4 5 6 7 8 9|9 8 7 6 5 4 3 2 1 : position indices
%e A279387 1 0 1 1 2 1 1 1 2|2 1 1 1 2 1 1 0 1 : row 45 of A262045
%e A279387 1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 : layer 1
%e A279387 1 1|1 1 : layer 2
%e A279387 1 1 1 0 1 1 0 0 1| : row 45 of A237048 (odd divisors)
%e A279387 + - + . + - . . +| : change in level ("." no change)
%e A279387 90 has 6 subparts and 3 layers (row length is 24).
%e A279387 1 2 3 4 5 6 7 8..10..12|.14..16..18..20..22..24 : position indices
%e A279387 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
%e A279387 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
%e A279387 1 1 1 1 1 1 1 1 1|1 1 1 1 1 1 1 1 1 : layer 2
%e A279387 1 1 1 | 1 1 1 : layer 3
%e A279387 1 0 1 1 1 0 0 0 1 0 0 1| : row 90 of A237048
%e A279387 + . + - + . . . + . . -| : change in level ("." no change)
%e A279387 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)
%e A279387 From _Omar E. Pol_, Nov 24 2020: (Start)
%e A279387 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:
%e A279387 . _ _
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 ._ | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . | | | |
%e A279387 . _ _ _ _| | _ _ _ _| |
%e A279387 . | _ _ _| | _ _ _ _|
%e A279387 . _| | _| | |
%e A279387 . _| _ _| _| _|_|
%e A279387 . _ _| _| _ _| _| 2
%e A279387 . | | 39 | _ _|
%e A279387 . | _ _| | |_ _|
%e A279387 . | | | | 2
%e A279387 . _ _ _ _ _ _ _ _ _| | _ _ _ _ _ _ _ _ _| |
%e A279387 . |_ _ _ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _ _ _|
%e A279387 . 35
%e A279387 .
%e A279387 . Figure 5. The symmetric Figure 6. After the dissection
%e A279387 . representation of sigma(18) of the symmetric representation
%e A279387 . has one part of size 39, so of sigma(18) into layers of
%e A279387 . A237271(18) = 1. width 1 we can see three "subparts".
%e A279387 . The first layer has one subpart of
%e A279387 . size 35. The second layer has
%e A279387 . two subparts of size 2, so
%e A279387 . the 18th row of this triangle is
%e A279387 . [1, 2], and the row sum is
%e A279387 . A001227(18) = 3.
%e A279387 (End)
%t A279387 (* function a341969[ ] is defined in A341969 *)
%t A279387 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}&];
%t A279387 cL[[ft]]++; partL=Join[Most[partL], sL]]; cL]
%t A279387 Flatten[a279387[74]] (* the first 74 rows of the table; _Hartmut F. W. Hoft_, Feb 24 2021 *)
%Y A279387 The sum of row n equals A001227(n).
%Y A279387 Hence, if n is odd, the sum of row n equals A000005(n).
%Y A279387 Row n has length A250068(n).
%Y A279387 Column 1 gives A237271.
%Y A279387 For more information about "subparts" see A279388 and A279391.
%Y A279387 Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A243982, A244050, A245092, A249223, A249351, A250070, A262045, A262611, A261699, A262626, A279693, A280850, A280851, A296508.
%Y A279387 Cf. A235791, A237048, A237270, A237591, A237593, A247687, A249223, A250070, A264102, A280851, A341969.
%K A279387 nonn,tabf
%O A279387 1,3
%A A279387 _Omar E. Pol_, Dec 12 2016
%E A279387 Definition edited by _Omar E. Pol_ and _N. J. A. Sloane_, Nov 25 2020