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 A279387 Irregular triangle read by rows: T(n,k) = number of subparts in the k-th layer of the symmetric representation of sigma(n), if such layer exists. 53
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS 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. The number of layers of width 1 in the symmetric representation of sigma(n) coincides with A250068(n). The number of subparts in the first layer of the symmetric representation of sigma(n) coincides with A237271(n). We can find the symmetric representation of sigma(n) as the terraces at the n-th level (starting from the top) of the step pyramid described in A245092. (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). The sum of row n equals the number of subparts in the symmetric representation of sigma(n). Conjecture: The number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n. From Hartmut F. W. Hoft, Dec 16 2016: (Start) Proof: 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) LINKS EXAMPLE Triangle begins (first 15 rows): 1; 1; 2; 1; 2; 1, 1; 2; 1; 3; 2; 2; 1, 1; 2; 2; 3, 1; ... 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: .                          _                                    _ .                         | |                                  | | .                         | |                                  | | .                         | |                                  | | .                         | |                                  | | .                         | |                                  | | .                    _ _ _| |                             _ _ _| | .              28  _|    _ _|                       23  _|  _ _ _| .                _|     |                             _|  _| | .               |      _|                            |  _|  _| .               |  _ _|                              | |_ _| .    _ _ _ _ _ _| |                       _ _ _ _ _ _| |      5 .   |_ _ _ _ _ _ _|                      |_ _ _ _ _ _ _| . .   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 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: .                                _                                  _ .                               | |                                | | .                               | |                                | | .                               | |                                | | .                               | |                                | | .                           8   | |                            8   | | .                               | |                                | | .                               | |                                | | .                          _ _ _|_|                           _ _ _|_| .                   8  _ _| |                          7  _ _| | .                     |    _|                            |  _ _| .                    _|  _|                             _| |_| .                   |_ _|                              |_ _|  1 .           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 we have the 359th 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) 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. For more information about "subparts" see A279388 and A279391. Cf. A000203, A196020, A235791, A236104, A237048, A237270, A237591, A237593, A239657, A243982, A244050, A245092, A249223, A249351, A250070, A262045, A262611, A261699, A262626, A279693. Sequence in context: A225743 A218828 A075117 * A309852 A029810 A321601 Adjacent sequences:  A279384 A279385 A279386 * A279388 A279389 A279390 KEYWORD nonn,tabf AUTHOR Omar E. Pol, Dec 12 2016 STATUS approved

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Last modified October 15 22:25 EDT 2019. Contains 328038 sequences. (Running on oeis4.)