A279387 - OEIS

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

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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) is given in A250068.

The number of subparts in the first layer of the symmetric representation of sigma(n) is equal to 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 stepped 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

Table of n, a(n) for n=1..97.

EXAMPLE

Triangle begins (first 18 rows):

1, 1;

3, 1;

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:

. _ _

. | | | |

. _ _ _| | _ _ _| |

. _| _ _| _| _ _ _|

. _| | _| _| |

. | _| | _| _|

. | _ _| | |_ _|

. _ _ _ _ _ _| | 28 _ _ _ _ _ _| | 5

. |_ _ _ _ _ _ _| |_ _ _ _ _ _ _|

. 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:

. _ _

. | | | |

. _ _ _|_| _ _ _|_|

. _ _| | 8 _ _| | 8

. | _| | _ _|

. _| _| _| |_|

. |_ _| 8 |_ _| 1

. | | 7

. _ _ _ _ _ _ _ _| _ _ _ _ _ _ _ _|

. |_ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _|

. 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:

. _ _

. | | | |

._ | | | |

. | | | |

. _ _ _ _| | _ _ _ _| |

. | _ _ _| | _ _ _ _|

. _| | _| | |

. _| _ _| _| _|_|

. _ _| _| _ _| _| 2

. | | 39 | _ _|

. | _ _| | |_ _|

. | | | | 2

. _ _ _ _ _ _ _ _ _| | _ _ _ _ _ _ _ _ _| |

. |_ _ _ _ _ _ _ _ _ _| |_ _ _ _ _ _ _ _ _ _|

. 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)

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 *)

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.