OFFSET
1,2
COMMENTS
T(n,k) is the number of cells in the k-th region of the n-th set of regions in a diagram of the symmetry of sigma(n), see example.
Row n is a palindromic composition of sigma(n).
Row sums give A000203.
Row n has length A237271(n).
In the row 2n-1 of triangle both the first term and the last term are equal to n.
If n is an odd prime then row n is [m, m], where m = (1 + n)/2.
The connection with A196020 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> this sequence.
For the boundary segments in an octant see A237591.
For the boundary segments in a quadrant see A237593.
For the boundary segments in the spiral see also A239660.
We can find the spiral on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016
T(n,k) is also the area of the k-th terrace, from left to right, at the n-th level, starting from the top, of the stepped pyramid described in A245092 (see Links section). - Omar E. Pol, Aug 14 2018
LINKS
Robert Price, Table of n, a(n) for n = 1..15542 (rows n = 1..5000, flattened)
Hartmut F. W. Hoft, Sample visual documentation for Mathematica code
Omar E. Pol, Perspective view of the stepped pyramid (16 levels)
Omar E. Pol, Perspective view of the stepped pyramid into four quadrants (11 levels). This is formed by combing four copies of the pyramid back-to-back (cf. A244050).
N. J. A. Sloane, Another drawing of the spiral
N. J. A. Sloane, Spiral showing only the outer boundary
EXAMPLE
Illustration of the first 27 terms as regions (or parts) of a spiral constructed with the first 15.5 rows of A239660:
.
. _ _ _ _ _ _ _ _
. | _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7
. | | |_ _ _ _ _ _ _|
. 12 _| | |
. |_ _| _ _ _ _ _ _ |_ _
. 12 _ _| | _ _ _ _ _|_ _ _ _ _ 5 |_
. _ _ _| | 9 _| | |_ _ _ _ _| |
. | _ _ _| 9 _|_ _| |_ _ 3 |_ _ _ 7
. | | _ _| | _ _ _ _ |_ | | |
. | | | _ _| 12 _| _ _ _|_ _ _ 3 |_|_ _ 5 | |
. | | | | _| | |_ _ _| | | | |
. | | | | | _ _| |_ _ 3 | | | |
. | | | | | | 3 _ _ | | | | | |
. | | | | | | | _|_ 1 | | | | | |
. _|_| _|_| _|_| _|_| |_| _|_| _|_| _|_| _
. | | | | | | | | | | | | | | | |
. | | | | | | |_|_ _ _| | | | | | | |
. | | | | | | 2 |_ _|_ _| _| | | | | | |
. | | | | |_|_ 2 |_ _ _|7 _ _| | | | | |
. | | | | 4 |_ _| _ _| | | | |
. | | |_|_ _ |_ _ _ _ | _| _ _ _| | | |
. | | 6 |_ |_ _ _ _|_ _ _ _| | 15 _| _ _| | |
. |_|_ _ _ |_ 4 |_ _ _ _ _| _| | _ _ _| |
. 8 | |_ _ | | _| | _ _ _|
. |_ | |_ _ _ _ _ _ | _ _|28 _| |
. |_ |_ |_ _ _ _ _ _|_ _ _ _ _ _| | _| _|
. 8 |_ _| 6 |_ _ _ _ _ _ _| _ _| _|
. | | _ _| 31
. |_ _ _ _ _ _ _ _ | |
. |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |
. 8 |_ _ _ _ _ _ _ _ _|
.
.
[For two other drawings of the spiral see the links. - N. J. A. Sloane, Nov 16 2020]
If the sequence does not contain negative terms then its terms can be represented in a quadrant. For the construction of the diagram we use the symmetric Dyck paths of A237593 as shown below:
---------------------------------------------------------------
Triangle Diagram of the symmetry of sigma (n = 1..24)
---------------------------------------------------------------
. _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1; |_| | | | | | | | | | | | | | | | | | | | | | | |
3; |_ _|_| | | | | | | | | | | | | | | | | | | | | |
2, 2; |_ _| _|_| | | | | | | | | | | | | | | | | | | |
7; |_ _ _| _|_| | | | | | | | | | | | | | | | | |
3, 3; |_ _ _| _| _ _|_| | | | | | | | | | | | | | | |
12; |_ _ _ _| _| | _ _|_| | | | | | | | | | | | | |
4, 4; |_ _ _ _| |_ _|_| _ _|_| | | | | | | | | | | |
15; |_ _ _ _ _| _| | _ _ _|_| | | | | | | | | |
5, 3, 5; |_ _ _ _ _| | _|_| | _ _ _|_| | | | | | | |
9, 9; |_ _ _ _ _ _| _ _| _| | _ _ _|_| | | | | |
6, 6; |_ _ _ _ _ _| | _| _| _| | _ _ _ _|_| | | |
28; |_ _ _ _ _ _ _| |_ _| _| _ _| | | _ _ _ _|_| |
7, 7; |_ _ _ _ _ _ _| | _ _| _| _| | | _ _ _ _|
12, 12; |_ _ _ _ _ _ _ _| | | | _|_| |* * * *
8, 8, 8; |_ _ _ _ _ _ _ _| | _ _| _ _|_| |* * * *
31; |_ _ _ _ _ _ _ _ _| | _ _| _| _ _|* * * *
9, 9; |_ _ _ _ _ _ _ _ _| | |_ _ _| _|* * * * * *
39; |_ _ _ _ _ _ _ _ _ _| | _ _| _|* * * * * * *
10, 10; |_ _ _ _ _ _ _ _ _ _| | | |* * * * * * * *
42; |_ _ _ _ _ _ _ _ _ _ _| | _ _ _|* * * * * * * *
11, 5, 5, 11; |_ _ _ _ _ _ _ _ _ _ _| | |* * * * * * * * * * *
18, 18; |_ _ _ _ _ _ _ _ _ _ _ _| |* * * * * * * * * * *
12, 12; |_ _ _ _ _ _ _ _ _ _ _ _| |* * * * * * * * * * *
60; |_ _ _ _ _ _ _ _ _ _ _ _ _|* * * * * * * * * * *
...
The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n), the sum of all divisors of all positive integers <= n, hence the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n).
For n = 9 the 9th row of A237593 is [5, 2, 2, 2, 2, 5] and the 8th row of A237593 is [5, 2, 1, 1, 2, 5] therefore between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5], so row 9 is [5, 3, 5].
The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the sum of the parts of the symmetric representation of sigma(9) is 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
For n = 24 the 24th row of A237593 is [13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13] and the 23rd row of A237593 is [12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12] therefore between both symmetric Dyck paths there are only one region (or part) of size 60, so row 24 is 60.
The sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = A000203(24) = 60. On the other hand the sum of the parts of the symmetric representation of sigma(24) is 60, equaling the sum of divisors of 24.
From Omar E. Pol, Nov 22 2020: (Start)
Also consider the infinite double-staircases diagram defined in A335616 (see the theorem).
For n = 15 the diagram with first 15 levels looks like this:
.
Level "Double-staircases" diagram
. _
1 _|1|_
2 _|1 _ 1|_
3 _|1 |1| 1|_
4 _|1 _| |_ 1|_
5 _|1 |1 _ 1| 1|_
6 _|1 _| |1| |_ 1|_
7 _|1 |1 | | 1| 1|_
8 _|1 _| _| |_ |_ 1|_
9 _|1 |1 |1 _ 1| 1| 1|_
10 _|1 _| | |1| | |_ 1|_
11 _|1 |1 _| | | |_ 1| 1|_
12 _|1 _| |1 | | 1| |_ 1|_
13 _|1 |1 | _| |_ | 1| 1|_
14 _|1 _| _| |1 _ 1| |_ |_ 1|_
15 |1 |1 |1 | |1| | 1| 1| 1|
.
Starting from A196020 and after the algorithm described in A280850 and A296508 applied to the above diagram we have a new diagram as shown below:
.
Level "Ziggurat" diagram
. _
6 |1|
7 _ | | _
8 _|1| _| |_ |1|_
9 _|1 | |1 1| | 1|_
10 _|1 | | | | 1|_
11 _|1 | _| |_ | 1|_
12 _|1 | |1 1| | 1|_
13 _|1 | | | | 1|_
14 _|1 | _| _ |_ | 1|_
15 |1 | |1 |1| 1| | 1|
.
The 15th row
of A249351 : [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]
The 15th row
of triangle: [ 8, 8, 8 ]
The 15th row
of A296508: [ 8, 7, 1, 0, 8 ]
The 15th row
of A280851 [ 8, 7, 1, 8 ]
.
More generally, for n >= 1, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n.
MATHEMATICA
T[n_, k_] := Ceiling[(n + 1)/k - (k + 1)/2] (* from A235791 *)
path[n_] := Module[{c = Floor[(Sqrt[8n + 1] - 1)/2], h, r, d, rd, k, p = {{0, n}}}, h = Map[T[n, #] - T[n, # + 1] &, Range[c]]; r = Join[h, Reverse[h]]; d = Flatten[Table[{{1, 0}, {0, -1}}, {c}], 1];
rd = Transpose[{r, d}]; For[k = 1, k <= 2c, k++, p = Join[p, Map[Last[p] + rd[[k, 2]] * # &, Range[rd[[k, 1]]]]]]; p]
segments[n_] := SplitBy[Map[Min, Drop[Drop[path[n], 1], -1] - path[n - 1]], # == 0 &]
a237270[n_] := Select[Map[Apply[Plus, #] &, segments[n]], # != 0 &]
Flatten[Map[a237270, Range[40]]] (* data *)
(* Hartmut F. W. Hoft, Jun 23 2014 *)
CROSSREFS
Cf. A000203, A004125, A023196, A024916, A153485, A196020, A221529, A231347, A235791, A235796, A236104, A236112, A236540, A237046, A237048, A237271, A237590, A237591, A237593, A239050, A239660, A239663, A239665, A239931, A239932, A239933, A239934, A240020, A240062, A244050, A245092, A249351, A262626, A280850, A280851, A296508, A335616, A340035.
KEYWORD
AUTHOR
Omar E. Pol, Feb 19 2014
EXTENSIONS
Drawing of the spiral extended by Omar E. Pol, Nov 22 2020
STATUS
approved