cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A345023 a(n) is the surface area of the symmetric tower described in A221529 which is a polycube whose successive terraces are the symmetric representation of sigma A000203(i) (from i = 1 to n) starting from the top and the levels of these terraces are the partition numbers A000041(h-1) (from h = 1 to n) starting from the base.

Original entry on oeis.org

6, 16, 32, 58, 90, 142, 202, 292, 406, 562, 754, 1034, 1370, 1822, 2410, 3176, 4136, 5402, 6982, 9026, 11598, 14838, 18894, 24034, 30396, 38312, 48136, 60288, 75220, 93624, 116104, 143598, 177090, 217770, 267106, 326820, 398804, 485472, 589644, 714564, 864000, 1042524, 1255308
Offset: 1

Views

Author

Omar E. Pol, Jun 05 2021

Keywords

Comments

The largest side of the base of the tower has length n.
The base of the tower is the symmetric representation of A024916(n).
The volume of the tower is equal to A066186(n).
The area of each lateral view of the tower is equal to A000070(n-1).
The growth of the volume of the tower represents the convolution of A000203 and A000041.
The above results are because the correspondence between divisors and partitions described in A338156 and A336812.
The tower is also a member of the family of the stepped pyramid described in A245092.
The equivalent sequence for the surface area of the stepped pyramid is A328366.

Examples

			For n = 7 we can see below some views of two associated polycubes called "prism of partitions" and "tower". Both objects contains the same number of cubes (that property is also valid for n >= 1).
     _ _ _ _ _ _ _
    |_ _ _ _      |                 7
    |_ _ _ _|_    |           4     3
    |_ _ _    |   |             5   2
    |_ _ _|_ _|_  |         3   2   2                                    _
    |_ _ _      | |               6 1                 1                 | |
    |_ _ _|_    | |         3     3 1                 1                 | |
    |_ _    |   | |           4   2 1                 1                 | |
    |_ _|_ _|_  | |       2   2   2 1                 1                _|_|
    |_ _ _    | | |             5 1 1               1 1               |   |
    |_ _ _|_  | | |         3   2 1 1               1 1              _|_ _|
    |_ _    | | | |           4 1 1 1             1 1 1             | |   |
    |_ _|_  | | | |       2   2 1 1 1             1 1 1            _|_|_ _|
    |_ _  | | | | |         3 1 1 1 1           1 1 1 1          _| |_ _ _|
    |_  | | | | | |       2 1 1 1 1 1         1 1 1 1 1      _ _|_ _|_ _ _|
    |_|_|_|_|_|_|_|     1 1 1 1 1 1 1     1 1 1 1 1 1 1     |_ _|_|_ _ _ _|
.
       Figure 1.           Figure 2.        Figure 3.           Figure 4.
   Front view of the      Partitions        Position          Lateral view
  prism of partitions.       of 7.         of the 1's.        of the tower.
.
.
                                                             _ _ _ _ _ _ _
                                                            |   | | | | |_|  1
                                                            |   | | |_|_ _|  2
                                                            |   |_|_  |_ _|  3
                                                            |_ _    |_ _ _|  4
                                                                |_  |_ _ _|  5
                                                                  |       |  6
                                                                  |_ _ _ _|  7
.
                                                               Figure 5.
                                                               Top view
                                                             of the tower.
.
Figure 1 is a two-dimensional diagram of the partitions of 7. The area of the diagram is A066186(7) = 105. Note that the diagram can be interpreted also as the front view of a right prism whose volumen is 1*7*A000041(7) = 1*7*15 = 105, equaling the volume of the tower that appears in the figures 4 and 5.
Figure 2 shows the partitions of 7 in accordance with the diagram.
Note that the shape and the area of the lateral view of the tower are the same as the shape and the area where the 1's are located in the diagram of partitions, see the figures 3 and 4. In this case the mentioned area equals A000070(7-1) = 30.
The connection between these two objects is a representation of the correspondence divisor/part described in A338156. See also A336812.
		

Crossrefs

Programs

  • Mathematica
    Accumulate @ Table[4 * PartitionsP[k-1] + 2 * DivisorSigma[1, k], {k, 1, 50}] (* Amiram Eldar, Jul 14 2021 *)

Formula

a(n) = 4*A000070(n-1) + 2*A024916(n).
a(n) = 4*A000070(n-1) + A327329(n).

A346530 a(n) is the number of faces of the polycube called "tower" described in A221529 where n is the longest side of its base.

Original entry on oeis.org

6, 6, 11, 14, 20, 27, 31, 38, 42, 51, 59
Offset: 1

Views

Author

Omar E. Pol, Jul 22 2021

Keywords

Comments

The tower is a geometric object associated to all partitions of n.
The height of the tower equals A000041(n-1).

Examples

			For n = 1 the tower is a cube, and a cube has 6 faces, so a(1) = 6.
		

Crossrefs

Cf. A000203 (area of the terraces), A000041 (height of the terraces), A066186 (volume), A345023 (surface area), A346531 (number of edges), A346532 (number of vertices).
Cf. A325300 (analog for the pyramid described in A245092).

Formula

a(n) = A346531(n) - A346532(n) + 2 (Euler's formula).

A346531 a(n) is the number of edges of the polycube called "tower" described in A221529 where n is the longest side of its base.

Original entry on oeis.org

12, 12, 27, 36, 51, 72, 84, 105, 117, 144, 165
Offset: 1

Views

Author

Omar E. Pol, Jul 22 2021

Keywords

Comments

The tower is a geometric object associated to all partitions of n.
The height of the tower equals A000041(n-1).

Examples

			For n = 1 the tower is a cube, and a cube has 12 edges, so a(1) = 12.
		

Crossrefs

Cf. A000203 (area of the terraces), A000041 (height of the terraces), A066186 (volume), A345023 (surface area), A346530 (number of faces), A346532 (number of vertices).
Cf. A325301 (analog for the pyramid described in A245092).

Formula

a(n) = A346530(n) + A346532(n) - 2 (Euler's formula).

A346532 a(n) is the number of vertices of the polycube called "tower" described in A221529 where n is the longest side of its base.

Original entry on oeis.org

8, 8, 18, 24, 33, 47, 55, 69, 77, 95, 108
Offset: 1

Views

Author

Omar E. Pol, Jul 22 2021

Keywords

Comments

The height of the tower equals A000041(n-1).

Examples

			For n = 1 the tower is a cube, and a cube has 8 vertices, so a(1) = 8.
		

Crossrefs

Cf. A000203 (area of the terraces), A000041 (height of the terraces), A066186 (volume), A345023 (surface area), A346530 (number of faces), A346531 (number of edges).
Cf. A325302 (analog for the pyramid described in A245092).

Formula

a(n) = A346531(n) - A346530(n) + 2 (Euler's formula).

A341148 Triangle read by rows: T(n,k) is number of cubes in the k-th vertical slice of the polycube called "tower" described in A221529 where n is the longest side of its base, 1 <= k <= n.

Original entry on oeis.org

1, 2, 2, 4, 3, 2, 7, 6, 4, 3, 12, 10, 7, 3, 3, 19, 17, 12, 9, 5, 4, 30, 26, 20, 13, 8, 4, 4, 45, 41, 31, 23, 16, 10, 5, 5, 67, 60, 48, 34, 25, 15, 11, 5, 5, 97, 89, 71, 55, 39, 28, 17, 12, 6, 6, 139, 127, 104, 78, 60, 40, 28, 17, 11, 6, 6, 195, 181, 149, 118, 89, 65, 45, 32, 21, 15, 7, 7
Offset: 1

Views

Author

Omar E. Pol, Feb 06 2021

Keywords

Comments

The row sums of triangle give A066186 because the correspondence divisor/part. For more information see A338156.
For further information about the tower see A221529.

Examples

			Triangle begins:
    1;
    2,   2;
    4,   3,   2;
    7,   6,   4,   3;
   12,  10,   7,   3,  3;
   19,  17,  12,   9,  5,  4;
   30,  26,  20,  13,  8,  4,  4;
   45,  41,  31,  23, 16, 10,  5,  5;
   67,  60,  48,  34, 25, 15, 11,  5,  5;
   97,  89,  71,  55, 39, 28, 17, 12,  6,  6;
  139, 127, 104,  78, 60, 40, 28, 17, 11,  6,  6;
  195, 181, 149, 118, 89, 65, 45, 32, 21, 15,  7,  7;
...
Illustration of initial terms:
              Top view
  n   k       of the tower       Heights        T(n,k)
               _
  1   1       |_|                1                 1
.              _ _
  2   1       |   |              1 1               2
  2   2       |_ _|              1 1               2
.              _ _ _
  3   1       |_|   |            2 1 1             4
  3   2       |    _|            1 1 1             3
  3   3       |_ _|              1 1               2
.              _ _ _ _
  4   1       |_| |   |          3 2 1 1           7
  4   2       |_ _|   |          2 2 1 1           6
  4   3       |      _|          1 1 1 1           4
  4   4       |_ _ _|            1 1 1             3
.              _ _ _ _ _
  5   1       |_| | |   |        5 3 2 1 1        12
  5   2       |_ _|_|   |        3 3 2 1 1        10
  5   3       |_ _|  _ _|        2 2 1 1 1         7
  5   4       |     |            1 1 1             3
  5   5       |_ _ _|            1 1 1             3
.              _ _ _ _ _ _
  6   1       |_| | | |   |      7 5 3 2 1 1      19
  6   2       |_ _|_| |   |      5 5 3 2 1 1      17
  6   3       |_ _|  _|   |      3 3 2 2 1 1      12
  6   4       |_ _ _|    _|      2 2 2 1 1 1       9
  6   5       |        _|        1 1 1 1 1         5
  6   6       |_ _ _ _|          1 1 1 1           4
.
The levels of the terraces of the tower are the partition numbers A000041 starting from the base.
Note that the top view of the tower is essentially the same as the top view of the stepped pyramid described in A245092 except that in the tower both the symmetric representation of sigma(n) and the symmetric representation of sigma(n-1) are unified in the level 1 of the structure because the first two partitions numbers A000041 are [1, 1].
		

Crossrefs

Column 1 gives A000070.
Leading diagonal gives A080513.
Row sums give A066186.

A237593 Triangle read by rows in which row n lists the elements of the n-th row of A237591 followed by the same elements in reverse order.

Original entry on oeis.org

1, 1, 2, 2, 2, 1, 1, 2, 3, 1, 1, 3, 3, 2, 2, 3, 4, 1, 1, 1, 1, 4, 4, 2, 1, 1, 2, 4, 5, 2, 1, 1, 2, 5, 5, 2, 2, 2, 2, 5, 6, 2, 1, 1, 1, 1, 2, 6, 6, 3, 1, 1, 1, 1, 3, 6, 7, 2, 2, 1, 1, 2, 2, 7, 7, 3, 2, 1, 1, 2, 3, 7, 8, 3, 1, 2, 2, 1, 3, 8, 8, 3, 2, 1, 1, 1, 1, 2, 3, 8
Offset: 1

Views

Author

Omar E. Pol, Feb 22 2014

Keywords

Comments

Row n is a palindromic composition of 2*n.
T(n,k) is also the length of the k-th segment in a Dyck path on the first quadrant of the square grid, connecting the x-axis with the y-axis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.
Conjecture 1: the area under the n-th Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.
If the conjecture is true then the n-th Dyck path represents the boundary segments after the alternating sum of the elements of the n-th row of A236104.
Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the n-th Dyck path and the (n-1)-st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.
The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> this sequence --> A239660 --> A237270 --> A237271.
PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000. - Michel Marcus, Mar 27 2014
Mathematica functions have been written that verified the 2 properties through n=30000. - Hartmut F. W. Hoft, Apr 07 2014
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
The symmetric representation of sigma, so defined, is row n of A237270. - Peter Munn, Jan 06 2025
It appears that, for the n-th set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n. - Michel Marcus, Jun 21 2014
Checked Michel Marcus's conjecture with two Mathematica functions up to n=100000, for more information see A240542. - Hartmut F. W. Hoft, Jul 17 2014
A003056(n) is also the number of peaks of the Dyck path related to the n-th row of triangle. - Omar E. Pol, Nov 03 2015
The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the n-th Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092. - Omar E. Pol, Dec 19 2016
The limit as n approaches infinity (area under the Dyck path described in the n-th row of triangle divided by n^2) equals Pi^2/12 = zeta(2)/2. (Cf. A072691.) - Omar E. Pol, Dec 18 2021
The connection between the isosceles triangle and the stepped pyramid is due to the fact that this object can also be interpreted as a pop-up card. - Omar E. Pol, Nov 09 2022

Examples

			Triangle begins:
   n
   1 |  1, 1;
   2 |  2, 2;
   3 |  2, 1, 1, 2;
   4 |  3, 1, 1, 3;
   5 |  3, 2, 2, 3;
   6 |  4, 1, 1, 1, 1, 4;
   7 |  4, 2, 1, 1, 2, 4;
   8 |  5, 2, 1, 1, 2, 5;
   9 |  5, 2, 2, 2, 2, 5;
  10 |  6, 2, 1, 1, 1, 1, 2, 6;
  11 |  6, 3, 1, 1, 1, 1, 3, 6;
  12 |  7, 2, 2, 1, 1, 2, 2, 7;
  13 |  7, 3, 2, 1, 1, 2, 3, 7;
  14 |  8, 3, 1, 2, 2, 1, 3, 8;
  15 |  8, 3, 2, 1, 1, 1, 1, 2, 3, 8;
  16 |  9, 3, 2, 1, 1, 1, 1, 2, 3, 9;
  17 |  9, 4, 2, 1, 1, 1, 1, 2, 4, 9;
  18 | 10, 3, 2, 2, 1, 1, 2, 2, 3, 10;
  19 | 10, 4, 2, 2, 1, 1, 2, 2, 4, 10;
  20 | 11, 4, 2, 1, 2, 2, 1, 2, 4, 11;
  21 | 11, 4, 3, 1, 1, 1, 1, 1, 1, 3, 4, 11;
  22 | 12, 4, 2, 2, 1, 1, 1, 1, 2, 2, 4, 12;
  23 | 12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12;
  24 | 13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13;
  ...
Illustration of rows 8 and 9 interpreted as Dyck paths in the first quadrant and the illustration of the symmetric representation of sigma(9) = 5 + 3 + 5 = 13, see below:
.
y                       y
.                       .
.                       ._ _ _ _ _                _ _ _ _ _ 5
._ _ _ _ _              .         |              |_ _ _ _ _|
.         |             .         |_ _                     |_ _ 3
.         |_            .             |                    |_  |
.           |_ _        .             |_ _                   |_|_ _ 5
.               |       .                 |                      | |
.   Area = 56   |       .    Area = 69    |          Area = 13   | |
.               |       .                 |                      | |
.               |       .                 |                      | |
. . . . . . . . | . x   . . . . . . . . . | . x                  |_|
.
.    Fig. 1                    Fig. 2                  Fig. 3
.
Figure 1. For n = 8 the 8th row of triangle is [5, 2, 1, 1, 2, 5] and the area under the symmetric Dyck path is equal to A024916(8) = 56.
Figure 2. For n = 9 the 9th row of triangle is [5, 2, 2, 2, 2, 5] and the area under the symmetric Dyck path is equal to A024916(9) = 69.
Figure 3. The symmetric representation of sigma(9): between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5].
The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the difference between the areas under the Dyck paths equals the sum of the parts of the symmetric representation of sigma(9) = 69 - 56 = 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
.
Illustration of initial terms as Dyck paths in the first quadrant:
(row n = 1..28)
.  _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
  |_ _ _ _ _ _ _ _ _ _ _ _ _ _  |
  |_ _ _ _ _ _ _ _ _ _ _ _ _ _| |
  |_ _ _ _ _ _ _ _ _ _ _ _ _  | |
  |_ _ _ _ _ _ _ _ _ _ _ _ _| | |
  |_ _ _ _ _ _ _ _ _ _ _ _  | | |_ _ _
  |_ _ _ _ _ _ _ _ _ _ _ _| | |_ _ _  |
  |_ _ _ _ _ _ _ _ _ _ _  | | |_ _  | |_
  |_ _ _ _ _ _ _ _ _ _ _| | |_ _ _| |_  |_
  |_ _ _ _ _ _ _ _ _ _  | |       |_ _|   |_
  |_ _ _ _ _ _ _ _ _ _| | |_ _    |_  |_ _  |_ _
  |_ _ _ _ _ _ _ _ _  | |_ _ _|     |_  | |_ _  |
  |_ _ _ _ _ _ _ _ _| | |_ _  |_      |_|_ _  | |
  |_ _ _ _ _ _ _ _  | |_ _  |_ _|_        | | | |_ _ _ _ _
  |_ _ _ _ _ _ _ _| |     |     | |_ _    | |_|_ _ _ _ _  |
  |_ _ _ _ _ _ _  | |_ _  |_    |_  | |   |_ _ _ _ _  | | |
  |_ _ _ _ _ _ _| |_ _  |_  |_ _  | | |_ _ _ _ _  | | | | |
  |_ _ _ _ _ _  | |_  |_  |_    | |_|_ _ _ _  | | | | | | |
  |_ _ _ _ _ _| |_ _|   |_  |   |_ _ _ _  | | | | | | | | |
  |_ _ _ _ _  |     |_ _  | |_ _ _ _  | | | | | | | | | | |
  |_ _ _ _ _| |_      | |_|_ _ _  | | | | | | | | | | | | |
  |_ _ _ _  |_ _|_    |_ _ _  | | | | | | | | | | | | | | |
  |_ _ _ _| |_  | |_ _ _  | | | | | | | | | | | | | | | | |
  |_ _ _  |_  |_|_ _  | | | | | | | | | | | | | | | | | | |
  |_ _ _|   |_ _  | | | | | | | | | | | | | | | | | | | | |
  |_ _  |_ _  | | | | | | | | | | | | | | | | | | | | | | |
  |_ _|_  | | | | | | | | | | | | | | | | | | | | | | | | |
  |_  | | | | | | | | | | | | | | | | | | | | | | | | | | |
  |_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|_|
.
n: 1 2 3 4 5 6 7 8 9 10..12..14..16..18..20..22..24..26..28
.
It appears that the total area (also the total number of cells) in the first n set of symmetric regions of the diagram is equal to A024916(n), the sum of all divisors of all positive integers <= n.
It appears that the total area (also the total number of cells) in the n-th set of symmetric regions of the diagram is equal to sigma(n) = A000203(n) (checked by hand up n = 128).
From _Omar E. Pol_, Aug 18 2015: (Start)
The above diagram is also the top view of the stepped pyramid described in A245092 and it is also the top view of the staircase described in A244580, in both cases the figure represents the first 28 levels of the structure. Note that the diagram contains (and arises from) a hidden pattern which is shown below.
.
Illustration of initial terms as an isosceles triangle:
Row                                 _ _
1                                 _|1|1|_
2                               _|2 _|_ 2|_
3                             _|2  |1|1|  2|_
4                           _|3   _|1|1|_   3|_
5                         _|3    |2 _|_ 2|    3|_
6                       _|4     _|1|1|1|1|_     4|_
7                     _|4      |2  |1|1|  2|      4|_
8                   _|5       _|2 _|1|1|_ 2|_       5|_
9                 _|5        |2  |2 _|_ 2|  2|        5|_
10              _|6         _|2  |1|1|1|1|  2|_         6|_
11            _|6          |3   _|1|1|1|1|_   3|          6|_
12          _|7           _|2  |2  |1|1|  2|  2|_           7|_
13        _|7            |3    |2 _|1|1|_ 2|    3|            7|_
14      _|8             _|3   _|1|2 _|_ 2|1|_   3|_             8|_
15    _|8              |3    |2  |1|1|1|1|  2|    3|              8|_
16   |9                |3    |2  |1|1|1|1|  2|    3|                9|
...
This diagram is the simpler representation of the sequence.
The number of horizontal line segments in the n-th level in each side of the diagram equals A001227(n), the number of odd divisors of n.
The number of horizontal line segments in the left side of the diagram plus the number of the horizontal line segment in the right side equals A054844(n).
The total number of vertical line segments in the n-th level of the diagram equals A131507(n).
Note that this symmetric pattern also emerges from the front view of the stepped pyramid described in A245092, which is related to sigma A000203, the sum-of-divisors function, and other related sequences. The diagram represents the first 16 levels of the pyramid. (End)
		

Crossrefs

Row n has length 2*A003056(n).
Row sums give A005843, n >= 1.
Column k starts in row A008805(k-1).
Column 1 = right border = A008619, n >= 1.
Bisections are in A259176, A259177.
For further information see A262626.

Programs

  • Mathematica
    row[n_]:=Floor[(Sqrt[8n+1]-1)/2]
    s[n_,k_]:=Ceiling[(n+1)/k-(k+1)/2]-Ceiling[(n+1)/(k+1)-(k+2)/2]
    f[n_,k_]:=If[k<=row[n],s[n,k],s[n,2 row[n]+1-k]]
    TableForm[Table[f[n,k],{n,1,50},{k,1,2 row[n]}]] (* Hartmut F. W. Hoft, Apr 08 2014 *)
  • PARI
    row(n) = {my(orow = row237591(n)); vector(2*#orow, i, if (i <= #orow, orow[i], orow[2*#orow-i+1]));}
    area(n) = {my(rown = row(n)); surf = 0; h = n; odd = 1; for (i=1, #row, if (odd, surf += h*rown[i], h -= rown[i];); odd = !odd;); surf;}
    heights(v, n) = {vh = vector(n); ivh = 1; h = n; odd = 1; for (i=1, #v, if (odd, for (j=1, v[i], vh[ivh] = h; ivh++), h -= v[i];); odd = !odd;); vh;}
    isabove(hb, ha) = {for (i=1, #hb, if (hb[i] < ha[i], return (0));); return (1);}
    chkcross(nn) = {hga = concat(heights(row(1), 1), 0); for (n=2, nn, hgb = heights(row(n), n); if (! isabove(hgb, hga), print("pb cross at n=", n)); hga = concat(hgb, 0););} \\ Michel Marcus, Mar 27 2014
    
  • Python
    from sympy import sqrt
    import math
    def row(n): return int(math.floor((sqrt(8*n + 1) - 1)/2))
    def s(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
    def T(n, k): return s(n, k) if k<=row(n) else s(n, 2*row(n) + 1 - k)
    for n in range(1, 11): print([T(n, k) for k in range(1, 2*row(n) + 1)]) # Indranil Ghosh, Apr 21 2017

Formula

Let j(n)= floor((sqrt(8n+1)-1)/2) then T(n,k) = A237591(n,k), if k <= j(n); otherwise T(n,k) = A237591(n,2*j(n)+1-k). - Hartmut F. W. Hoft, Apr 07 2014 (corrected by Omar E. Pol, May 31 2015)

Extensions

A minor edit to the definition. - N. J. A. Sloane, Jul 31 2025

A237270 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(n).

Original entry on oeis.org

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, 13, 5, 13, 21, 21, 14, 6, 6, 14, 56, 15, 15, 72, 16, 16, 63, 17, 7, 7, 17, 27, 27, 18, 12, 18, 91, 19, 19, 30, 30, 20, 8, 8, 20, 90
Offset: 1

Views

Author

Omar E. Pol, Feb 19 2014

Keywords

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.
For the parts in every quadrant of the spiral see A239931, A239932, A239933, A239934.
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

Examples

			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.
Note that the number of *'s in the diagram is 24^2 - A024916(24) = 576 - 491 = A004125(24) = 85.
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.
For the definition of subparts see A239387 and also A296508, A280851. (End)
		

Crossrefs

Programs

  • 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 *)

Formula

T(n, k) = (A384149(n, k) + A384149(n, m+1-k))/2, where m = A237271(n) is the row length. (conjectured) - Peter Munn, Jun 01 2025

Extensions

Drawing of the spiral extended by Omar E. Pol, Nov 22 2020

A237271 Number of parts in the symmetric representation of sigma(n).

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 1, 3, 2, 4, 1, 2, 1, 2, 1, 4, 2, 3, 1, 2, 2, 4, 1, 2, 1, 2, 2, 3, 2, 2, 1, 3, 3, 4, 2, 2, 1, 4, 1, 4, 2, 2, 1, 2, 2, 5, 1, 4, 1, 2, 2, 4, 3, 2, 1, 2, 2, 4, 2, 3, 2, 2, 1, 5, 2, 2, 1, 4, 2, 4, 1, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Feb 25 2014

Keywords

Comments

The diagram of the symmetry of sigma has been via A196020 --> A236104 --> A235791 --> A237591 --> A237593.
For more information see A237270.
a(n) is also the number of terraces at n-th level (starting from the top) of the stepped pyramid described in A245092. - Omar E. Pol, Apr 20 2016
a(n) is also the number of subparts in the first layer of the symmetric representation of sigma(n). For the definion of "subpart" see A279387. - Omar E. Pol, Dec 08 2016
Note that the number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n. (See the second example). - Omar E. Pol, Dec 20 2016
From Hartmut F. W. Hoft, Dec 26 2016: (Start)
Using odd prime number 3, observe that the 1's in the 3^k-th row of the irregular triangle of A237048 are at index positions
3^0 < 2*3^0 < 3^1 < 2*3^1 < ... < 2*3^((k-1)/2) < 3^(k/2) < ...
the last being 2*3^((k-1)/2) when k is odd and 3^(k/2) when k is even. Since odd and even index positions alternate, each pair (3^i, 2*3^i) specifies one part in the symmetric representation with a center part present when k is even. A straightforward count establishes that the symmetric representation of 3^k, k>=0, has k+1 parts. Since this argument is valid for any odd prime, every positive integer occurs infinitely many times in the sequence. (End)
a(n) = number of runs of consecutive nonzero terms in row n of A262045. - N. J. A. Sloane, Jan 18 2021
Indices of odd terms give A071562. Indices of even terms give A071561. - Omar E. Pol, Feb 01 2021
a(n) is also the number of prisms in the three-dimensional version of the symmetric representation of k*sigma(n) where k is the height of the prisms, with k >= 1. - Omar E. Pol, Jul 01 2021
With a(1) = 0; a(n) is also the number of parts in the symmetric representation of A001065(n), the sum of aliquot parts of n. - Omar E. Pol, Aug 04 2021
The parity of this sequence is also the characteristic function of numbers that have middle divisors. - Omar E. Pol, Sep 30 2021
a(n) is also the number of polycubes in the 3D-version of the ziggurat of order n described in A347186. - Omar E. Pol, Jun 11 2024
Conjecture 1: a(n) is the number of odd divisors of n except the "e" odd divisors described in A005279. Thus a(n) is the length of the n-th row of A379288. - Omar E. Pol, Dec 21 2024
The conjecture 1 was checked up n = 10000 by Amiram Eldar. - Omar E. Pol, Dec 22 2024
The conjecture 1 is true. For a proof see A379288. - Hartmut F. W. Hoft, Jan 21 2025
From Omar E. Pol, Jul 31 2025: (Start)
Conjecture 2: a(n) is the number of 2-dense sublists of divisors of n.
We call "2-dense sublists of divisors of n" to the maximal sublists of divisors of n whose terms increase by a factor of at most 2.
In a 2-dense sublist of divisors of n the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of n.
Example: for n = 10 the list of divisors of 10 is [1, 2, 5, 10]. There are two 2-dense sublists of divisors of 10, they are [1, 2], [5, 10], so a(10) = 2.
The conjecture 2 is essentially the same as the second conjecture in the Comments of A384149. See also Peter Munn's formula in A237270.
The indices where a(n) = 1 give A174973 (2-dense numbers). See the proof there. (End)
Conjecture 3: a(n) is the number of divisors p of n such that p is greater than twice the adjacent previous divisor of n. The divisors p give the n-th row of A379288. - Omar E. Pol, Aug 02 2025

Examples

			Illustration of initial terms (n = 1..12):
---------------------------------------------------------
n   A000203  A237270    a(n)            Diagram
---------------------------------------------------------
.                               _ _ _ _ _ _ _ _ _ _ _ _
1       1      1         1     |_| | | | | | | | | | | |
2       3      3         1     |_ _|_| | | | | | | | | |
3       4      2+2       2     |_ _|  _|_| | | | | | | |
4       7      7         1     |_ _ _|    _|_| | | | | |
5       6      3+3       2     |_ _ _|  _|  _ _|_| | | |
6      12      12        1     |_ _ _ _|  _| |  _ _|_| |
7       8      4+4       2     |_ _ _ _| |_ _|_|    _ _|
8      15      15        1     |_ _ _ _ _|  _|     |
9      13      5+3+5     3     |_ _ _ _ _| |      _|
10     18      9+9       2     |_ _ _ _ _ _|  _ _|
11     12      6+6       2     |_ _ _ _ _ _| |
12     28      28        1     |_ _ _ _ _ _ _|
...
For n = 9 the sum of divisors of 9 is 1+3+9 = A000203(9) = 13. On the other hand the 9th set of symmetric regions of the diagram is formed by three regions (or parts) with 5, 3 and 5 cells, so the total number of cells is 5+3+5 = 13, equaling the sum of divisors of 9. There are three parts: [5, 3, 5], so a(9) = 3.
From _Omar E. Pol_, Dec 21 2016: (Start)
Illustration of the diagram of subparts (n = 1..12):
---------------------------------------------------------
n   A000203  A279391  A001227           Diagram
---------------------------------------------------------
.                               _ _ _ _ _ _ _ _ _ _ _ _
1       1      1         1     |_| | | | | | | | | | | |
2       3      3         1     |_ _|_| | | | | | | | | |
3       4      2+2       2     |_ _|  _|_| | | | | | | |
4       7      7         1     |_ _ _|  _ _|_| | | | | |
5       6      3+3       2     |_ _ _| |_|  _ _|_| | | |
6      12      11+1      2     |_ _ _ _|  _| |  _ _|_| |
7       8      4+4       2     |_ _ _ _| |_ _|_|  _ _ _|
8      15      15        1     |_ _ _ _ _|  _|  _| |
9      13      5+3+5     3     |_ _ _ _ _| |  _|  _|
10     18      9+9       2     |_ _ _ _ _ _| |_ _|
11     12      6+6       2     |_ _ _ _ _ _| |
12     28      23+5      2     |_ _ _ _ _ _ _|
...
For n = 6 the symmetric representation of sigma(6) has two subparts: [11, 1], so A000203(6) = 12 and A001227(6) = 2.
For n = 12 the symmetric representation of sigma(12) has two subparts: [23, 5], so A000203(12) = 28 and A001227(12) = 2. (End)
From _Hartmut F. W. Hoft_, Dec 26 2016: (Start)
Two examples of the general argument in the Comments section:
Rows 27 in A237048 and A249223 (4 parts)
i:  1  2 3 4 5 6 7 8 9 . . 12
27: 1  1 1 0 0 1                           1's in A237048 for odd divisors
    1 27 3     9                           odd divisors represented
27: 1  0 1 1 1 0 0 1 1 1 0 1               blocks forming parts in A249223
Rows 81 in A237048 and A249223 (5 parts)
i:  1  2 3 4 5 6 7 8 9 . . 12. . . 16. . . 20. . . 24
81: 1  1 1 0 0 1 0 0 1 0 0 0                          1's in A237048 f.o.d
    1 81 3    27     9                                odd div. represented
81: 1  0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1  blocks fp in A249223
(End)
		

Crossrefs

Programs

  • Mathematica
    a237271[n_] := Length[a237270[n]] (* code defined in A237270 *)
    Map[a237271, Range[90]] (* data *)
    (* Hartmut F. W. Hoft, Jun 23 2014 *)
    a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, 1 + Count[d, ?(OddQ[#[[2]]] && #[[2]] >= 2*#[[1]] &)]]; Array[a, 100] (* _Amiram Eldar,  Dec 22 2024 *)
  • PARI
    fill(vcells, hga, hgb) = {ic = 1; for (i=1, #hgb, if (hga[i] < hgb[i], for (j=hga[i], hgb[i]-1, cell = vector(4); cell[1] = i - 1; cell[2] = j; vcells[ic] = cell; ic ++;););); vcells;}
    findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
    findxy(vcells, x, y) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[1]==x) && (vcelli[2]==y) && (vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
    findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == iz) && (vcelli[4] == 0), return (i)); ); return (0);}
    zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i][3]);); nbz;}
    docell(vcells, ic, iz) = {x = vcells[ic][1]; y = vcells[ic][2]; if (icdo = findxy(vcells, x-1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x+1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y-1), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y+1), vcells[icdo][3] = iz); vcells[ic][4] = 1; vcells;}
    docells(vcells, ic, iz) = {vcells[ic][3] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz);); vcells;}
    nbzb(n, hga, hgb) = {vcells = vector(sigma(n)); vcells = fill(vcells, hga, hgb); iz = 1; while (ic = findfree(vcells), vcells = docells(vcells, ic, iz); iz++;); zcount(vcells);}
    lista(nn) = {hga = concat(heights(row237593(0), 0), 0); for (n=1, nn, hgb = heights(row237593(n), n); nbz = nbzb(n, hga, hgb); print1(nbz, ", "); hga = concat(hgb, 0););} \\ with heights() also defined in A237593; \\ Michel Marcus, Mar 28 2014
    
  • Python
    from sympy import divisors
    def a(n: int) -> int:
        divs = list(divisors(n))
        d = [divs[i:i+2] for i in range(len(divs) - 1)]
        s = sum(1 for pair in d if len(pair) == 2 and pair[1] % 2 == 1 and pair[1] >= 2 * pair[0])
        return s + 1
    print([a(n) for n in range(1, 80)])  # Peter Luschny, Aug 05 2025

Formula

a(n) = A001227(n) - A239657(n). - Omar E. Pol, Mar 23 2014
a(p^k) = k + 1, where p is an odd prime and k >= 0. - Hartmut F. W. Hoft, Dec 26 2016
Theorem: a(n) <= number of odd divisors of n (cf. A001227). The differences are in A239657. - N. J. A. Sloane, Jan 19 2021
a(n) = A340846(n) - A340833(n) + 1 (Euler's formula). - Omar E. Pol, Feb 01 2021
a(n) = A000005(n) - A243982(n). - Omar E. Pol, Aug 02 2025

A066186 Sum of all parts of all partitions of n.

Original entry on oeis.org

0, 1, 4, 9, 20, 35, 66, 105, 176, 270, 420, 616, 924, 1313, 1890, 2640, 3696, 5049, 6930, 9310, 12540, 16632, 22044, 28865, 37800, 48950, 63336, 81270, 104104, 132385, 168120, 212102, 267168, 334719, 418540, 520905, 647172, 800569, 988570, 1216215, 1493520
Offset: 0

Views

Author

Wouter Meeussen, Dec 15 2001

Keywords

Comments

Sum of the zeroth moments of all partitions of n.
Also the number of one-element transitions from the integer partitions of n to the partitions of n-1 for labeled parts with the assumption that any part z is composed of labeled elements of amount 1, i.e., z = 1_1 + 1_2 + ... + 1_z. Then one can take from z a single element in z different ways. E.g., for n=3 to n=2 we have A066186(3) = 9 and [111] --> [11], [111] --> [11], [111] --> [11], [12] --> [111], [12] --> [111], [12] --> [2], [3] --> 2, [3] --> 2, [3] --> 2. For the unlabeled case, one can take a single element from z in only one way. Then the number of one-element transitions from the integer partitions of n to the partitions of n-1 is given by A000070. E.g., A000070(3) = 4 and for the transition from n=3 to n=2 one has [111] --> [11], [12] --> [11], [12] --> [2], [3] --> [2]. - Thomas Wieder, May 20 2004
Also sum of all parts of all regions of n (Cf. A206437). - Omar E. Pol, Jan 13 2013
From Omar E. Pol, Jan 19 2021: (Start)
Apart from initial zero this is also as follows:
Convolution of A000203 and A000041.
Convolution of A024916 and A002865.
For n >= 1, a(n) is also the number of cells in a symmetric polycube in which the terraces are the symmetric representation of sigma(k), for k = n..1, (cf. A237593) starting from the base and located at the levels A000041(0)..A000041(n-1) respectively. The polycube looks like a symmetric tower (cf. A221529). A dissection is a three-dimensional spiral whose top view is described in A239660. The growth of the volume of the polycube represents each convolution mentioned above. (End)
From Omar E. Pol, Feb 04 2021: (Start)
a(n) is also the sum of all divisors of all positive integers in a sequence with n blocks where the m-th block consists of A000041(n-m) copies of m, with 1 <= m <= n. The mentioned divisors are also all parts of all partitions of n.
Apart from initial zero this is also the convolution of A340793 and A000070. (End)

Examples

			a(3)=9 because the partitions of 3 are: 3, 2+1 and 1+1+1; and (3) + (2+1) + (1+1+1) = 9.
a(4)=20 because A000041(4)=5 and 4*5=20.
		

Crossrefs

Cf. A000041, A093694, A000070, A132825, A001787 (same for ordered partitions), A277029, A000203, A221529, A237593, A239660.
First differences give A138879. - Omar E. Pol, Aug 16 2013

Programs

  • Haskell
    a066186 = sum . concat . ps 1 where
       ps _ 0 = [[]]
       ps i j = [t:ts | t <- [i..j], ts <- ps t (j - t)]
    -- Reinhard Zumkeller, Jul 13 2013
    
  • Maple
    with(combinat): a:= n-> n*numbpart(n): seq(a(n), n=0..50); # Zerinvary Lajos, Apr 25 2007
  • Mathematica
    PartitionsP[ Range[0, 60] ] * Range[0, 60]
  • PARI
    a(n)=numbpart(n)*n \\ Charles R Greathouse IV, Mar 10 2012
    
  • Python
    from sympy import npartitions
    def A066186(n): return n*npartitions(n) # Chai Wah Wu, Oct 22 2023
  • Sage
    [n*Partitions(n).cardinality() for n in range(41)] # Peter Luschny, Jul 29 2014
    

Formula

a(n) = n * A000041(n). - Omar E. Pol, Oct 10 2011
G.f.: x * (d/dx) Product_{k>=1} 1/(1-x^k), i.e., derivative of g.f. for A000041. - Jon Perry, Mar 17 2004 (adjusted to match the offset by Geoffrey Critzer, Nov 29 2014)
Equals A132825 * [1, 2, 3, ...]. - Gary W. Adamson, Sep 02 2007
a(n) = A066967(n) + A066966(n). - Omar E. Pol, Mar 10 2012
a(n) = A207381(n) + A207382(n). - Omar E. Pol, Mar 13 2012
a(n) = A006128(n) + A196087(n). - Omar E. Pol, Apr 22 2012
a(n) = A220909(n)/2. - Omar E. Pol, Jan 13 2013
a(n) = Sum_{k=1..n} A000203(k)*A000041(n-k), n >= 1. - Omar E. Pol, Jan 20 2013
a(n) = Sum_{k=1..n} k*A036043(n,n-k+1). - L. Edson Jeffery, Aug 03 2013
a(n) = Sum_{k=1..n} A024916(k)*A002865(n-k), n >= 1. - Omar E. Pol, Jul 13 2014
a(n) ~ exp(Pi*sqrt(2*n/3))/(4*sqrt(3)) * (1 - (sqrt(3/2)/Pi + Pi/(24*sqrt(6))) / sqrt(n)). - Vaclav Kotesovec, Oct 24 2016
a(n) = Sum_{k=1..n} A340793(k)*A000070(n-k), n >= 1. - Omar E. Pol, Feb 04 2021

Extensions

a(0) added by Franklin T. Adams-Watters, Jul 28 2014

A336811 Irregular triangle read by rows T(n,k) in which the length of row n equals the partition number A000041(n-1) and every column k gives the positive integers A000027, with n >= 1 and k >= 1.

Original entry on oeis.org

1, 2, 3, 1, 4, 2, 1, 5, 3, 2, 1, 1, 6, 4, 3, 2, 2, 1, 1, 7, 5, 4, 3, 3, 2, 2, 1, 1, 1, 1, 8, 6, 5, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 9, 7, 6, 5, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 10, 8, 7, 6, 6, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Nov 20 2020

Keywords

Comments

In other words: row n lists A028310(n-1) blocks where the m-th block consists of A187219(m) copies of n - m + [m=1], with n >= 1 and m >= 1, where [] is the Iverson bracket. [Corrected by Paolo Xausa, Feb 10 2023]
All divisors of all terms in row n are also all parts in the last section of the set of partitions of n.
Thus all divisors of all terms of the first n rows of triangle are also all parts of all partitions of n. In other words: all divisors of the first A000070(n-1) terms of the sequence are also all parts of all partitions of n. - Omar E. Pol, Jun 19 2021
From Omar E. Pol, Jul 31 2021: (Start)
The number of k's in row n is equal to A002865(n-k), 1 <= k <= n.
The number of terms >= k in row n is equal to A000041(n-k), 1 <= k <= n.
The number of k's in the first n rows (or in the first A000070(n-1) terms of the sequence) is equal to A000041(n-k), 1 <= k <= n.
The number of terms >= k in the first n rows (or in the first A000070(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.
First n rows of triangle (or first A000070(n-1) terms of the sequence) in nonincreasing order give the n-th row of A176206. (End)

Examples

			Triangle begins:
1;
2;
3, 1;
4, 2, 1;
5, 3, 2, 1, 1;
6, 4, 3, 2, 2, 1, 1;
7, 5, 4, 3, 3, 2, 2, 1, 1, 1, 1;
8, 6, 5, 4, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1;
9, 7, 6, 5, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
...
For n = 6, by definition the length of row 6 is A000041(6-1) = A000041(5) = 7, so the row 6 of triangle has seven terms. Since every column lists the positive integers A000027 so the row 6 is [6, 4, 3, 2, 2, 1, 1].
Then we have that the divisors of the numbers of the 6th row are:
.
6th row of the triangle ---------->   6 4 3 2 2 1 1
                                      3 2 1 1 1
                                      2 1
                                      1
.
There are seven 1's, four 2's, two 3's, one 4 and one 6.
In total there are 7 + 4 + 2 + 1 + 1 = 15 divisors.
On the other hand the last section of the set of the partitions of 6 can be represented in several ways, five of them as shown below:
._ _ _ _ _ _
|_ _ _      |       6    6                  6                       6
|_ _ _|_    |     3 3    3 3              3   3                     3   3
|_ _    |   |     4 2    4 2            4       2                     4     2
|_ _|_ _|_  |   2 2 2    2 2 2        2   2       2                 2 2   2
          | |       1      1                        1                           1
          | |       1        1                        1                       1
          | |       1        1                          1                   1
          | |       1          1                          1               1
          | |       1          1                            1           1
          | |       1            1                            1       1
          |_|       1              1                            1   1
.
   Figure 1.  Figure 2.  Figure 3.        Figure 4.                   Figure 5.
.
In every figure there are seven 1's, four 2's, two 3's, one 4 and one 6, as shown also the 6th row of A182703.
In total there are 7 + 4 + 2 + 1 + 1 = A138137(6) = 15 parts in every figure.
Figure 5 is an arrangement that shows the correspondence between divisors and parts since the columns give the divisors of the terms of 6th row of triangle.
Finally we can see that all divisors of all numbers in the 6th row of the triangle are the same positive integers as all parts in the last section of the set of the partitions of 6.
Example edited by _Omar E. Pol_, Aug 10 2021
		

Crossrefs

Row sums give A000070.
Row n has length A000041(n-1).
Every column k gives A000027.
Companion of A176206.

Programs

  • Mathematica
    A336811[row_]:=Flatten[Table[ConstantArray[row-m,PartitionsP[m]-PartitionsP[m-1]],{m,0,row-1}]];
    Array[A336811,10] (* Generates 10 rows *) (* Paolo Xausa, Feb 10 2023 *)
  • PARI
    f(n) = numbpart(n-1);
    T(n, k) = {if (k > f(n), error("invalid k")); if (k==1, return (n)); my(s=0); while (k <= f(n-1), s++; n--;); 1+s;}
    tabf(nn) = {for (n=1, nn, for (k=1, f(n), print1(T(n,k), ", ");); print;);} \\ Michel Marcus, Jan 13 2021
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