A339106 -id:A339106 - cp's OEIS frontend

A346533 Irregular triangle read by rows in which row n lists the first n - 2 terms of A000203 together with the sum of A000203(n-1) and A000203(n), with a(1) = 1.

1, 4, 1, 7, 1, 3, 11, 1, 3, 4, 13, 1, 3, 4, 7, 18, 1, 3, 4, 7, 6, 20, 1, 3, 4, 7, 6, 12, 23, 1, 3, 4, 7, 6, 12, 8, 28, 1, 3, 4, 7, 6, 12, 8, 15, 31, 1, 3, 4, 7, 6, 12, 8, 15, 13, 30, 1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 40, 1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 42, 1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 38

Offset: 1

Omar E. Pol, Jul 22 2021

T(n,k) is the total area (or number of cells) of the terraces that are in the k-th level that contains terraces starting from the top of the symmetric tower (a polycube) described in A221529.
The height of the tower equals A000041(n-1).
The terraces of the tower are the symmetric representation of sigma.
The terraces are in the levels that are the partition numbers A000041 starting from the base.
Note that for n >= 2 there are n - 1 terraces because the lower terrace of the tower is formed by two symmetric representations of sigma in the same level.

			Triangle begins:
  1;
  4;
  1, 7;
  1, 3, 11;
  1, 3,  4, 13;
  1, 3,  4,  7, 18;
  1, 3,  4,  7,  6, 20;
  1, 3,  4,  7,  6, 12, 23;
  1, 3,  4,  7,  6, 12,  8, 28;
  1, 3,  4,  7,  6, 12,  8, 15, 31;
  1, 3,  4,  7,  6, 12,  8, 15, 13, 30;
  1, 3,  4,  7,  6, 12,  8, 15, 13, 18, 40;
  1, 3,  4,  7,  6, 12,  8, 15, 13, 18, 12, 42;
  1, 3,  4,  7,  6, 12,  8, 15, 13, 18, 12, 28, 38;
  ...
For n = 7, sigma(7) = 1 + 7 = 8 and sigma(6) = 1 + 2 + 3 + 6 = 12, and 8 + 12 = 20, so the last term of row 7 is T(7,6) = 20. The other terms in row 7 are the first five terms of A000203, so the 7th row of the triangle is [1, 3, 4, 7, 6, 20].
For n = 7 we can see below the top view and the lateral view of the pyramid described in A245092 (with seven levels) and the top view and the lateral view of the tower described in A221529 (with 11 levels).
                                           _
                                          | |
                                          | |
                                          | |
        _                                 |_|_
       |_|_                               |   |
       |_ _|_                             |_ _|_
       |_ _|_|_                           |   | |
       |_ _ _| |_                         |_ _|_|_
       |_ _ _|_ _|_                       |_ _ _| |_
       |_ _ _ _| | |_                     |_ _ _|_ _|_ _
       |_ _ _ _|_|_ _|                    |_ _ _ _|_|_ _|
.
         Figure 1.                           Figure 2.
        Lateral view                       Lateral view
       of the pyramid.                     of the tower.
.
.       _ _ _ _ _ _ _                      _ _ _ _ _ _ _
       |_| | | | | | |                    |_| | | | |   |
       |_ _|_| | | | |                    |_ _|_| | |   |
       |_ _|  _|_| | |                    |_ _|  _|_|   |
       |_ _ _|    _|_|                    |_ _ _|    _ _|
       |_ _ _|  _|                        |_ _ _|  _|
       |_ _ _ _|                          |       |
       |_ _ _ _|                          |_ _ _ _|
.
          Figure 3.                          Figure 4.
          Top view                           Top view
       of the pyramid.                     of the tower.
.
Both polycubes have the same base which has an area equal to A024916(7) = 41 equaling the sum of the 7th row of triangle.
Note that in the top view of the tower the symmetric representation of sigma(6) and the symmetric representation of sigma(7) appear unified in the level 1 of the structure as shown above in the figure 4 (that is due the first two partition numbers A000041 are [1, 1]), so T(7,6) = sigma(7) + sigma(6) = 8 + 12 = 20.
.
Illustration of initial terms:
   Row 1    Row 2      Row 3      Row 4        Row 5          Row 6
.
    1        4         1 7        1 3 11       1 3 4 13       1 3 4 7 18
.   _        _ _       _ _ _      _ _ _ _      _ _ _ _ _      _ _ _ _ _ _
   |_|      |   |     |_|   |    |_| |   |    |_| | |   |    |_| | | |   |
            |_ _|     |    _|    |_ _|   |    |_ _|_|   |    |_ _|_| |   |
                      |_ _|      |      _|    |_ _|  _ _|    |_ _|  _|   |
                                 |_ _ _|      |     |        |_ _ _|    _|
                                              |_ _ _|        |        _|
                                                             |_ _ _ _|
.

Paolo Xausa, Table of n, a(n) for n = 1..11176 (rows 1..150 of the triangle, flattened)
Omar E. Pol, Illustration of the partitions of 10, prism and tower

Mirror of A340584.
The length of row n is A028310(n-1).
Row sums give A024916.
Leading diagonal gives A092403.
Other diagonals give A000203.
Companion of A346562.
Cf. A175254 (volume of the pyramid).
Cf. A066186 (volume of the tower).
Cf. A000041, A221529, A237270, A237593, A245092, A245093 (similar), A336811, A336812, A338156, A339106, A340035.

A346533row[n_]:=If[n==1,{1},Join[DivisorSigma[1,Range[n-2]],{Total[DivisorSigma[1,{n-1,n}]]}]];Array[A346533row,15] (* Paolo Xausa, Oct 23 2023 *)

A340524 Triangle read by rows: T(n,k) = A000005(n-k+1)*A002865(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 2, 0, 2, 0, 1, 3, 0, 2, 1, 2, 0, 2, 2, 2, 4, 0, 3, 2, 4, 2, 2, 0, 2, 3, 4, 4, 4, 4, 0, 4, 2, 6, 4, 8, 4, 3, 0, 2, 4, 4, 6, 8, 8, 7, 4, 0, 4, 2, 8, 4, 12, 8, 14, 8, 2, 0, 3, 4, 4, 8, 8, 12, 14, 16, 12, 6, 0, 4, 3, 8, 4, 16, 8, 21, 16, 24, 14, 2, 0, 2, 4, 6, 8, 8, 16, 14, 24, 24, 28, 21

Offset: 1

Omar E. Pol, Jan 10 2021

Conjecture: the sum of row n equals A138137(n), the total number of parts in the last section of the set of partitions of n.

			Triangle begins:
1;
2, 0;
2, 0, 1;
3, 0, 2, 1;
2, 0, 2, 2, 2;
4, 0, 3, 2, 4, 2;
2, 0, 2, 3, 4, 4,  4;
4, 0, 4, 2, 6, 4,  8,  4;
3, 0, 2, 4, 4, 6,  8,  8,  7;
4, 0, 4, 2, 8, 4, 12,  8, 14,  8;
2, 0, 3, 4, 4, 8,  8, 12, 14, 16, 12;
6, 0, 4, 3, 8, 4, 16,  8, 21, 16, 24, 14;
2, 0, 2, 4, 6, 8,  8, 16, 14, 24, 24, 28, 21;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k   A002865         T(6,k)
--------------------------
1      1   *   4   =   4
2      0   *   2   =   0
3      1   *   3   =   3
4      1   *   2   =   2
5      2   *   2   =   4
6      2   *   1   =   2
.           A000005
--------------------------
The sum of row 6 is 4 + 0 + 3 + 2 + 4 + 2 = 15, equaling A138137(6) = 15.

Row sums give A138137 (conjectured).
Columns 1, 3 and 4 are A000005.
Column 2 gives A000004.
Columns 5 and 6 give A062011.
Columns 7 and 8 give A145154, n >= 1.
Leading diagonal gives A002865.
Cf. A339304 (irregular or expanded version).
Cf. A135010, A138121, A221531, A336811, A339106, A340424, A340426.

f(n) = if (n==0, 1, numbpart(n) - numbpart(n-1)); \\ A002865
T(n, k) = numdiv(n-k+1) * f(k-1); \\ Michel Marcus, Jan 13 2021

A340527 Triangle read by rows: T(n,k) = A024916(n-k+1)*A000041(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 4, 1, 8, 4, 2, 15, 8, 8, 3, 21, 15, 16, 12, 5, 33, 21, 30, 24, 20, 7, 41, 33, 42, 45, 40, 28, 11, 56, 41, 66, 63, 75, 56, 44, 15, 69, 56, 82, 99, 105, 105, 88, 60, 22, 87, 69, 112, 123, 165, 147, 165, 120, 88, 30, 99, 87, 138, 168, 205, 231, 231, 225, 176, 120, 42, 127, 99, 174

Offset: 1

Omar E. Pol, Jan 10 2021

Conjecture 1: T(n,k) is the sum of divisors of the terms that are in the k-th blocks of the first n rows of triangle A176206.
Conjecture 2: the sum of row n equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
Conjecture 3: T(n,k) is also the volume (or number of cubes) of the k-th block of a symmetric tower in which the terraces are the symmetric representation of sigma (n..1) starting from the base respectively (cf. A237270, A237593), hence the total area of the terraces is A024916(n), the same as the area of the base.
The levels of the terraces starting from the base are the first n terms of A000070, that is A000070(0)..A000070(n-1). Hence the differences between levels give the partition numbers A000041, that is A000041(0)..A000041(n-1).
This symmetric tower has the property that its volume (or total number of cubes) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
For another symmetric tower of the same family and whose volume equals A066186(n) see A339106 and A221529.
The above three conjectures are connected due to the correspondence between divisors and partitions (cf. A336811).

			Triangle begins:
   1;
   4,   1;
   8,   4,   2;
  15,   8,   8,   3;
  21,  15,  16,  12,   5;
  33,  21,  30,  24,  20,   7;
  41,  33,  42,  45,  40,  28,  11;
  56,  41,  66,  63,  75,  56,  44,  15;
  69,  56,  82,  99, 105, 105,  88,  60,  22;
  87,  69, 112, 123, 165, 147, 165, 120,  88,  30;
  99,  87, 138, 168, 205, 231, 231, 225, 176, 120,  42;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k   A000041         T(6,k)
1      1  *  33   =   33
2      1  *  21   =   21
3      2  *  15   =   30
4      3  *   8   =   24
5      5  *   4   =   20
6      7  *   1   =    7
.          A024916
--------------------------
The sum of row 6 is 33 + 21 + 30 + 24 + 20 + 7 = 135, equaling A182738(6).

Columns 1 and 2 give A024916.
Column 3 gives A327329.
Leading diagonal gives A000041.
Row sums give A182738.
Cf. A000070, A066186, A176206, A221529, A221531, A237270, A237593, A336811, A336812, A338156, A339106, A340035,  A340424, A340425, A340426, A340524, A340526.

A340525 Triangle read by rows: T(n,k) = A006218(n-k+1)*A002865(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 3, 0, 5, 0, 1, 8, 0, 3, 1, 10, 0, 5, 3, 2, 14, 0, 8, 5, 6, 2, 16, 0, 10, 8, 10, 6, 4, 20, 0, 14, 10, 16, 10, 12, 4, 23, 0, 16, 14, 20, 16, 20, 12, 7, 27, 0, 20, 16, 28, 20, 32, 20, 21, 8, 29, 0, 23, 20, 32, 28, 40, 32, 35, 24, 12, 35, 0, 27, 23, 40, 32, 56, 40, 56, 40, 36, 14

Offset: 1

Omar E. Pol, Jan 10 2021

Conjecture: the sum of row n equals A006128(n), the total number of parts in all partitions of n.

			Triangle begins:
   1;
   3,  0;
   5,  0,  1;
   8,  0,  3,  1;
  10,  0,  5,  3,  2;
  14,  0,  8,  5,  6,  2;
  16,  0, 10,  8, 10,  6,  4;
  20,  0, 14, 10, 16, 10, 12,  4;
  23,  0, 16, 14, 20, 16, 20, 12,  7;
  27,  0, 20, 16, 28, 20, 32, 20, 21,  8;
  29,  0, 23, 20, 32, 28, 40, 32, 35, 24, 12;
  35,  0, 27, 23, 40, 32, 56, 40, 56, 40, 36, 14;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k   A002865         T(6,k)
--------------------------
1      1   *  14   =  14
2      0   *  10   =   0
3      1   *   8   =   8
4      1   *   5   =   5
5      2   *   3   =   6
6      2   *   1   =   2
.           A006218
--------------------------
The sum of row 6 is 14 + 0 + 8 + 5 + 6 + 2 = 35, equaling A006128(6).

Mirror of A245095.
Row sums give A006128 (conjectured).
Columns 1, 3 and 4 are A006218.
Column 2 gives A000004.
Leading diagonal gives A002865.
Cf. A135010, A138121, A221531, A336811, A339106, A340424, A340524, A340426.

A340526 Triangle read by rows: T(n,k) = A006218(n-k+1)*A000041(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 3, 1, 5, 3, 2, 8, 5, 6, 3, 10, 8, 10, 9, 5, 14, 10, 16, 15, 15, 7, 16, 14, 20, 24, 25, 21, 11, 20, 16, 28, 30, 40, 35, 33, 15, 23, 20, 32, 42, 50, 56, 55, 45, 22, 27, 23, 40, 48, 70, 70, 88, 75, 66, 30, 29, 27, 46, 60, 80, 98, 110, 120, 110, 90, 42, 35, 29, 54, 69, 100, 112, 154, 150, 176, 150, 126, 56

Offset: 1

Omar E. Pol, Jan 10 2021

Conjecture 1: T(n,k) is the total number of divisors of the terms that are in the k-th blocks of the first n rows of triangle A176206.
Conjecture 2: the sum of row n equals A284870, the total number of parts in all partitions of all positive integers <= n.
The above conjectures are connected due to the correspondence between divisors and partitions (cf. A336811).

			Triangle begins:
   1;
   3,  1;
   5,  3,  2;
   8,  5,  6,  3;
  10,  8, 10,  9,   5;
  14, 10, 16, 15,  15,   7;
  16, 14, 20, 24,  25,  21,  11;
  20, 16, 28, 30,  40,  35,  33,  15;
  23, 20, 32, 42,  50,  56,  55,  45,  22;
  27, 23, 40, 48,  70,  70,  88,  75,  66,  30;
  29, 27, 46, 60,  80,  98, 110, 120, 110,  90,  42;
  35, 29, 54, 69, 100, 112, 154, 150, 176, 150, 126,  56;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k   A000041         T(6,k)
1      1  *  14   =   14
2      1  *  10   =   10
3      2  *   8   =   16
4      3  *   5   =   15
5      5  *   3   =   15
6      7  *   1   =    7
.          A006218
--------------------------
The sum of row 6 is 14 + 10 + 16 + 15 + 15 + 7 = 77, equaling A284870(6).

Columns 1 and 2 give A006218.
Leading diagonal gives A000041.
Row sums give A284870.
Cf. A176206, A221531, A339106, A340424, A340425, A340524, A340426, A340527, A336811.

f(n) = sum(k=1, n, n\k); \\ A006218
T(n,k) = f(n-k+1)*numbpart(k-1); \\ Michel Marcus, Jan 15 2021

A340579 Triangle read by rows: T(n,k) = A000203(n-k+1)*A000070(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 3, 2, 4, 6, 4, 7, 8, 12, 7, 6, 14, 16, 21, 12, 12, 12, 28, 28, 36, 19, 8, 24, 24, 49, 48, 57, 30, 15, 16, 48, 42, 84, 76, 90, 45, 13, 30, 32, 84, 72, 133, 120, 135, 67, 18, 26, 60, 56, 144, 114, 210, 180, 201, 97, 12, 36, 52, 105, 96, 228, 180, 315, 268, 291, 139, 28, 24, 72, 91

Offset: 1

Omar E. Pol, Jan 12 2021

Consider a symmetric tower (a polycube) in which the terraces are the symmetric representation of sigma (n..1) respectively starting from the base (cf. A237270, A237593). The total area of the terraces equals A024916(n), the same as the area of the base.
The levels of the terraces starting from the base are the first n terms of A000070, that is A000070(0)..A000070(n-1), hence the differences between two successive levels give the partition numbers A000041, that is A000041(0)..A000041(n-1).
T(n,k) is the total volume (or total number of cubes) exactly below the symmetric representation of sigma(n-k+1). In other words: T(n,k) is the total volume (the total number of cubes) exactly below the terraces that are in the k-th level that contains terraces starting from the base.
This symmetric tower has the property that its volume (the total number of cubes) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n. That is due to the correspondence between divisors and partitions (cf. A336811).
The growth of the volume represents the convolution of A000203 and A000070.
The symmetric tower is a member of the family of the pyramid described in A245092.
For another symmetric tower of the same family and whose volume equals A066186(n) see A221529 and A339106.

			Triangle begins:
   1;
   3,   2;
   4,   6,   4;
   7,   8,  12,   7;
   6,  14,  16,  21,  12;
  12,  12,  28,  28,  36,  19;
   8,  24,  24,  49,  48,  57,  30;
  15,  16,  48,  42,  84,  76,  90,  45;
  13,  30,  32,  84,  72, 133, 120, 135,  67;
  18,  26,  60,  56, 144, 114, 210, 180, 201,  97;
  12,  36,  52, 105,  96, 228, 180, 315, 268, 291, 139;
...
For n = 6 the calculation of every term of row 6 is as follows:
-------------------------
k   A000070        T(6,k)
1      1  *  12  =   12
2      2  *  6   =   12
3      4  *  7   =   28
4      7  *  4   =   28
5     12  *  3   =   36
6     19  *  1   =   19
.         A000203
-------------------------
The sum of row 6 is 12 + 12 + 28 + 28 + 36 + 19 = 135, equaling A182738(6).

Row sums give A182738.

Cf. A000070, A000203, A024916, A221529, A221531, A237593, A339106, A340424, A340426, A340524, A340525, A340526, A340527, A340531.

row(n) = vector(n, k, sigma(n-k+1)*sum(i=0, k-1, numbpart(i))); \\ Michel Marcus, Jul 23 2021

A341149 Irregular triangle read by rows T(n,k) in which row n lists n blocks where the m-th block consists of A000203(m) copies of A000041(n-m), with 1 <= m <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 5, 5, 5, 3, 3, 3, 3, 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, 11, 7, 7, 7, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3

Offset: 1

Omar E. Pol, Feb 06 2021

In the n-th row of the triangle the values of the m-th block are the number of cubes that are exactly below every cell of the symmetric representation of sigma(m) in the tower described in A221529 (see figure 5 in the example here).

			Triangle begins:
  1;
  1,1,1,1;
  2,1,1,1,1,1,1,1;
  3,2,2,2,1,1,1,1,1,1,1,1,1,1,1;
  5,3,3,3,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1;
  7,5,5,5,3,3,3,3,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
  ...
For n = 6 we have that:
                                 Row 6                    Row 6 of
m    A000203(m)  A000041(n-m)   block(m)                  A221529
1        1           7          [7]                           7
2        3           5          [5,5,5]                      15
3        4           3          [3,3,3,3]                    12
4        7           2          [2,2,2,2,2,2,2]              14
5        6           1          [1,1,1,1,1,1]                 6
6       12           1          [1,1,1,1,1,1,1,1,1,1,1,1]    12
.
so the 6th row of triangle is [7,5,5,5,3,3,3,3,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1] and the row sums equals A066186(6) = 66.
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). For further information about these two associated objects see A221529.
       _ _ _ _ _ _
  11  |_ _ _      |              6
      |_ _ _|_    |        3     3
      |_ _    |   |          4   2
      |_ _|_ _|_  |      2   2   2      _
   7  |_ _ _    | |            5 1     | |
      |_ _ _|_  | |        3   2 1     |_|_
   5  |_ _    | | |          4 1 1     |   |
      |_ _|_  | | |      2   2 1 1     |_ _|_
   3  |_ _  | | | |        3 1 1 1     |_ _|_|_
   2  |_  | | | | |      2 1 1 1 1     |_ _ _|_|_ _
   1  |_|_|_|_|_|_|    1 1 1 1 1 1     |_ _ _ _|_|_|
.
        Figure 1.        Figure 2.       Figure 3.
       Front view       Partitions     Lateral view
      of the prism         of 6.       of the tower.
      of partitions.
.
                                                                      Row 6 of
                                        _ _ _ _ _ _                    A341148
                                    1  |_| | | |   |    7 5 3 2 1 1       19
                                    2  |_ _|_| |   |    5 5 3 2 1 1       17
                                    3  |_ _|  _|   |    3 3 2 2 1 1       12
                                    4  |_ _ _|    _|    2 2 2 1 1 1        9
                                    5  |        _|      1 1 1 1 1          5
                                    6  |_ _ _ _|        1 1 1 1            4
.
                                         Figure 4.       Figure 5.
                                         Top view         Heights
                                       of the tower.      in the
                                                         top view.
.
Figure 5 shows the heights of the terraces of the tower, or in other words the number of cubes in the column exactly below every cell of the top view. For example: in the 6th row of triangle the first block is [7] because there are seven cubes exactly below the symmetric representation of sigma(1) = 1. The second block is [5, 5, 5] because there are five cubes exactly below every cell of the symmetric representation of sigma(2) = 3. The third block is [3, 3, 3, 3] because there are three cubes exactly below every cell of the symmetric representation of sigma(3) = 4, and so on.
Note that the terraces that are the symmetric representation of sigma(5) and the terraces that are the symmetric representation of sigma(6) both are unified in level 1 of the structure. That is because the first two partition numbers A000041 are [1, 1].

Paolo Xausa, Table of n, a(n) for n = 1..12451 (rows 1..35 of triangle, flattened)

Every column gives A000041.
Row lengths give A024916.
Row sums give the nonzero terms of A066186.
Cf. A000203, A221529, A236104, A237270, A237271, A237593, A337209, A339106, A340584, A341148, A345023.

A341149row[n_]:=Flatten[Array[ConstantArray[PartitionsP[n-#],DivisorSigma[1,#]]&,n]];
nrows=7;Array[A341149row,nrows] (* Paolo Xausa, Jun 20 2022 *)

cp's OEIS Frontend

A346533 Irregular triangle read by rows in which row n lists the first n - 2 terms of A000203 together with the sum of A000203(n-1) and A000203(n), with a(1) = 1.

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A340524 Triangle read by rows: T(n,k) = A000005(n-k+1)*A002865(k-1), 1 <= k <= n.

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PARI

A340527 Triangle read by rows: T(n,k) = A024916(n-k+1)*A000041(k-1), 1 <= k <= n.

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A340525 Triangle read by rows: T(n,k) = A006218(n-k+1)*A002865(k-1), 1 <= k <= n.

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A340526 Triangle read by rows: T(n,k) = A006218(n-k+1)*A000041(k-1), 1 <= k <= n.

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PARI

A340579 Triangle read by rows: T(n,k) = A000203(n-k+1)*A000070(k-1), 1 <= k <= n.

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PARI

A341149 Irregular triangle read by rows T(n,k) in which row n lists n blocks where the m-th block consists of A000203(m) copies of A000041(n-m), with 1 <= m <= n.

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Mathematica