A336811 -id:A336811 - cp's OEIS frontend

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

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.

A340531 Irregular triangle read by rows T(n,k), (n >= 1, k >= 1), in which row n has length is A000070(n-1) and every column k is A024916, the sum of all divisors of all numbers <= n.

Original entry on oeis.org

1, 4, 1, 8, 4, 1, 1, 15, 8, 4, 4, 1, 1, 1, 21, 15, 8, 8, 4, 4, 4, 1, 1, 1, 1, 1, 33, 21, 15, 15, 8, 8, 8, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 41, 33, 21, 21, 15, 15, 15, 8, 8, 8, 8, 8, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 56, 41, 33, 33, 21, 21, 21, 15, 15, 15, 15, 15

Offset: 1

Omar E. Pol, Jan 10 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 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 volume (the number of cells) in the k-th level starting from the base.
This polycube has the property that the volume (the total number of cells) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
A dissection of the symmetric tower is a three-dimensional spiral whose top view is described in A239660.
Other triangles related to the volume of this polycube are A340527 and A340579.
The symmetric tower is a member of the family of the stepped pyramid described in A245092.
For another symmetric tower of the same family and whose volume equals A066186(n) see A340423.
The sum of row n of triangle equals A182738(n). That property is due to the correspondence between divisors and parts. For more information see A336811.

			Triangle begins:
   1;
   4,  1;
   8,  4,  1,  1;
  15,  8,  4,  4, 1, 1, 1;
  21, 15,  8,  8, 4, 4, 4, 1, 1, 1, 1, 1;
  33, 21, 15, 15, 8, 8, 8, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1;
...
For n = 5 the length of row 5 is A000070(4) = 12.
The sum of row 5 is 21 + 15 + 8 + 8 + 4 + 4 + 4 + 1 + 1 + 1 + 1 + 1 = 69, equaling A182738(5).

Row sums give A182738.
Cf. A340527 (a regular version).
Members of the same family are: A176206, A337209, A339258, A340530.
Cf. A221529, A239660, A339278, A339304, A340423, A340529.
Cf. A000070, A024916, A237593, A336811, A338156.

a(m) = A024916(A176206(m)), assuming A176206 has offset 1.

T(n,k) = A024916(A176206(n,k)), assuming A176206 has offset 1.

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

A340529 Irregular triangle read by rows T(n,k), (n >= 1, k >= 1), in which row n has length A000041(n-1) and every column k is A006218.

Original entry on oeis.org

1, 3, 5, 1, 8, 3, 1, 10, 5, 3, 1, 1, 14, 8, 5, 3, 3, 1, 1, 16, 10, 8, 5, 5, 3, 3, 1, 1, 1, 1, 20, 14, 10, 8, 8, 5, 5, 3, 3, 3, 3, 1, 1, 1, 1, 23, 16, 14, 10, 10, 8, 8, 5, 5, 5, 5, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 27, 20, 16, 14, 14, 10, 10, 8, 8, 8, 8, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Omar E. Pol, Jan 10 2021

			Triangle begins:
   1;
   3;
   5,  1;
   8,  3,  1;
  10,  5,  3,  1,  1;
  14,  8,  5,  3,  3, 1, 1;
  16, 10,  8,  5,  5, 3, 3, 1, 1, 1, 1;
  20, 14, 10,  8,  8, 5, 5, 3, 3, 3, 3, 1, 1, 1, 1;
  23, 16, 14, 10, 10, 8, 8, 5, 5, 5, 5, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
For n = 6, the length of row 6 is A000041(5) = 7.
The sum of row 6 is 14 + 8 + 5 + 3 + 3 + 1 + 1 = 35, equaling A006128(6).

Row sums give A006128.
Cf. A340525 (a regular version).
Members of the same family are: A336811, A339278, A339304, A340423.
Cf. A337209, A339258, A340530, A340531.
Cf. A000041, A006218.

a(m) = A006218(A336811(m)).

T(n,k) = A006218(A336811(n,k)).

A340581 Irregular triangle read by rows in which row n has length A014153(n-1) and every column k lists the positive integers A000027, n >= 1, k >= 1.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 2, 1, 1, 1, 1, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 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

Offset: 1

Omar E. Pol, Jan 14 2021

Row n lists in nonincreasing order the first A014153(n-1) terms of A176206.
In other words: row n lists in nonincreasing order the terms of the first n rows of triangle A176206.
Conjecture: all divisors of all terms in row n are also all parts of all partitions of all positive integers <= n.
The conjecture is in accordance with the conjectures in A336811 and in A176206.
A336811 contains the most elementary conjecture about the correspondence divisors/partitions.
The connection with A336811 (the main sequence) is as follows: A336811 --> A176206 --> this sequence.

			Triangle begins:
1;
2, 1, 1;
3, 2, 2, 1, 1, 1, 1;
4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
...
For n = 4, by definition the length of row 6 is A014153(4-1) = A014153(3) = 14, so the row 4 of triangle has 14 terms. Since every column lists the positive integers A000027 so the row 4 is [4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1].
Then we have that the divisors of the numbers of the 4th row are:
.
4th row of the triangle ----------> 4  3  3  2  2  2  2  1  1  1  1  1  1  1
                                    2  1  1  1  1  1  1
                                    1
.
There are fourteen 1's, five 2's, two 3's and one 4.
In total there are 14 + 5 + 2 + 1 = 22 divisors.
On the other hand all partitions of all positive integers <= 4 are as shown below:
.
.    Partition   Partitions    Partitions     Partitions
.       of 1        of 2          of 3           of 4
.
.                                             4
.                                             2  2
.                               3             3  1
.                   2           2  1          2  1  1
.        1          1  1        1  1  1       1  1  1  1
.
In these partitions there are fourteen 1's, five 2's, two 3's and one 4.
In total there are 14 + 5 + 2 + 1 = A284870(4) = 22 parts.
Finally in accordance with the conjecture we can see that all divisors of all numbers in the 4th row of the triangle are the same positive integers as all parts of all partitions of all positive integers <= 4.

Cf. A000041, A000070, A027750, A014153, A176206, A284870, A336811.

A340530 Irregular triangle read by rows T(n,k) in which row n has length is A000070(n-1) and every column k is A006218, (n >= 1, k >= 1).

Original entry on oeis.org

1, 3, 1, 5, 3, 1, 1, 8, 5, 3, 3, 1, 1, 1, 10, 8, 5, 5, 3, 3, 3, 1, 1, 1, 1, 1, 14, 10, 8, 8, 5, 5, 5, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 16, 14, 10, 10, 8, 8, 8, 5, 5, 5, 5, 5, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 20, 16, 14, 14, 10, 10, 10, 8, 8, 8, 8, 8, 5, 5, 5, 5, 5, 5, 5

Offset: 1

Omar E. Pol, Jan 10 2021

The sum of row n equals A284870(n), the total number of parts in all partitions of all positive integers <= n. It is conjectured that this property is due to the correspondence between divisors and partitions. For more information see A336811.

			Triangle begins:
   1;
   3,  1;
   5,  3,  1,  1;
   8,  5,  3,  3, 1, 1, 1;
  10,  8,  5,  5, 3, 3, 3, 1, 1, 1, 1, 1;
  14, 10,  8,  8, 5, 5, 5, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
For n = 5 the length of row 5 is A000070(4) = 12.
The sum of row 5 is 10 + 8 + 5 + 5 + 3 + 3 + 3 + 1 + 1 + 1 + 1 + 1 = 42, equaling A284870(5).

Row sums give A284870.
Cf. A340526 (a regular version).
Members of the same family are: A176206, A337209, A339258, A340531.
Cf. A339278, A339304, A340423, A340529.
Cf. A000070, A006128, A336811.

a(m) = A006218(A176206(m)), assuming A176206 has offset 1.

T(n,k) = A006218(A176206(n,k)), assuming A176206 has offset 1.

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

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

Omar E. Pol, Jul 22 2021

The tower is a geometric object associated to all partitions of n.

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

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

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).
Cf. A221529, A236104, A237270, A237271, A237593, A336811, A338156, A340035, A340584.

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

Omar E. Pol, Jul 22 2021

The tower is a geometric object associated to all partitions of n.

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

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

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).
Cf. A221529, A236104, A237270, A237271, A237593, A336811, A338156, A340035, A340584.

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

cp's OEIS Frontend

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

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A340531 Irregular triangle read by rows T(n,k), (n >= 1, k >= 1), in which row n has length is A000070(n-1) and every column k is A024916, the sum of all divisors of all numbers <= 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

A340529 Irregular triangle read by rows T(n,k), (n >= 1, k >= 1), in which row n has length A000041(n-1) and every column k is A006218.

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A340581 Irregular triangle read by rows in which row n has length A014153(n-1) and every column k lists the positive integers A000027, n >= 1, k >= 1.

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A340530 Irregular triangle read by rows T(n,k) in which row n has length is A000070(n-1) and every column k is A006218, (n >= 1, k >= 1).

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

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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.

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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.

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