A340426 -id:A340426 - cp's OEIS frontend

A138879 Sum of all parts of the last section of the set of partitions of n.

1, 3, 5, 11, 15, 31, 39, 71, 94, 150, 196, 308, 389, 577, 750, 1056, 1353, 1881, 2380, 3230, 4092, 5412, 6821, 8935, 11150, 14386, 17934, 22834, 28281, 35735, 43982, 55066, 67551, 83821, 102365, 126267, 153397, 188001, 227645, 277305, 334383

Offset: 1

Omar E. Pol, Apr 30 2008

nonn

Row sums of the triangles A135010, A138121, A138151 and others related to the section model of partitions (see A135010 and A138121).
From Omar E. Pol, Jan 20 2021: (Start)
Convolution of A000203 and A002865.
Convolution of A340793 and A000041.
Row sums of triangles A339278, A340426, A340583. (End)
a(n) is also the sum of all divisors of all terms of n-th row of A336811. These divisors are also all parts in the last section of the set of partitions of n. - Omar E. Pol, Jul 27 2021
Row sums of A336812. - Omar E. Pol, Aug 03 2021

			a(6)=31 because the parts of the last section of the set of partitions of 6 are (6), (3,3), (4,2), (2,2,2), (1), (1), (1), (1), (1), (1), (1), so the sum is a(6) = 6 + 3 + 3 + 4 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 31.
From _Omar E. Pol_, Aug 13 2013: (Start)
Illustration of initial terms:
.                                           _ _ _ _ _ _
.                                          |_ _ _ _ _ _|
.                                          |_ _ _|_ _ _|
.                                          |_ _ _ _|_ _|
.                               _ _ _ _ _  |_ _|_ _|_ _|
.                              |_ _ _ _ _|           |_|
.                     _ _ _ _  |_ _ _|_ _|           |_|
.                    |_ _ _ _|         |_|           |_|
.             _ _ _  |_ _|_ _|         |_|           |_|
.       _ _  |_ _ _|       |_|         |_|           |_|
.   _  |_ _|     |_|       |_|         |_|           |_|
.  |_|   |_|     |_|       |_|         |_|           |_|
.
.   1    3      5        11         15           31
.
(End)
On the other hand for n = 6 the 6th row of triangle A336811 is [6, 4, 3, 2, 2, 1, 1] and the sum of all divisors of these terms is [1 + 2 + 3 + 6] + [1 + 2 + 4] + [1 + 3] + [1 + 2] + [1 + 2] + [1] + [1] = 31, so a(6) = 31. - _Omar E. Pol_, Jul 27 2021

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

Cf. A000041, A000203, A002865, A066186, A133041, A135010, A138121, A138135 - A138138, A138151, A138880, A139100, A237593, A336811, A336812, A338156, A339278, A340035, A340426, A340583, A340793.

A066186 := proc(n) n*combinat[numbpart](n) ; end proc:
A138879 := proc(n) A066186(n)-A066186(n-1) ; end proc:
seq(A138879(n),n=1..80) ; # R. J. Mathar, Jan 27 2011

Table[PartitionsP[n]*n - PartitionsP[n-1]*(n-1), {n, 1, 50}] (* Vaclav Kotesovec, Oct 21 2016 *)

for(n=1, 50, print1(numbpart(n)*n - numbpart(n - 1)*(n - 1),", ")) \\ Indranil Ghosh, Mar 19 2017

from sympy.ntheory import npartitions
print([npartitions(n)*n - npartitions(n - 1)*(n - 1) for n in range(1, 51)]) # Indranil Ghosh, Mar 19 2017

a(n) = A000041(n)*n - A000041(n-1)*(n-1) = A138880(n) + A000041(n-1).
a(n) = A066186(n) - A066186(n-1), for n>=1.
a(n) ~ exp(Pi*sqrt(2*n/3)) * Pi/(12*sqrt(2*n)) * (1 - (72 + 13*Pi^2) / (24*Pi*sqrt(6*n)) + (7/12 + 3/(2*Pi^2) + 217*Pi^2/6912)/n - (15*sqrt(3/2)/(16*Pi) + 115*Pi/(288*sqrt(6)) + 4069*Pi^3/(497664*sqrt(6)))/n^(3/2)). - Vaclav Kotesovec, Oct 21 2016, extended Jul 06 2019
G.f.: x*(1 - x)*f'(x), where f(x) = Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Apr 13 2017

a(34) corrected by R. J. Mathar, Jan 27 2011

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.

A340583 Triangle read by rows: T(n,k) = A002865(n-k)*A000203(k), 1 <= k <= n.

Original entry on oeis.org

1, 0, 3, 1, 0, 4, 1, 3, 0, 7, 2, 3, 4, 0, 6, 2, 6, 4, 7, 0, 12, 4, 6, 8, 7, 6, 0, 8, 4, 12, 8, 14, 6, 12, 0, 15, 7, 12, 16, 14, 12, 12, 8, 0, 13, 8, 21, 16, 28, 12, 24, 8, 15, 0, 18, 12, 24, 28, 28, 24, 24, 16, 15, 13, 0, 12, 14, 36, 32, 49, 24, 48, 16, 30, 13, 18, 0, 28

Offset: 1

Omar E. Pol, Jan 15 2021

T(n,k) is the total number of cubic cells added at n-th stage to the right prisms whose bases are the parts of the symmetric representation of sigma(k) in the polycube described in A221529.

Partial sums of column k gives the column k of A221529.

			Triangle begins:
   1;
   0,  3;
   1,  0,  4;
   1,  3,  0,  7;
   2,  3,  4,  0,  6;
   2,  6,  4,  7,  0, 12;
   4,  6,  8,  7,  6,  0,  8;
   4, 12,  8, 14,  6, 12,  0, 15;
   7, 12, 16, 14, 12, 12,  8,  0, 13;
   8, 21, 16, 28, 12, 24,  8, 15,  0, 18;
  12, 24, 28, 28, 24, 24, 16, 15, 13,  0, 12;
  14, 36, 32, 49, 24, 48, 16, 30, 13, 18,  0, 28;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k   A000203         T(6,k)
--------------------------
1      1   *   2  =    2
2      3   *   2   =   6
3      4   *   1   =   4
4      7   *   1   =   7
5      6   *   0   =   0
6     12   *   1   =  12
.           A002865
--------------------------
The sum of row 6 is 2 + 6 + 4 + 7 + 0 + 12 = 31, equaling A138879(6).

Row sums give A138879.
Column 1 gives A002865.
Diagonals 1, 3 and 4 give A000203.
Diagonal 2 gives A000004.
Diagonals 5 and 6 give A074400.
Diagonals 7 and 8 give A239050.
Diagonal 9 gives A319527.
Diagonal 10 gives A319528.
Cf. A221529 (partial column sums).
Cf. A340426 (mirror).
Cf. A135010, A221530, A245095, A245099, A339278.

A340583[n_, k_] := (PartitionsP[n - k] - PartitionsP[(n - k) - 1])*
   DivisorSigma[1, k];
Table[A340583[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Robert P. P. McKone, Jan 25 2021 *)

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

cp's OEIS Frontend

A138879 Sum of all parts of the last section of the set of partitions of n.

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

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A340583 Triangle read by rows: T(n,k) = A002865(n-k)*A000203(k), 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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A340579 Triangle read by rows: T(n,k) = A000203(n-k+1)*A000070(k-1), 1 <= k <= n.

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