A284870 -id:A284870 - cp's OEIS frontend

A341062 Sequence whose partial sums give A000005.

1, 1, 0, 1, -1, 2, -2, 2, -1, 1, -2, 4, -4, 2, 0, 1, -3, 4, -4, 4, -2, 0, -2, 6, -5, 1, 0, 2, -4, 6, -6, 4, -2, 0, 0, 5, -7, 2, 0, 4, -6, 6, -6, 4, 0, -2, -2, 8, -7, 3, -2, 2, -4, 6, -4, 4, -4, 0, -2, 10, -10, 2, 2, 1, -3, 4, -6, 4, -2, 4, -6, 10, -10, 2, 2, 0, -2, 4, -6, 8, -5, -1, -2, 10, -8, 0, 0, 4, -6, 10

Offset: 1

Omar E. Pol, Feb 04 2021

Essentially a duplicate of A051950.
Convolved with A000041 gives A138137.
Convolved with A000027 gives the nonzero terms of A006218.
Convolved with A000070 gives the nonzero terms of A006128.
Convolved with A014153 gives the nonzero terms of A284870.
Convolved with A036469 gives the nonzero terms of A305082.
Convolved with the nonzero terms of A006218 gives A055507.
Convolved with the nonzero terms of A000217 gives the nonzero terms of A078567.

1 together with A051950.
Cf. A000005 (partial sums).
Cf. A000027, A000041, A000070, A000217, A006128, A006218, A014153, A036469, A055507, A078567, A138137, A284870, A305082, A340793.

Join[{1}, Differences[Table[DivisorSigma[0, n], {n, 1, 90}]]] (* Amiram Eldar, Feb 06 2021 *)

a(n) = A051950(n) for n > 1.

A299779 Triangle read by rows: T(n,k) is the total number of cliques of size k in all partitions of all positive integers <= n.

Original entry on oeis.org

1, 2, 1, 5, 1, 1, 9, 3, 1, 1, 17, 5, 2, 1, 1, 28, 9, 4, 2, 1, 1, 47, 14, 7, 3, 2, 1, 1, 73, 24, 10, 6, 3, 2, 1, 1, 114, 35, 17, 9, 5, 3, 2, 1, 1, 170, 55, 25, 14, 8, 5, 3, 2, 1, 1, 253, 80, 38, 20, 13, 7, 5, 3, 2, 1, 1, 365, 118, 55, 31, 18, 12, 7, 5, 3, 2, 1, 1, 525, 167, 80, 44, 27, 17, 11, 7, 5, 3, 2, 1, 1

Offset: 1

Omar E. Pol, Apr 04 2018

Column k gives the partial sums of the k-th column of triangle A197126.

			Triangle begins:
    1;
    2,   1;
    5,   1,  1;
    9,   3,  1,  1;
   17,   5,  2,  1,  1;
   28,   9,  4,  2,  1,  1;
   47,  14,  7,  3,  2,  1,  1;
   73,  24, 10,  6,  3,  2,  1,  1;
  114,  35, 17,  9,  5,  3,  2,  1,  1;
  170,  55, 25, 14,  8,  5,  3,  2,  1,  1;
  253,  80, 38, 20, 13,  7,  5,  3,  2,  1,  1;
  365, 118, 55, 31, 18, 12,  7,  5,  3,  2,  1,  1;
  525, 167, 80, 44, 27, 17, 11,  7,  5,  3,  2,  1,  1;
...

Alois P. Heinz, Rows n = 1..150, flattened

Column 1 gives A000097.
Row sums give A014153.
Cf. A000041, A000070, A197126, A284870.

b:= proc(n, p, k) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
      add((l->`if`(m=k, l+[0, l[1]], l))(b(n-p*m, p-1, k)), m=0..n/p)))
    end:
T:= proc(n, k) option remember;
      b(n$2, k)[2]+`if`(nAlois P. Heinz, Apr 27 2018

b[n_, p_, k_] := b[n, p, k] = If[n == 0, {1, 0}, If[p < 1, {0, 0}, Sum[ Function[l, If[m==k, l+{0, l[[1]]}, l]][b[n-p*m, p-1, k]], {m, 0, n/p}]]];
T[n_, k_] := b[n, n, k][[2]] + If[n < k, 0, T[n-1, k]];
Table[Table[T[n, k], {k, 1, n}], {n, 1, 20}] // Flatten (* Jean-François Alcover, Dec 07 2019, after Alois P. Heinz *)

T(n,k) = Sum_{j=k..n} A197126(j,k).
T(2n+1,n+1) = A000041(n). - Alois P. Heinz, Apr 27 2018
Sum_{k=1..n} k * T(n,k) = A284870(n). - Alois P. Heinz, May 14 2018

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

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.

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A341062 Sequence whose partial sums give A000005.

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A299779 Triangle read by rows: T(n,k) is the total number of cliques of size k in all partitions of all positive integers <= 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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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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