A176206 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n has length A000070(n-1) and every column k gives the positive integers.

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

Offset: 1

Author

Alford Arnold, Apr 11 2010

Keywords

• easy
• nonn
• tabf

Comments

The original definition was: An irregular table: Row n begins with n, counts down to 1 and repeats the intermediate numbers as often as given by the partition numbers.

Row n contains a decreasing sequence where n-k is repeated A000041(k) times, k = 0..n-1.

From Omar E. Pol, Nov 23 2020: (Start)

Row n lists in nonincreasing order the first A000070(n-1) terms of A336811.

In other words: row n lists in nonincreasing order the terms from the first n rows of triangle A336811.

Conjecture: all divisors of all terms in row n are also all parts of all partitions of n.

For more information see the example and A336811 which contains the most elementary conjecture about the correspondence divisors/partitions.

Row sums give A014153.

A338156 lists the divisors of every term of this sequence.

The n-th row of A340581 lists in nonincreasing order the terms of the first n rows of this triangle.

For a regular triangle with the same row sums see A141157. (End)

From Omar E. Pol, Jul 31 2021: (Start)

The number of k's in row n is equal to A000041(n-k), 1 <= k <= n.

The number of terms >= k in row n is equal to A000070(n-k), 1 <= k <= n.

The number of k's in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.

The number of terms >= k in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A014153(n-k), 1 <= k <= n. (End)

Examples

			Triangle begins:
  1;
  2, 1;
  3, 2, 1, 1;
  4, 3, 2, 2, 1, 1, 1;
  5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1;
  6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
  7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, ...
  ... Extended by _Omar E. Pol_, Nov 23 2020
From _Omar E. Pol_, Jan 25 2020: (Start)
For n = 5, by definition the length of row 5 is A000070(5-1) = A000070(4) = 12, so the row 5 of triangle has 12 terms. Since every column lists the positive integers A000027 so the row 5 is [5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1].
Then we have that the divisors of the numbers of the 5th row are:
.
5th row of triangle -----> 5  4  3  3  2  2  2  1  1  1  1  1
                           1  2  1  1  1  1  1
                              1
.
There are twelve 1's, four 2's, two 3's, one 4 and one 5.
In total there are 12 + 4 + 2 + 1 + 1 = 20 divisors.
On the other hand the partitions of 5 are as shown below:
.
.      5
.      3  2
.      4  1
.      2  2  1
.      3  1  1
.      2  1  1  1
.      1  1  1  1  1
.
There are twelve 1's, four 2's, two 3's, one 4 and one 5, as shown also in the 5th row of triangle A066633.
In total there are 12 + 4 + 2 + 1 + 1 = A006128(5) = 20 parts.
Finally in accordance with the conjecture we can see that all divisors of all numbers in the 5th row of the triangle are the same positive integers as all parts of all partitions of 5. (End)
		

Links

• Paolo Xausa, Table of n, a(n) for n = 1..10980 (rows 1..21 of triangle, flattened)

Crossrefs

Cf. A000027 (columns), A000070 (row lengths), A338156 (divisors), A340061 (mirror).

Cf. A000041, A006128, A014153, A027750, A066633, A141157, A176207, A336811, A338156, A340581.

Programs

Mathematica

Table[Flatten[Table[ConstantArray[n-k,PartitionsP[k]],{k,0,n-1}]],{n,10}] (* Paolo Xausa, May 30 2022 *)

Extensions

New name, changed offset, edited and more terms from Omar E. Pol, Nov 22 2020