A359350 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n is constructed by replacing A336811(n,k) with the largest partition into consecutive parts of A000217(A336811(n,k)).

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

Offset: 1

Author

Omar E. Pol, Dec 27 2022

Keywords

• nonn
• tabf

Comments

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

The terms of row n listed in nonincreasing order give the n-th row of A176206.

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.

Row n is constructed replacing A336811(n,k) with the largest partition into consecutive parts of A359279(n,k).

Examples

			Triangle begins:
  1;
  2, 1;
  3, 2, 1, 1;
  4, 3, 2, 1, 2, 1, 1;
  5, 4, 3, 2, 1, 3, 2, 1, 2, 1, 1, 1;
  6, 5, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1, 2, 1, 2, 1, 1, 1;
  ...
Or also the triangle begins:
  [1];
  [2, 1];
  [3, 2, 1],          [1];
  [4, 3, 2, 1],       [2, 1],       [1];
  [5, 4, 3, 2, 1],    [3, 2, 1],    [2, 1],    [1],    [1];
  [6, 5, 4, 3, 2, 1], [4, 3, 2, 1], [3, 2, 1], [2, 1], [2, 1], [1], [1];
  ...
For n = 3 the third row is [3, 2, 1, 1]. The divisors of these terms are [1, 3], [1, 2], [1], [1]. These six divisors are also all parts of all partitions of 3. They are [3], [2, 1], [1, 1, 1].
		

Links

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

Crossrefs

Row sums give A014153 (convolution of A000041 and A000027).

Row lengths give A000070.

Row n has A000041(n-1) blocks.

This triangle has the same row sums as A176206, A299779 and A359279.

Cf. A000217, A336811, A336812, A338156, A359280.

Programs

Mathematica

A359350row[n_]:=Flatten[Table[ConstantArray[Range[n-m,1,-1],PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]];Array[A359350row,10] (* Paolo Xausa, Sep 01 2023 *)