A359350 - OEIS

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

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OFFSET

1,2

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).

LINKS

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

EXAMPLE

Triangle begins:

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].

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 *)

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.

Sequence in context: A365805 A334749 A266640 * A065120 A176206 A232890

Adjacent sequences: A359347 A359348 A359349 * A359351 A359352 A359353

KEYWORD

nonn,tabf

AUTHOR

Omar E. Pol, Dec 27 2022

STATUS

approved