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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)).
2
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
1,2
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:
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].
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.
Sequence in context: A365805 A334749 A266640 * A065120 A176206 A232890
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Dec 27 2022
STATUS
approved