A210950 Triangle read by rows: T(n,k) = number of parts in the k-th column of the partitions of n but with the partitions aligned to the right margin.

1, 1, 2, 1, 2, 3, 1, 2, 4, 5, 1, 2, 4, 6, 7, 1, 2, 4, 7, 10, 11, 1, 2, 4, 7, 11, 14, 15, 1, 2, 4, 7, 12, 17, 21, 22, 1, 2, 4, 7, 12, 18, 25, 29, 30, 1, 2, 4, 7, 12, 19, 28, 36, 41, 42, 1, 2, 4, 7, 12, 19, 29, 40, 50, 55, 56, 1, 2, 4, 7, 12, 19, 30, 43

Offset: 1

Author

Omar E. Pol, Apr 22 2012

Keywords

• nonn
• tabl

Comments

Index of the first partition of n that has k parts, when the partitions of n are listed in reverse lexicographic order, as in Mathematica's IntegerPartitions[n]. - Clark Kimberling, Oct 16 2023

Examples

			For n = 6 the partitions of 6 aligned to the right margin look like this:
.
.                      6
.                  3 + 3
.                  4 + 2
.              2 + 2 + 2
.                  5 + 1
.              3 + 2 + 1
.              4 + 1 + 1
.          2 + 2 + 1 + 1
.          3 + 1 + 1 + 1
.      2 + 1 + 1 + 1 + 1
.  1 + 1 + 1 + 1 + 1 + 1
.
The number of parts in columns 1-6 are
.  1,  2,  4,  7, 10, 11, the same as the 6th row of triangle.
Triangle begins:
  1;
  1, 2;
  1, 2, 3;
  1, 2, 4, 5;
  1, 2, 4, 6, 7;
  1, 2, 4, 7, 10, 11;
  1, 2, 4, 7, 11, 14, 15;
  1, 2, 4, 7, 12, 17, 21, 22;
  1, 2, 4, 7, 12, 18, 25, 29, 30;
  1, 2, 4, 7, 12, 19, 28, 36, 41, 42;
  1, 2, 4, 7, 12, 19, 29, 40, 50, 55, 56;
  1, 2, 4, 7, 12, 19, 30, 43, 58, 70, 76, 77;
		

Crossrefs

Mirror of A058399. Row sums give A006128. Right border gives A000041, n >= 1. Rows converge to A000070.

Cf. A135010, A194714, A210945, A210951, A210952, A210953, A210970.

Programs

Mathematica

m[n_, k_] := Length[IntegerPartitions[n][[k]]]; c[n_] := PartitionsP[n];
t[n_, h_] := Select[Range[c[n]], m[n, #] == h &, 1];
Column[Table[t[n, h], {n, 1, 20}, {h, 1, n}]]
 (* Clark Kimberling, Oct 16 2023 *)

Formula

T(n,k) = Sum_{j=1..n} A210951(j,k).