login
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.
3

%I #27 Oct 22 2023 17:05:38

%S 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,

%T 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,

%U 12,19,29,40,50,55,56,1,2,4,7,12,19,30,43

%N 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.

%C 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

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

%e For n = 6 the partitions of 6 aligned to the right margin look like this:

%e .

%e . 6

%e . 3 + 3

%e . 4 + 2

%e . 2 + 2 + 2

%e . 5 + 1

%e . 3 + 2 + 1

%e . 4 + 1 + 1

%e . 2 + 2 + 1 + 1

%e . 3 + 1 + 1 + 1

%e . 2 + 1 + 1 + 1 + 1

%e . 1 + 1 + 1 + 1 + 1 + 1

%e .

%e The number of parts in columns 1-6 are

%e . 1, 2, 4, 7, 10, 11, the same as the 6th row of triangle.

%e Triangle begins:

%e 1;

%e 1, 2;

%e 1, 2, 3;

%e 1, 2, 4, 5;

%e 1, 2, 4, 6, 7;

%e 1, 2, 4, 7, 10, 11;

%e 1, 2, 4, 7, 11, 14, 15;

%e 1, 2, 4, 7, 12, 17, 21, 22;

%e 1, 2, 4, 7, 12, 18, 25, 29, 30;

%e 1, 2, 4, 7, 12, 19, 28, 36, 41, 42;

%e 1, 2, 4, 7, 12, 19, 29, 40, 50, 55, 56;

%e 1, 2, 4, 7, 12, 19, 30, 43, 58, 70, 76, 77;

%t m[n_, k_] := Length[IntegerPartitions[n][[k]]]; c[n_] := PartitionsP[n];

%t t[n_, h_] := Select[Range[c[n]], m[n, #] == h &, 1];

%t Column[Table[t[n, h], {n, 1, 20}, {h, 1, n}]]

%t (* _Clark Kimberling_, Oct 16 2023 *)

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

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

%K nonn,tabl

%O 1,3

%A _Omar E. Pol_, Apr 22 2012