A238606 Array: t(n,k) = number of partitions p of n such that the principal antidiagonal of the Ferrers matrix of p has k 1's.
1, 2, 2, 1, 4, 1, 4, 3, 8, 2, 1, 10, 3, 2, 14, 7, 1, 20, 5, 5, 30, 5, 6, 1, 36, 15, 2, 3, 52, 16, 6, 3, 70, 13, 15, 3, 94, 22, 12, 7, 122, 32, 8, 13, 1, 160, 45, 12, 10, 4, 206, 51, 26, 8, 6, 276, 49, 44, 9, 7, 350, 75, 30, 28, 7, 448, 108, 22, 38, 11, 566
Offset: 1
Examples
First 17 rows: 1 2 2 .... 1 4 .... 1 4 .... 3 8 .... 2 .... 1 10 ... 3 .... 2 14 ... 7 .... 1 20 ... 5 .... 5 30 ... 5 .... 6 ... 1 36 ... 15 ... 2 ... 3 52 ... 16 ... 6 ... 3 70 ... 13 ... 15 .. 3 94 ... 22 ... 12 .. 7 122 ... 32 .. 8 ... 13 .. 1 160 ... 45 .. 12 .. 10 .. 4 Row 5 counts 4 antidiagonals that have exactly one 1 and 3 antidiagonals that have exactly two 1's. The Ferrers matrix for each of the latter three cases are as shown below. For the partition 32: 1 1 1 1 1 0 0 0 0 (antidiagonal, from row 1: 1,1,0) For the partition 311: 1 1 1 1 0 0 1 0 0 (antidiagonal, from row 1: 1,0,1,) For the partition 221: 1 1 0 1 1 0 1 0 0 (antidiagonal, from row 1: 0,1,1)
Links
- Clark Kimberling and Peter J. C. Moses, Ferrers Matrices and Related Partitions of Integers
Programs
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Mathematica
z = 30; ferrersMatrix[list_] := PadRight[Map[Table[1, {#}] &, #], {#, #} &[Max[#, Length[#]]]] &[list]; diagAntidiagDots[list_] := {Total[Diagonal[#]], Total[Diagonal[Reverse[#]]]} &[ferrersMatrix[list]]; u[n_, k_] := Length[Select[ Map[diagAntidiagDots, IntegerPartitions[n]], #[[2]] == k &]]; t[n_] := t[n] = Floor[(-1 + Sqrt[1 + 8 n])/2]; w = Table[u[n, k], {n, 1, z}, {k, 1, t[n]}]; y = Flatten[w] (* A238606 *) (* Peter J. C. Moses, Mar 01 2014 *)
Comments