OFFSET
1,2
COMMENTS
Suppose that p is a partition of n, let F(p) be its Ferrers matrix, as defined at A237981, and let mXm be the size of F(p). The numbers of 1s in each of the 2m-1 diagonals of F(p) form a partition of n. Any partition which is associated with a partition of n in this manner is introduced here as a diagonal partition of n. A000041(n) = sum of the numbers in row n; A003114(n) = number of terms in row n. Every diagonal partition is an antidiagonal partition, as in A238325 (but not conversely).
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
EXAMPLE
The Mathematica ordering of the 3 antidiagonal partitions of 6 follows: 2211, 21111, 111111.) Frequencies of these among the 11 partitions of 6 are given in reverse Mathematica ordering as follows: 111111 occurs 6 times, 21111 occurs 3 times, and 2211 occurs 2 times, so that row 6 of the array is 6 3 2.
...
First 9 rows:
1
2
3
4 1
5 2
6 3 2
7 4 4
8 5 6 3
9 6 8 6 1
MATHEMATICA
z = 20; ferrersMatrix[list_] := PadRight[Map[Table[1, {#}] &, #], {#, #} &[Max[#, Length[#]]]] &[list]; diagPartSE[list_] := Module[{m = ferrersMatrix[list]}, Map[Diagonal[m, #] &, Range[-#, #] &[Length[m] - 1]]]; Tally[Map[ DeleteCases[Reverse[Sort[Map[Count[#, 1] &, diagPartSE[#]]]], 0] &, IntegerPartitions[z]]]; a1[n_] := Last[Transpose[Tally[Map[DeleteCases[Reverse[Sort[Map[Count[#, 1] &, diagPartSE[#]]]], 0] &, IntegerPartitions[n]]]]]; t = Table[a1[n], {n, 1, z}]; u = Flatten[t]
Map[Last[Transpose[Tally[Map[DeleteCases[Reverse[Sort[Map[Count[#, 1] &, diagPartSE[#]]]], 0] &, IntegerPartitions[#]]]]] &, Range[z]] // TableForm
(* Peter J. C. Moses, Feb 25 2014 *)
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Clark Kimberling and Peter J. C. Moses, Feb 25 2014
STATUS
approved