OFFSET
1,2
COMMENTS
Suppose that p is a partition of n. Let m X m be the size of its Ferrers matrix, f(p), defined at A237981. Then f(p) consists of ceiling(m/2) concentric squares, where the innermost square is a single point if m is odd. The square partition of p is introduced here as the partition [x(1), x(2), ..., x(k)], where x(i) is the number of 1s in the i-th concentric square, where the squares are taken in order starting with the outermost.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
Clark Kimberling and Peter J. C. Moses, Ferrers Matrices and Related Partitions of Integers
EXAMPLE
The 7 square partitions of 12 are as follows: [12], [11,1], [10,2], [9,3], [8,4], [8,3,1], [7,4,1]. The Ferrers matrix of the partition [4,3,3,1,1] of 12 is shown here:
...
1 . 1 . 1 . 1 . 0
1 . 1 . 1 . 0 . 0
1 . 1 . 1 . 0 . 0
1 . 0 . 0 . 0 . 0
1 . 0 . 0 . 0 . 0.
The outermost square has 8 1s, the next has 3 1s, and the innermost, 1 1, so that [8,3,1] is a square partition of 12. The first 9 rows of the array:
1
2
3
4
5 4 1
6 5 1
7 6 1 5 2
8 7 1 6 2
9 8 1 7 2 6 3
MATHEMATICA
z=20;
ferrersMatrix[list_]:=PadRight[Map[Table[1, {#}]&, #], {#, #}&[Max[#, Length[#]]]]&[list];
sqPart[list_]:=DeleteCases[Total[{Total[LowerTriangularize[#]+Transpose[UpperTriangularize[#, 1]]]&[Reverse[LowerTriangularize[#]]], Reverse[Total[Transpose[LowerTriangularize[#]]+UpperTriangularize[#, 1]]]&[Reverse[UpperTriangularize[#, 1]]]}&[ferrersMatrix[list]]], 0];
sqParts[n_]:=#[[Reverse[Ordering[PadRight[#]]]]]&[DeleteDuplicates[Map[sqPart, IntegerPartitions[n]]]]
Flatten[sq=Map[sqParts[#]&, Range[z]]] (*A237985*)
Map[Length, sq] (*A237980*)
(* Peter J. C. Moses, Feb 19 2014 *)
CROSSREFS
KEYWORD
nonn,tabf,easy
AUTHOR
Clark Kimberling and Peter J. C. Moses, Feb 25 2014
STATUS
approved