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%I #16 Jan 28 2022 01:23:39
%S 1,1,1,1,2,2,3,3,4,5,6,7,9,10,13,15,18,21,25,28,32,36,44,49,60,66,80,
%T 89,103,115,132,147,168,188,212,236,269,301,344,385,437,485,549,606,
%U 678,751,837,926,1031,1133,1263,1389,1541,1696,1889,2068,2306,2529
%N Array: row n gives the number of distinct square partitions of n; see Comments.
%C 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.
%e The 7 square partitions of 12 are as follows: [12], [11,1], [10,2], [9,3], [8,3,1], [8,4], [7,4,1]. The Ferrers matrix of the partition [4,3,3,1,1] of 12 is shown here:
%e 1 . 1 . 1 . 1 . 0
%e 1 . 1 . 1 . 0 . 0
%e 1 . 1 . 1 . 0 . 0
%e 1 . 0 . 0 . 0 . 0
%e 1 . 0 . 0 . 0 . 0.
%e 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.
%t z=20;
%t ferrersMatrix[list_]:=PadRight[Map[Table[1,{#}]&,#],{#,#}&[Max[#,Length[#]]]]&[list];
%t sqPart[list_]:=DeleteCases[Total[{Total[LowerTriangularize[#]+ Transpose[UpperTriangularize[#,1]]]&[Reverse[LowerTriangularize[#]]],Reverse[Total[Transpose[ LowerTriangularize[#]]+UpperTriangularize[#,1]]]&[Reverse[UpperTriangularize[#,1]]]}&[ferrersMatrix[list]]],0];
%t sqParts[n_]:=#[[Reverse[Ordering[PadRight[#]]]]]&[DeleteDuplicates[Map[sqPart,IntegerPartitions[n]]]]
%t Flatten[sq=Map[sqParts[#]&,Range[z]]] (*A237985*)
%t Map[Length,sq] (*A237980*)
%t (* _Peter J. C. Moses_, Feb 19 2014 *)
%Y Cf. A237985.
%K nonn,tabf,easy
%O 1,5
%A _Clark Kimberling_ and _Peter J. C. Moses_, Feb 25 2014
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