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A237980
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Array: row n gives the number of distinct square partitions of n; see Comments.
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2
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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, 89, 103, 115, 132, 147, 168, 188, 212, 236, 269, 301, 344, 385, 437, 485, 549, 606, 678, 751, 837, 926, 1031, 1133, 1263, 1389, 1541, 1696, 1889, 2068, 2306, 2529
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OFFSET
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1,5
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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:
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.
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MATHEMATICA
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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*)
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CROSSREFS
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KEYWORD
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nonn,tabf,easy
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AUTHOR
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STATUS
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approved
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