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A115623
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Irregular triangle read by rows: row n lists numbers of distinct parts of partitions of n in Mathematica order.
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23
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0, 1, 1, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 1, 3, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 3, 2, 1, 3, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 3, 3, 3, 2, 1, 3, 2, 2, 3, 3, 2, 2, 2, 2, 2, 1, 1, 2, 2, 2, 2, 3, 2, 2
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OFFSET
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0,6
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COMMENTS
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The row length sequence of this table is p(n)=A000041(n) (number of partitions).
In order to count distinct parts of a partition consider the partition as a set instead of a multiset. E.g., n=6: read [3,1,1,1] as {1,3} and count the elements, here 2.
Rows are the same as the rows of A103921, but in reverse order.
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LINKS
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FORMULA
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a(n, m) = number of distinct parts of the m-th partition of n in Mathematica order; n >= 0, m = 1..p(n) = A000041(n).
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EXAMPLE
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Triangle starts:
0
1
1, 1
1, 2, 1
1, 2, 1, 2, 1
1, 2, 2, 2, 2, 2, 1
1, 2, 2, 2, 1, 3, 2, 1, 2, 2, 1
1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1
1, 2, 2, 2, 2, 3, 2, 1, 3, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 2, 2, 1
1, 2, 2, 2, 2, 3, 2, 2, ...
a(5,4)=2 from the fourth partition of 5 in the mentioned order, i.e., [3,1^2], which has two distinct parts, namely 1 and 3.
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MATHEMATICA
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Table[Length /@ Union /@ IntegerPartitions[n], {n, 0, 8}] // Flatten (* Robert Price, Jun 11 2020 *)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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