

A115623


Irregular triangle read by rows: row n lists numbers of distinct parts of partitions of n in Mathematica order.


23



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

0,6


COMMENTS

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.


LINKS



FORMULA

a(n, m) = number of distinct parts of the mth partition of n in Mathematica order; n >= 0, m = 1..p(n) = A000041(n).


EXAMPLE

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.


MATHEMATICA

Table[Length /@ Union /@ IntegerPartitions[n], {n, 0, 8}] // Flatten (* Robert Price, Jun 11 2020 *)


CROSSREFS



KEYWORD

nonn,tabf


AUTHOR



EXTENSIONS



STATUS

approved



