login
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
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
Robert Price, Table of n, a(n) for n = 0..9295 (25 rows).
FORMULA
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).
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
Sequence in context: A239228 A346080 A103921 * A279044 A134265 A182858
KEYWORD
nonn,tabf
AUTHOR
EXTENSIONS
Edited and corrected by Franklin T. Adams-Watters, May 29 2006
STATUS
approved