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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 (list; graph; refs; listen; history; text; internal format)
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

Cf. A080577, A000041, A103921, A115622, row sums A000070.

Sequence in context: A239228 A346080 A103921 * A279044 A134265 A182858

Adjacent sequences:  A115620 A115621 A115622 * A115624 A115625 A115626

KEYWORD

nonn,tabf

AUTHOR

Franklin T. Adams-Watters, Jan 25 2006

EXTENSIONS

Edited and corrected by Franklin T. Adams-Watters, May 29 2006

STATUS

approved

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Last modified November 28 05:30 EST 2021. Contains 349401 sequences. (Running on oeis4.)