|
Search: id:A115623
|
|
|
| A115623 |
|
Table of number of distinct parts of partitions of n in Mathematica order. |
|
+0
3
|
|
| 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; listen)
|
|
|
OFFSET
|
0,6
|
|
|
COMMENT
|
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 number of elements, here 2.
Rows are the same as the rows of A103921, but in reverse order.
|
|
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
|
0; 1; 1,1; 1,2,1; 1,2,1,2,1; 1,2,2,2,2,2,1; ...
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.
|
|
CROSSREFS
|
Cf. A080577, A000041, A103921, A115622, row sums A000070.
Sequence in context: A095771 A007421 A103921 this_sequence A134265 A001030 A071709
Adjacent sequences: A115620 A115621 A115622 this_sequence A115624 A115625 A115626
|
|
KEYWORD
|
nonn,tabf
|
|
AUTHOR
|
Franklin T. Adams-Watters (FrankTAW(AT)Netscape.net), Jan 25 2006
|
|
EXTENSIONS
|
Edited and corrected by Franklin T. Adams-Watters, May 29 2006
|
|
|
Search completed in 0.002 seconds
|
