A115623 - OEIS

%I #16 Jun 11 2020 16:47:22

%S 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,

%T 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,

%U 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

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

%C The row length sequence of this table is p(n)=A000041(n) (number of partitions).

%C 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.

%C Rows are the same as the rows of A103921, but in reverse order.

%H Robert Price, <a href="/A115623/b115623.txt">Table of n, a(n) for n = 0..9295</a> (25 rows).

%F 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).

%e Triangle starts:

%e 0

%e 1

%e 1, 1

%e 1, 2, 1

%e 1, 2, 1, 2, 1

%e 1, 2, 2, 2, 2, 2, 1

%e 1, 2, 2, 2, 1, 3, 2, 1, 2, 2, 1

%e 1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1

%e 1, 2, 2, 2, 2, 3, 2, 1, 3, 2, 3, 2, 2, 2, 3, 3, 2, 1, 2, 2, 2, 1

%e 1, 2, 2, 2, 2, 3, 2, 2, ...

%e 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.

%t Table[Length /@ Union /@ IntegerPartitions[n], {n, 0, 8}] // Flatten  (* _Robert Price_, Jun 11 2020 *)

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

%K nonn,tabf

%O 0,6

%A _Franklin T. Adams-Watters_, Jan 25 2006

%E Edited and corrected by _Franklin T. Adams-Watters_, May 29 2006