A118457 Table of partitions of n into distinct parts, in Mathematica ordering.
1, 2, 3, 2, 1, 4, 3, 1, 5, 4, 1, 3, 2, 6, 5, 1, 4, 2, 3, 2, 1, 7, 6, 1, 5, 2, 4, 3, 4, 2, 1, 8, 7, 1, 6, 2, 5, 3, 5, 2, 1, 4, 3, 1, 9, 8, 1, 7, 2, 6, 3, 6, 2, 1, 5, 4, 5, 3, 1, 4, 3, 2, 10, 9, 1, 8, 2, 7, 3, 7, 2, 1, 6, 4, 6, 3, 1, 5, 4, 1, 5, 3, 2, 4, 3, 2, 1, 11, 10, 1, 9, 2, 8, 3, 8, 2, 1, 7, 4, 7, 3, 1, 6, 5
Offset: 1
Examples
The partitions of 5 into distinct parts are [5], [4,1] and [3,2], so row 5 is 5,4,1,3,2. 1; 2; 3; 2,1; 4; 3,1; 5; 4,1; 3,2; 6; 5,1; 4,2; 3,2,1; 7; 6,1; 5,2; 4,3; 4,2,1; 8; 7,1; 6,2; 5,3; 5,2,1; 4,3,1; 9; 8,1; 7,2; 6,3; 6,2,1; 5,4; 5,3,1; 4,3,2; 10; 9,1; 8,2; 7,3; 7,2,1; 6,4; 6,3,1; 5,4,1; 5,3,2; 4,3,2,1; 11; 10,1; 9,2; 8,3; 8,2,1; 7,4; 7,3,1; 6,5; 6,4,1; 6,3,2; 5,4,2; 5,3,2,1;
Links
- Alois P. Heinz, Rows n = 1..32, flattened
Crossrefs
Programs
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Mathematica
d[n_] := Select[IntegerPartitions[n], Max[Length /@ Split@ #] == 1 &]; Flatten[Table[d[n], {n, 15}]] (* Clark Kimberling, Mar 11 2012 *)
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SageMath
def StrictPartitions(n): return [partition for partition in Partitions(n) if Set(partition.to_exp()).issubset(Set([0,1]))] def A118457row(n): return [p for parts in StrictPartitions(n) for p in parts] for n in (1..9): print(A118457row(n)) # Peter Luschny, Apr 11 2020
Comments