A364613 a(n) = number of partitions of n whose sum multiset is free of duplicates; see Comments.
1, 1, 2, 2, 3, 3, 5, 5, 7, 8, 10, 12, 15, 18, 20, 26, 29, 36, 38, 50, 53, 67, 69, 89, 95, 115, 122, 151, 161, 195, 201, 247, 266, 312, 330, 386, 419, 487, 520, 600, 641, 742, 793, 901, 979, 1088, 1186, 1331, 1454, 1605, 1730, 1925, 2102, 2311, 2525, 2741, 3001
Offset: 0
Keywords
Examples
The partitions of 8 are [8], [7,1], [6,2], [6,1,1], [5,3], [5,2,1], [5,1,1,1], [4,4], [4,3,1], [4,2,2], [4,2,1,1], [4,1,1,1,1], [3,3,2], [3,3,1,1], [3,2,2,1], [3,2,1,1,1], [3,1,1,1,1,1], [2,2,2,2], [2,2,2,1,1], [2,2,1,1,1,1], [2,1,1,1,1,1,1], [1,1,1,1,1,1,1,1]. The 7 partitions whose sum multiset is duplicate-free are [8], [7,1], [6,2], [5,3], [5,2,1], [4,4], [4,3,1].
Programs
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Mathematica
s[n_, k_] := s[n, k] = Subsets[IntegerPartitions[n][[k]], {2}]; g[n_, k_] := g[n, k] = DuplicateFreeQ[Map[Total, s[n, k]]]; t[n_] := Table[g[n, k], {k, 1, PartitionsP[n]}]; a[n_] := Count[t[n], True] Table[a[n], {n, 1, 40}]
Formula
a(n) = A325877(n) - (1 - n mod 2) for n > 0. - Andrew Howroyd, Sep 17 2023
Extensions
More terms from Alois P. Heinz, Sep 17 2023
Comments