A363994 a(n) is the number of partitions of n whose difference multiset has no duplicates; see Comments.
1, 1, 2, 2, 3, 3, 4, 5, 7, 6, 10, 11, 11, 15, 18, 18, 25, 29, 28, 38, 44, 47, 54, 67, 68, 84, 88, 102, 114, 137, 132, 167, 180, 204, 214, 261, 264, 315, 328, 377, 414, 476, 473, 564, 603, 677, 708, 820, 846, 969, 1028, 1131, 1214, 1364, 1414, 1596, 1701, 1858
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 difference 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[Differences, s[n, k]]] t[n_] := t[n] = Table[g[n, k], {k, 1, PartitionsP[n]}]; a[n_] := Count[t[n], True]; Table[a[n], {n, 1, 20}] -
Python
from collections import Counter from itertools import combinations from sympy.utilities.iterables import partitions def A363994(n): return sum(1 for p in partitions(n) if max(list(Counter(abs(d[0]-d[1]) for d in combinations(list(Counter(p).elements()),2)).values()),default=1)==1) # Chai Wah Wu, Sep 17 2023
Formula
a(n) = A325876(n) - (1 - n mod 2) for n > 0. - Andrew Howroyd, Sep 17 2023
Extensions
More terms from Alois P. Heinz, Sep 12 2023
Comments