A364612 a(n) = number of partitions of n whose difference multiset has at least one duplicate; see Comments.
0, 0, 0, 1, 2, 4, 7, 10, 15, 24, 32, 45, 66, 86, 117, 158, 206, 268, 357, 452, 583, 745, 948, 1188, 1507, 1874, 2348, 2908, 3604, 4428, 5472, 6675, 8169, 9939, 12096, 14622, 17713, 21322, 25687, 30808, 36924, 44107, 52701, 62697, 74572, 88457, 104850, 123934
Offset: 0
Keywords
Examples
The partitions of 8 are [8], [7,1], [6,2], [6,1,1], [5,3], [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 15 partitions whose difference multiset includes at least one duplicate are all the 22 partitions of 8 except these 7: [8], [7,1], [6,2], [5,3], [6,2], [5,3], [5,2,1], [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], False]; Table[a[n], {n, 1, 30}] -
Python
from collections import Counter from itertools import combinations from sympy.utilities.iterables import partitions def A364612(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
Extensions
More terms from Alois P. Heinz, Sep 12 2023
Comments