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A366856
a(n) = number of partitions p of n such that (Ferrers conjugate of p) = contraconjugate of p.
0
1, 2, 3, 5, 7, 7, 8, 6, 10, 8, 8, 8, 8, 8, 8, 8, 8, 8, 12, 6, 10, 10, 8, 6, 6, 8, 12, 8, 12, 8, 10, 6, 8, 12, 10, 6, 10, 6, 8, 6, 8, 12, 8, 8, 8, 10, 8, 14, 14, 16, 12, 6, 8, 8, 10, 6, 12, 8, 8, 8
OFFSET
1,2
COMMENTS
See A366745 for the definition of contraconjugate.
EXAMPLE
The partitions of 6 in reverse lexicographic order:
[6], [5,1], [4,2], [4,1,1], [3,3], [3,2,1], [3,1,1,1], [2,2,2], [2,2,1,1], [2,1,1,1,1], [1,1,1,1,1,1]
Reversed (i.e., the contraconjugates of the partitions of 6, respectively)::
[1,1,1,1,1,1], [2,1,1,1,1], [2,2,1,1], [2,2,2], [3,1,1,1], [3,2,1], [3,3], [4,1,1], [4,2], [5,1], [6]
Ferrers conjugates:
[1,1,1,1,1,1], [2,1,1,1,1], [2,2,1,1], [3,1,1,1], [2,2,2], [3,2,1], [4,1,1], [3,3], [4,2], [5,1], [6]
Comparing the 2nd and 3rd lists shows that 11-4 = 7 partitions of 6 have identical Ferrers conjugate and contraconjugate, so that a(6) = 7.
MATHEMATICA
c[n_] := PartitionsP[n]
p[n_] := p[n] = IntegerPartitions[n];
r[n_] := r[n] = Reverse[p[n]]
q1[u_] := q1[u] = Table[Count[#, _?(# >= i &)], {i, First[#]}] &[u];
(* q1[u]=conjugate of partition u *)
q[n_] := q[n] = Table[q1[p[n][[k]]], {k, 1, c[n]}]
s[n_] := s[n] = Select[Range[c[n]], r[n][[#]] == q[n][[#]] &]
Table[Length[s[n]], {n, 1, 20}]
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Clark Kimberling, Dec 05 2023
STATUS
approved