A366856 a(n) = number of partitions p of n such that (Ferrers conjugate of p) = contraconjugate of p.

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

Author

Clark Kimberling, Dec 05 2023

Keywords

• nonn
• more

Comments

See A366745 for the definition of contraconjugate.

Examples

			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.
		

Crossrefs

Cf. A000041, A366745, A366746.

Programs

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}]