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

1,2

Comments

See A366745 for the definition of contraconjugate.

Links

Table of n, a(n) for n=1..60.

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

Cf. A000041, A366745, A366746.

Sequence in context: A318954 A296375 A306923 * A137750 A196365 A195882

Adjacent sequences: A366853 A366854 A366855 * A366857 A366858 A366859

Keyword

nonn,more

Author

Clark Kimberling, Dec 05 2023

Status

approved