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a(n) = number of partitions p of n such that (Ferrers conjugate of p) = contraconjugate of p.
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%I #4 Dec 23 2023 14:41:47

%S 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,

%T 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

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

%C See A366745 for the definition of contraconjugate.

%e The partitions of 6 in reverse lexicographic order:

%e [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]

%e Reversed (i.e., the contraconjugates of the partitions of 6, respectively)::

%e [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]

%e Ferrers conjugates:

%e [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]

%e 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.

%t c[n_] := PartitionsP[n]

%t p[n_] := p[n] = IntegerPartitions[n];

%t r[n_] := r[n] = Reverse[p[n]]

%t q1[u_] := q1[u] = Table[Count[#, _?(# >= i &)], {i, First[#]}] &[u];

%t (* q1[u]=conjugate of partition u *)

%t q[n_] := q[n] = Table[q1[p[n][[k]]], {k, 1, c[n]}]

%t s[n_] := s[n] = Select[Range[c[n]], r[n][[#]] == q[n][[#]] &]

%t Table[Length[s[n]], {n, 1, 20}]

%Y Cf. A000041, A366745, A366746.

%K nonn,more

%O 1,2

%A _Clark Kimberling_, Dec 05 2023