A230025 Triangular array: t(n, k) = number of occurrences of k as the number of outliers in all the partitions of n.
1, 0, 2, 1, 0, 2, 1, 2, 0, 2, 1, 2, 2, 0, 2, 1, 2, 4, 2, 0, 2, 1, 4, 2, 4, 2, 0, 2, 2, 2, 6, 2, 6, 2, 0, 2, 2, 6, 2, 8, 2, 6, 2, 0, 2, 2, 4, 12, 2, 8, 2, 8, 2, 0, 2, 2, 8, 6, 14, 2, 10, 2, 8, 2, 0, 2, 3, 6, 14, 8, 18, 2, 10, 2, 10, 2, 0, 2, 3, 10, 10, 20, 10
Offset: 1
Examples
The first 9 rows: 1 0 2 1 0 2 1 2 0 2 1 2 2 0 2 1 2 4 2 0 2 1 4 2 4 2 0 2 2 2 6 2 6 2 0 2 2 6 2 8 2 6 2 0 2 The Ferrers graph of the partition p = [4,4,1,1] of 10 follows: 1 1 1 1 1 1 1 1 1 1 The self-conjugate portion of p is 1 1 1 1 1 1 1 1 so that the number of outliers of p is 2.
Programs
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Mathematica
ferrersMatrix[list_] := PadRight[Map[Table[1, {#}] &, #], {#, #} &[Max[#, Length[#]]]] &[list]; conjugatePartition[part_] := Table[Count[#, ?(# >= i &)], {i, First[#]}] &[part]; selfConjugatePortion[list] := ferrersMatrix[#]*ferrersMatrix[conjugatePartition[#]] &[list]; outliers[list_] := Count[Flatten[ferrersMatrix[#] - selfConjugatePortion[#] &[list]], 1]; a[n_] := Map[outliers, IntegerPartitions[n]]; t = Table[Count[a[n], k], {n, 1, 13}, {k, 0, n - 1}] u = Flatten[t] (* Peter J. C. Moses, Feb 21 2014 *)
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