A240181 Array: t(n,k) is the number of partitions p of n such that the number of distinct numbers in the intersection of p and its conjugate is k, for k >= 0, n >= 1.
0, 1, 2, 2, 0, 1, 2, 3, 2, 4, 1, 4, 6, 0, 1, 4, 8, 3, 8, 8, 5, 1, 10, 9, 11, 10, 22, 8, 1, 1, 14, 22, 17, 3, 18, 34, 19, 5, 1, 18, 50, 21, 12, 26, 60, 34, 13, 2, 30, 74, 52, 19, 0, 1, 36, 105, 57, 29, 4, 44, 120, 93, 34, 5, 1, 60, 144, 128, 40, 13, 64, 186
Offset: 1
Examples
First 15 rows: 0 ... 1 2 2 ... 0 ... 1 2 ... 3 2 ... 4 ... 1 4 ... 6 ... 0 ...1 4 ... 8 ... 3 8 ... 8 ... 5 ... 1 10 .. 9 ... 11 10 .. 22 .. 8 ... 1 ... 1 14 .. 22 .. 17 .. 3 18 .. 34 .. 19 .. 5 ... 1 18 .. 50 .. 21 .. 12 26 .. 60 .. 34 .. 13 .. 2 30 .. 74 .. 52 .. 19 .. 0 .. 1 In the following table, p and c(p) denote a partition of 6 and its conjugate: p ........ c(p) 6 ........ 111111 51 ....... 21111 42 ....... 2211 411 ...... 3111 33 ....... 222 321 ...... 321 3111 ..... 411 222 ...... 33 2211 ..... 42 21111 .... 51 111111 ... 6 Let I(p) be number of numbers in the intersection of c and c(p); Then I(p) = 0 for 4 choices of p, I(p) = 1 for 6 choices, I(p) = 2 for 0 choices, and I(p) = 3 for 1 choice. Thus, row 6 is 4 6 0 1.
Links
- Clark Kimberling, Table of n, a(n) for n = 1..200
Programs
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Mathematica
z = 30; conjugatePartition[part_] := Table[Count[#, _?(# >= i &)], {i, First[#]}] &[part]; c = Map[BinCounts[#, {0, 1 + Max[#]}] &[Map[Length, Map[Intersection[#, conjugatePartition[#]] &, IntegerPartitions[#]]]] &, Range[z]]; Flatten[c] (* this sequence *) Table[Length[c[[n]]], {n, 1, z}] (* A240450 *) (* Peter J. C. Moses, Apr 10 2014 *)
Extensions
Name corrected by Clark Kimberling, Sep 28 2023
Comments