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%I #19 Sep 28 2023 18:32:26
%S 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,
%T 22,17,3,18,34,19,5,1,18,50,21,12,26,60,34,13,2,30,74,52,19,0,1,36,
%U 105,57,29,4,44,120,93,34,5,1,60,144,128,40,13,64,186
%N 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.
%C First two columns are A240674 and A240675. Sum of numbers in row n is A000041(n), for n >= 1. Number of numbers in row n is A240450(n).
%H Clark Kimberling, <a href="/A240181/b240181.txt">Table of n, a(n) for n = 1..200</a>
%e First 15 rows:
%e 0 ... 1
%e 2
%e 2 ... 0 ... 1
%e 2 ... 3
%e 2 ... 4 ... 1
%e 4 ... 6 ... 0 ...1
%e 4 ... 8 ... 3
%e 8 ... 8 ... 5 ... 1
%e 10 .. 9 ... 11
%e 10 .. 22 .. 8 ... 1 ... 1
%e 14 .. 22 .. 17 .. 3
%e 18 .. 34 .. 19 .. 5 ... 1
%e 18 .. 50 .. 21 .. 12
%e 26 .. 60 .. 34 .. 13 .. 2
%e 30 .. 74 .. 52 .. 19 .. 0 .. 1
%e In the following table, p and c(p) denote a partition of 6 and its conjugate:
%e p ........ c(p)
%e 6 ........ 111111
%e 51 ....... 21111
%e 42 ....... 2211
%e 411 ...... 3111
%e 33 ....... 222
%e 321 ...... 321
%e 3111 ..... 411
%e 222 ...... 33
%e 2211 ..... 42
%e 21111 .... 51
%e 111111 ... 6
%e 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.
%t 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 *)
%t Table[Length[c[[n]]], {n, 1, z}] (* A240450 *) (* _Peter J. C. Moses_, Apr 10 2014 *)
%Y Cf. A240674, A240675, A240450.
%K nonn,easy,tabf
%O 1,3
%A _Clark Kimberling_, Apr 12 2014
%E Name corrected by _Clark Kimberling_, Sep 28 2023