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Triangular array: T(n,k) = number of strict partitions P of n into positive parts such that P includes a partition of k.
7

%I #21 Nov 22 2023 12:25:59

%S 1,0,1,1,1,2,1,0,1,2,1,1,1,1,3,2,2,1,2,2,4,2,2,2,2,2,2,5,3,2,3,1,3,2,

%T 3,6,3,3,4,3,3,4,3,3,8,5,4,5,4,3,4,5,4,5,10,5,5,5,5,5,5,5,5,5,5,12,7,

%U 6,7,7,7,4,7,7,7,6,7,15,8,7,8,8,8,8,8

%N Triangular array: T(n,k) = number of strict partitions P of n into positive parts such that P includes a partition of k.

%H Clark Kimberling, <a href="/A237194/b237194.txt">Table of n, a(n) for n = 1..1000</a>

%F T(n,k) = T(n,n-k) for k=1..n-1, n >= 2.

%e First 13 rows:

%e 1

%e 0 1

%e 1 1 2

%e 1 0 1 2

%e 1 1 1 1 3

%e 2 2 1 2 2 4

%e 2 2 2 2 2 2 5

%e 3 2 3 1 3 2 3 6

%e 3 3 4 3 3 4 3 3 8

%e 5 4 5 4 3 4 5 4 5 10

%e 5 5 5 5 5 5 5 5 5 5 12

%e 7 6 7 7 7 4 7 7 7 6 7 15

%e 8 7 8 8 8 8 8 8 8 8 7 8 18

%e T(12,4) = 7 counts these partitions: [8,4], [8,3,1], [7,4,1], [6,4,2], [6,3,2,1], [5,4,3], [5,4,2,1].

%t Table[theTotals = Map[{#, Map[Total, Subsets[#]]} &, Select[IntegerPartitions[nn], # == DeleteDuplicates[#] &]]; Table[Length[Map[#[[1]] &, Select[theTotals, Length[Position[#[[2]], sumTo]] >= 1 &]]], {sumTo, nn}], {nn, 45}] // TableForm

%t u = Flatten[%] (* _Peter J. C. Moses_, Feb 04 2014 *)

%t Table[Length[Select[IntegerPartitions[n], UnsameQ@@#&&MemberQ[Total/@Subsets[#], k]&]], {n,6}, {k,n}] (* _Gus Wiseman_, Nov 16 2023 *)

%Y Column k = n is A000009.

%Y Column k = 2 is A015744.

%Y Column k = 1 is A025147.

%Y The non-strict complement is obtained by adding zeros after A046663.

%Y Diagonal n = 2k is A237258.

%Y Row sums are A284640.

%Y For subsets instead of partitions we have A365381.

%Y The non-strict version is obtained by removing column k = 0 from A365543.

%Y Including column k = 0 gives A365661.

%Y The complement is obtained by adding zeros after A365663.

%Y Cf. A002219, A006827, A108796, A108917, A122768, A275972, A299701, A304792, A364272.

%K nonn,tabl,easy

%O 1,6

%A _Clark Kimberling_, Feb 05 2014