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%I #9 Sep 17 2023 12:08:33
%S 1,1,1,2,1,2,3,2,2,3,5,3,3,3,5,7,5,5,5,5,7,11,7,8,6,8,7,11,15,11,11,
%T 11,11,11,11,15,22,15,17,15,14,15,17,15,22,30,22,23,23,22,22,23,23,22,
%U 30,42,30,33,30,33,25,33,30,33,30,42
%N Triangle read by rows where T(n,k) is the number of integer partitions of n with a submultiset summing to k.
%C Rows are palindromic.
%e Triangle begins:
%e 1
%e 1 1
%e 2 1 2
%e 3 2 2 3
%e 5 3 3 3 5
%e 7 5 5 5 5 7
%e 11 7 8 6 8 7 11
%e 15 11 11 11 11 11 11 15
%e 22 15 17 15 14 15 17 15 22
%e 30 22 23 23 22 22 23 23 22 30
%e 42 30 33 30 33 25 33 30 33 30 42
%e 56 42 45 44 44 43 43 44 44 45 42 56
%e 77 56 62 58 62 56 53 56 62 58 62 56 77
%e Row n = 6 counts the following partitions:
%e (6) (51) (42) (33) (42) (51) (6)
%e (51) (411) (411) (321) (411) (411) (51)
%e (42) (321) (321) (3111) (321) (321) (42)
%e (411) (3111) (3111) (2211) (3111) (3111) (411)
%e (33) (2211) (222) (21111) (222) (2211) (33)
%e (321) (21111) (2211) (111111) (2211) (21111) (321)
%e (3111) (111111) (21111) (21111) (111111) (3111)
%e (222) (111111) (111111) (222)
%e (2211) (2211)
%e (21111) (21111)
%e (111111) (111111)
%t Table[Length[Select[IntegerPartitions[n],MemberQ[Total/@Subsets[#],k]&]],{n,0,15},{k,0,n}]
%Y Columns k = 0 and k = n are A000041.
%Y Central column n = 2k is A002219.
%Y The complement is counted by A046663, strict A365663.
%Y Row sums are A304792.
%Y For subsets instead of partitions we have A365381.
%Y The strict case is A365661.
%Y A000009 counts subsets summing to n.
%Y A000124 counts distinct possible sums of subsets of {1..n}.
%Y A364272 counts sum-full strict partitions, sum-free A364349.
%Y Cf. A088809, A093971, A122768, A108917, A299701, A364911, A365541, A365658.
%K nonn,tabl
%O 0,4
%A _Gus Wiseman_, Sep 16 2023
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