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A326034
Number of knapsack partitions of n with largest part 3.
8
0, 0, 0, 1, 1, 2, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 3, 1, 3, 2
OFFSET
0,6
COMMENTS
An integer partition is knapsack if every distinct submultiset has a different sum.
Appears to repeat the terms (2,2,2,3,1,3) ad infinitum.
I computed terms a(n) for n = 0..5000 and (2,2,2,3,1,3) is repeated continuously starting at a(8). - Fausto A. C. Cariboni, May 14 2021
EXAMPLE
The initial values count the following partitions:
3: (3)
4: (3,1)
5: (3,2)
5: (3,1,1)
6: (3,3)
7: (3,3,1)
7: (3,2,2)
8: (3,3,2)
8: (3,3,1,1)
9: (3,3,3)
9: (3,2,2,2)
10: (3,3,3,1)
10: (3,3,2,2)
11: (3,3,3,2)
11: (3,3,3,1,1)
11: (3,2,2,2,2)
12: (3,3,3,3)
13: (3,3,3,3,1)
13: (3,3,3,2,2)
13: (3,2,2,2,2,2)
14: (3,3,3,3,2)
14: (3,3,3,3,1,1)
15: (3,3,3,3,3)
15: (3,2,2,2,2,2,2)
MATHEMATICA
sums[ptn_]:=sums[ptn]=If[Length[ptn]==1, ptn, Union@@(Join[sums[#], sums[#]+Total[ptn]-Total[#]]&/@Union[Table[Delete[ptn, i], {i, Length[ptn]}]])];
kst[n_]:=Select[IntegerPartitions[n, All, {1, 2, 3}], Length[sums[Sort[#]]]==Times@@(Length/@Split[#]+1)-1&];
Table[Length[Select[kst[n], Max@@#==3&]], {n, 0, 30}]
KEYWORD
nonn
AUTHOR
Gus Wiseman, Jun 04 2019
STATUS
approved