A326016
Number of knapsack partitions of n such that no addition of one part up to the maximum is knapsack.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 1, 0, 3, 0, 0, 0, 1, 0, 8, 0, 8, 4, 3, 0, 11, 5, 3, 2, 5, 0, 29, 2, 9, 8, 20, 2
Offset: 1
The initial terms count the following partitions:
15: (5,4,3,3)
21: (7,6,5,3)
21: (7,5,3,3,3)
24: (8,7,6,3)
25: (7,5,5,4,4)
27: (9,8,7,3)
27: (9,7,6,5)
27: (8,7,3,3,3,3)
31: (10,8,6,6,1)
33: (11,9,7,3,3)
33: (11,8,5,5,4)
33: (11,7,6,6,3)
33: (11,7,3,3,3,3,3)
33: (11,5,5,4,4,4)
33: (10,9,8,3,3)
33: (10,8,6,6,3)
33: (10,8,3,3,3,3,3)
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sums[ptn_]:=sums[ptn]=If[Length[ptn]==1,ptn,Union@@(Join[sums[#],sums[#]+Total[ptn]-Total[#]]&/@Union[Table[Delete[ptn,i],{i,Length[ptn]}]])];
ksQ[y_]:=Length[sums[Sort[y]]]==Times@@(Length/@Split[Sort[y]]+1)-1;
maxks[n_]:=Select[IntegerPartitions[n],ksQ[#]&&Select[Table[Sort[Append[#,i]],{i,Range[Max@@#]}],ksQ]=={}&];
Table[Length[maxks[n]],{n,30}]
A325867
Number of maximal subsets of {1..n} containing n such that every subset has a different sum.
Original entry on oeis.org
1, 1, 2, 2, 4, 8, 10, 12, 17, 34, 45, 77, 99, 136, 166, 200, 238, 328, 402, 660, 674, 1166, 1331, 1966, 2335, 3286, 3527, 4762, 5383, 6900, 7543, 9087, 10149, 12239, 13569, 16452, 17867, 22869, 23977, 33881, 33820, 43423, 48090, 68683, 67347, 95176, 97917, 131666, 136205
Offset: 1
The a(1) = 1 through a(8) = 12 subsets:
{1} {1,2} {1,3} {1,2,4} {1,2,5} {1,2,6} {1,2,7} {1,3,8}
{2,3} {2,3,4} {1,3,5} {1,3,6} {1,3,7} {1,5,8}
{2,4,5} {1,4,6} {1,4,7} {5,7,8}
{3,4,5} {2,3,6} {1,5,7} {1,2,4,8}
{2,5,6} {2,3,7} {1,4,6,8}
{3,4,6} {2,4,7} {2,3,4,8}
{3,5,6} {2,6,7} {2,4,5,8}
{4,5,6} {4,5,7} {2,4,7,8}
{4,6,7} {3,4,6,8}
{3,5,6,7} {3,6,7,8}
{4,5,6,8}
{4,6,7,8}
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fasmax[y_]:=Complement[y,Union@@(Most[Subsets[#]]&)/@y];
Table[Length[fasmax[Select[Subsets[Range[n]],MemberQ[#,n]&&UnsameQ@@Plus@@@Subsets[#]&]]],{n,15}]
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def f(p0, n, m, cm):
full, t, p = True, 0, p0
while p>k)&1)==0 and ((m<Bert Dobbelaere, Mar 07 2021
A059519
Number of partitions of n all of whose subpartitions sum to distinct values. Partition(n) = [a, b, c...] where 2n = 2^a + 2^b + 2^c + ...
Original entry on oeis.org
1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 14, 16, 17, 18, 19, 20, 21, 24, 26, 28, 32, 33, 34, 35, 36, 37, 38, 40, 41, 44, 48, 50, 52, 56, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 80, 81, 84, 88, 96, 98, 100, 104, 112, 116, 128, 129, 130, 131, 132, 133, 134, 136, 137, 138, 139, 140
Offset: 1
14=2+4+8 so Partition(14) = [2,3,4], whose sub-sums are 0,2,3,4,5,6,7 and 14.
Other sequences classifying numbers by their binary indices:
A291166 (relatively prime),
A295235 (arithmetic progression),
A326669 (integer average),
A326675 (pairwise coprime).
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bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],UnsameQ@@Total/@Subsets[bpe[#]]&] (* Gus Wiseman, Jul 22 2019 *)
A326034
Number of knapsack partitions of n with largest part 3.
Original entry on oeis.org
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
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)
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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}]
A326018
Heinz numbers of knapsack partitions such that no addition of one part up to the maximum is knapsack.
Original entry on oeis.org
1925, 12155, 20995, 23375, 37145
Offset: 1
The sequence of terms together with their prime indices begins:
1925: {3,3,4,5}
12155: {3,5,6,7}
20995: {3,6,7,8}
23375: {3,3,3,5,7}
37145: {3,7,8,9}
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ksQ[y_]:=UnsameQ@@Total/@Union[Subsets[y]];
Select[Range[2,200],With[{phm=If[#==1,{},Flatten[Cases[FactorInteger[#],{p_,k_}:>Table[PrimePi[p],{k}]]]]},ksQ[phm]&&Select[Table[Sort[Append[phm,i]],{i,Max@@phm}],ksQ]=={}]&]
A326035
Number of uniform knapsack partitions of n.
Original entry on oeis.org
1, 1, 2, 3, 4, 4, 6, 6, 9, 10, 12, 12, 17, 16, 20, 25, 27, 29, 35, 39, 44, 57, 53, 66, 75, 84, 84, 114, 112, 131, 133, 162, 167, 209, 192, 242, 250, 289, 279, 363, 348, 417, 404, 502, 487, 608, 557, 706, 682, 835, 773, 1004, 922, 1149, 1059, 1344, 1257, 1595
Offset: 0
The a(1) = 1 through a(8) = 9 partitions:
(1) (2) (3) (4) (5) (6) (7) (8)
(11) (21) (22) (32) (33) (43) (44)
(111) (31) (41) (42) (52) (53)
(1111) (11111) (51) (61) (62)
(222) (421) (71)
(111111) (1111111) (521)
(2222)
(3311)
(11111111)
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sums[ptn_]:=sums[ptn]=If[Length[ptn]==1,ptn,Union@@(Join[sums[#],sums[#]+Total[ptn]-Total[#]]&/@Union[Table[Delete[ptn,i],{i,Length[ptn]}]])];
ks[n_]:=Select[IntegerPartitions[n],Length[sums[Sort[#]]]==Times@@(Length/@Split[#]+1)-1&];
Table[Length[Select[ks[n],SameQ@@Length/@Split[#]&]],{n,30}]
A326033
Number of knapsack partitions of n such that no addition of one part equal to an existing part is knapsack.
Original entry on oeis.org
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 2, 0, 0, 1, 1, 0, 3, 0, 0, 1, 1, 0, 8, 0, 8, 4, 3, 0, 11, 5, 3, 4, 5, 0, 30, 2, 9, 9, 20, 3, 37, 6, 18, 16, 37, 20, 71, 12, 37, 40
Offset: 1
The partition (10,8,6,6) is counted under a(30) because (10,10,8,6,6), (10,8,8,6,6), and (10,8,6,6,6) are not knapsack.
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sums[ptn_]:=sums[ptn]=If[Length[ptn]==1,ptn,Union@@(Join[sums[#],sums[#]+Total[ptn]-Total[#]]&/@Union[Table[Delete[ptn,i],{i,Length[ptn]}]])];
ksQ[y_]:=Length[sums[Sort[y]]]==Times@@(Length/@Split[Sort[y]]+1)-1;
maxks[n_]:=Select[IntegerPartitions[n],ksQ[#]&&Select[Table[Sort[Append[#,i]],{i,Union[#]}],ksQ]=={}&];
Table[Length[maxks[n]],{n,30}]
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