A381872 Number of multisets that can be obtained by taking the sum of each block of a multiset partition of the prime indices of n into blocks having a common sum.
1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1
Offset: 1
Keywords
Examples
The prime indices of 144 are {1,1,1,1,2,2}, with the following 4 multiset partitions having common block sum: {{1,1,1,1,2,2}} {{2,2},{1,1,1,1}} {{1,1,2},{1,1,2}} {{2},{2},{1,1},{1,1}} with sums: 8, 4, 4, 2, of which 3 are distinct, so a(144) = 3. The prime indices of 1296 are {1,1,1,1,2,2,2,2}, with the following 7 multiset partitions having common block sum: {{1,1,1,1,2,2,2,2}} {{2,2,2},{1,1,1,1,2}} {{1,1,2,2},{1,1,2,2}} {{2,2},{2,2},{1,1,1,1}} {{2,2},{1,1,2},{1,1,2}} {{1,2},{1,2},{1,2},{1,2}} {{2},{2},{2},{2},{1,1},{1,1}} with sums: 12, 6, 6, 4, 4, 3, 2, of which 5 are distinct, so a(1296) = 5.
Crossrefs
Programs
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Mathematica
prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}]; mps[mset_]:=Union[Sort[Sort/@(#/.x_Integer:>mset[[x]])]&/@sps[Range[Length[mset]]]]; Table[Length[Union[Sort[Total/@#]&/@Select[mps[prix[n]],SameQ@@Total/@#&]]],{n,100}]
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