A381634 Number of multisets that can be obtained by taking the sum of each block of a set multipartition (multiset of sets) of the prime indices of n with distinct block-sums.
1, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 1, 1, 2, 2, 0, 1, 1, 1, 1, 2, 2, 1, 0, 0, 2, 0, 1, 1, 4, 1, 0, 2, 2, 2, 1, 1, 2, 2, 0, 1, 5, 1, 1, 1, 2, 1, 0, 0, 1, 2, 1, 1, 0, 2, 0, 2, 2, 1, 3, 1, 2, 1, 0, 2, 5, 1, 1, 2, 4, 1, 0, 1, 2, 1, 1, 2, 5, 1, 0, 0, 2, 1, 4, 2, 2, 2
Offset: 1
Keywords
Examples
The prime indices of 120 are {1,1,2,3}, with 3 ways: {{1},{1,2,3}} {{1,2},{1,3}} {{1},{2},{1,3}} with block-sums: {1,6}, {3,4}, {1,2,4}, so a(120) = 3. The prime indices of 210 are {1,2,3,4}, with 12 ways: {{1,2,3,4}} {{1},{2,3,4}} {{2},{1,3,4}} {{3},{1,2,4}} {{4},{1,2,3}} {{1,2},{3,4}} {{1,3},{2,4}} {{1},{2},{3,4}} {{1},{3},{2,4}} {{1},{4},{2,3}} {{2},{3},{1,4}} {{1},{2},{3},{4}} with block-sums: {10}, {1,9}, {2,8}, {3,7}, {4,6}, {3,7}, {4,6}, {1,2,7}, {1,3,6}, {1,4,5}, {2,3,5}, {1,2,3,4}, of which 10 are distinct, so a(210) = 10.
Crossrefs
Programs
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Mathematica
hwt[n_]:=Total[Cases[FactorInteger[n],{p_,k_}:>PrimePi[p]*k]]; sfacs[n_]:=If[n<=1,{{}},Join@@Table[(Prepend[#,d]&)/@Select[sfacs[n/d],Min@@#>=d&],{d,Select[Rest[Divisors[n]],SquareFreeQ]}]]; Table[Length[Union[Sort[hwt/@#]&/@Select[sfacs[n],UnsameQ@@hwt/@#&]]],{n,100}]
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