A370817 Greatest number of multisets that can be obtained by choosing a prime factor of each factor in an integer factorization of n into unordered factors > 1.
1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 4, 1, 2, 2, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 2
Offset: 1
Keywords
Examples
For the factorizations of 60 we have the following choices (using prime indices {1,2,3} instead of prime factors {2,3,5}): (2*2*3*5): {{1,1,2,3}} (2*2*15): {{1,1,2},{1,1,3}} (2*3*10): {{1,1,2},{1,2,3}} (2*5*6): {{1,1,3},{1,2,3}} (3*4*5): {{1,2,3}} (2*30): {{1,1},{1,2},{1,3}} (3*20): {{1,2},{2,3}} (4*15): {{1,2},{1,3}} (5*12): {{1,3},{2,3}} (6*10): {{1,1},{1,2},{1,3},{2,3}} (60): {{1},{2},{3}} So a(60) = 4.
Links
- Max Alekseyev, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]]; Table[Max[Length[Union[Sort/@Tuples[If[#==1,{},First/@FactorInteger[#]]&/@#]]]&/@facs[n]],{n,100}]
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