A319841 Number of distinct positive integers that can be obtained by iteratively adding or multiplying together parts of the integer partition with Heinz number n until only one part remains.
0, 1, 1, 2, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 2, 4, 1, 3, 1, 4, 2, 2, 1, 5, 2, 2, 2, 4, 1, 5, 1, 6, 2, 2, 2, 6, 1, 2, 2, 7, 1, 6, 1, 4, 4, 2, 1, 8, 2, 5, 2, 4, 1, 6, 2, 8, 2, 2, 1, 7, 1, 2, 4, 9, 2, 6, 1, 4, 2, 6, 1, 8, 1, 2, 6, 4, 2, 6, 1, 9, 4, 2, 1, 10, 2, 2, 2
Offset: 1
Keywords
Examples
60 is the Heinz number of (3,2,1,1) and 5 = (3+2)*1*1 6 = 3*2*1*1 7 = 3+2+1+1 8 = (3+1)*2*1 9 = 3*(2+1)*1 10 = (3+2)*(1+1) 12 = (3+1)*(2+1) so we have a(60) = 7. It is not possible to obtain 11 by adding or multiplying together the parts of (3,2,1,1).
Crossrefs
Programs
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Mathematica
ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]]; Table[Length[Select[ReplaceListRepeated[{If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]],{n,100}]
Formula
a(2^n) = A048249(n).