A050326 Number of factorizations of n into distinct squarefree numbers > 1.
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, 5, 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, 4, 1, 2, 1, 0, 2, 5, 1, 1, 2, 5, 1, 0, 1, 2, 1, 1, 2, 5, 1, 0, 0, 2, 1, 4, 2, 2, 2, 0, 1, 4, 2, 1, 2, 2, 2, 0, 1, 1, 1, 1, 1, 5, 1
Offset: 1
Keywords
Examples
The a(30) = 5 factorizations are: 2*3*5, 2*15, 3*10, 5*6, 30. The a(180) = 5 factorizations are: 2*3*5*6, 2*3*30, 2*6*15, 3*6*10, 6*30. - _Gus Wiseman_, Oct 10 2017
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Haskell
import Data.List (subsequences, genericIndex) a050326 n = genericIndex a050326_list (n-1) a050326_list = 1 : f 2 where f x = (if x /= s then a050326 s else length $ filter (== x) $ map product $ subsequences $ tail $ a206778_row x) : f (x + 1) where s = a046523 x -- Reinhard Zumkeller, May 03 2013
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Maple
N:= 1000: # to get a(1)..a(N) A:= Vector(N): A[1]:= 1: for n from 2 to N do if numtheory:-issqrfree(n) then S:= [$1..N/n]; T:= n*S; A[T]:= A[T]+A[S] fi; od: convert(A,list); # Robert Israel, Oct 10 2017
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Mathematica
sqfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[sqfacs[n/d],Min@@#>d&]],{d,Select[Rest[Divisors[n]],SquareFreeQ]}]]; Table[Length[sqfacs[n]],{n,100}] (* Gus Wiseman, Oct 10 2017 *)
Formula
Dirichlet g.f.: prod{n is squarefree and > 1}(1+1/n^s).
Comments