A348379 Number of factorizations of n with an alternating permutation.
1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 6, 1, 2, 2, 4, 1, 5, 1, 5, 2, 2, 2, 8, 1, 2, 2, 6, 1, 5, 1, 4, 4, 2, 1, 10, 1, 4, 2, 4, 1, 6, 2, 6, 2, 2, 1, 11, 1, 2, 4, 6, 2, 5, 1, 4, 2, 5, 1, 15, 1, 2, 4, 4, 2, 5, 1, 10, 3, 2, 1, 11, 2
Offset: 1
Keywords
Examples
The a(270) = 19 factorizations: (2*3*3*15) (2*3*45) (2*135) (270) (2*3*5*9) (2*5*27) (3*90) (3*3*5*6) (2*9*15) (5*54) (3*3*30) (6*45) (3*5*18) (9*30) (3*6*15) (10*27) (3*9*10) (15*18) (5*6*9)
Links
- Wikipedia, Alternating permutation
Crossrefs
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]]}]]; wigQ[y_]:=Or[Length[y]==0,Length[Split[y]]==Length[y]&&Length[Split[Sign[Differences[y]]]]==Length[y]-1]; Table[Length[Select[facs[n],Select[Permutations[#],wigQ]!={}&]],{n,100}]
Formula
a(2^n) = A345170(n).
Comments