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A050291
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Number of double-free subsets of {1, 2, ..., n}.
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31
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1, 2, 3, 6, 10, 20, 30, 60, 96, 192, 288, 576, 960, 1920, 2880, 5760, 9360, 18720, 28080, 56160, 93600, 187200, 280800, 561600, 898560, 1797120, 2695680, 5391360, 8985600, 17971200, 26956800, 53913600, 87091200, 174182400, 261273600, 522547200, 870912000
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OFFSET
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0,2
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COMMENTS
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A set is double-free if it does not contain both x and 2x.
So these are equally "half-free" subsets. - Gus Wiseman, Jul 08 2019
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REFERENCES
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Wang, E. T. H. ``On Double-Free Sets of Integers.'' Ars Combin. 28, 97-100, 1989.
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LINKS
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FORMULA
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a(n) = a(n-1)*Fibonacci(b(2n)+2)/Fibonacci(b(2n)+1), Fibonacci = A000045, b = A007814.
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EXAMPLE
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The a(0) = 1 through a(5) = 20 double-free subsets:
{} {} {} {} {} {}
{1} {1} {1} {1} {1}
{2} {2} {2} {2}
{3} {3} {3}
{1,3} {4} {4}
{2,3} {1,3} {5}
{1,4} {1,3}
{2,3} {1,4}
{3,4} {1,5}
{1,3,4} {2,3}
{2,5}
{3,4}
{3,5}
{4,5}
{1,3,4}
{1,3,5}
{1,4,5}
{2,3,5}
{3,4,5}
{1,3,4,5}
(End)
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MAPLE
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a:= proc(n) option remember; `if`(n=0, 1, (F-> (p-> a(n-1)*F(p+3)
/F(p+2))(padic[ordp](n, 2)))(j-> (<<0|1>, <1|1>>^j)[1, 2]))
end:
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MATHEMATICA
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a[n_] := a[n] = (b = IntegerExponent[2n, 2]; a[n-1]*Fibonacci[b+2]/Fibonacci[b+1]); a[1]=2; Table[a[n], {n, 1, 34}] (* Jean-François Alcover, Oct 10 2012, from first formula *)
Table[Length[Select[Subsets[Range[n]], Intersection[#, #/2]=={}&]], {n, 0, 10}] (* Gus Wiseman, Jul 08 2019 *)
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PROG
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(PARI) first(n)=my(v=vector(n)); v[1]=2; for(k=2, n, v[k]=v[k-1]*fibonacci(valuation(k, 2)+3)/fibonacci(valuation(k, 2)+2)); v \\ Charles R Greathouse IV, Feb 07 2017
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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