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A286281
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a(n) = floor the elevator is on at the n-th stage of Ken Knowlton's elevator problem, version 2.
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4
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1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 2, 1, 2, 1, 2, 3, 4
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OFFSET
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1,2
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COMMENTS
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An elevator steps up or down a floor at a time. It starts at floor 1, and always goes up from floor 1. From each floor m, it steps up every m-th time it stops there (except that stops when the elevator is going down don't count), otherwise down.
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REFERENCES
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Ken Knowlton, Email to R. L. Graham and N. J. A. Sloane, May 04 2017
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LINKS
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MAPLE
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hit:=Array(1..50, 0);
hit[1]:=1; a:=[1]; dir:=1; f:=1;
for s from 2 to 1000 do
if dir>0 or f=1 then f:=f+1; hit[f]:=hit[f]+1; dir:=1; else f:=f-1; dir:=-1; fi;
a:=[op(a), f];
if (dir=1) and ((hit[f] mod f) = 0) then dir:=1; else dir:=-1; fi;
od:
a;
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MATHEMATICA
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f[n_, m_: 20] := Block[{a = {}, r = ConstantArray[0, m], f = 1, d = 0}, Do[AppendTo[a, f]; If[d == 1, r = MapAt[# + 1 &, r, f]]; If[Or[And[ Divisible[r[[f]], f], d == 1], f == 1], f++; d = 1, f--; d = -1], {i, n}]; a]; f@ 100 (* Michael De Vlieger, May 10 2017 *)
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CROSSREFS
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See A285200 for the first version of the elevator problem.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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