A286281 a(n) = floor the elevator is on at the n-th stage of Ken Knowlton's elevator problem, version 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, 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

Offset: 1

Author

N. J. A. Sloane, May 09 2017

Keywords

• nonn

Comments

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.

References

• Ken Knowlton, Email to R. L. Graham and N. J. A. Sloane, May 04 2017

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..20000
• Ken Knowlton, Illustration of initial terms showing floors the two versions of the elevator are on. Top: version 1 (A285200), bottom: version 2 (A286281)

Crossrefs

For records see A286282.

See A285200 for the first version of the elevator problem.

Programs

Maple

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;

Mathematica

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 *)