A286281 -id:A286281 - cp's OEIS frontend

A285200 a(n) = floor the elevator is on at the n-th stage of Ken Knowlton's elevator problem, version 1.

1, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1, 2, 3, 2, 1, 2, 3, 4, 3, 2, 1

Offset: 1

N. J. A. Sloane, May 02 2017

nonn

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, otherwise down.

See A285202 for an alternative way to display this sequence.

Ken Knowlton, Email to R. L. Graham, Apr 26 2017

N. J. A. Sloane, Table of n, a(n) for n = 1..20000
Ken Knowlton, Illustration showing what floor the elevator is on for the first 49 stages

Cf. A285201, A285202, A285203.

See A286281 for a second version of the elevator problem.

hit:=Array(1..50,0);
hit[1]:=1; a:=[1]; dir:=1; f:=1;
for s from 2 to 1000 do
if dir>0 then f:=f+1; else f:=f-1; fi;
hit[f]:=hit[f]+1; a:=[op(a),f];
if (hit[f] mod f) = 0 then dir:=1; else dir:=-1; fi;
od:
a;

A285201 Stage at which Ken Knowlton's elevator (version 1) reaches floor n for the first time.

Original entry on oeis.org

1, 2, 5, 14, 45, 174, 825, 4738, 32137, 251338, 2224157, 21952358, 238962581, 2843085270, 36696680241, 510647009850, 7619901954001, 121367981060434, 2055085325869813, 36861997532438654, 698193329457246653, 13924819967953406654, 291683979376372766697, 6402385486361598687666, 146948520147021794869977

Offset: 1

R. L. Graham, May 02 2017

nonn

Indices of records in A285200.

When prefixed by a(0)=0, the first differences give A111063. - N. J. A. Sloane, May 03 2017

N. J. A. Sloane, Table of n, a(n) for n = 1..400
Ken Knowlton, Illustration showing what floor the elevator is on for the first 49 stages

Cf. A285200, A285203, A111063, A098558, A052849, A286281, A286282.

a:= proc(n) option remember; `if`(n<3, n, ((n-1)^2*a(n-1)
      -(n-2)*(2*n-3)*a(n-2)+(n-1)*(n-3)*a(n-3))/(n-2))
    end:
seq(a(n), n=1..25);  # Alois P. Heinz, Jul 11 2018

a[n_] := 2 - n + 2 Sum[k!/j!, {k, 0, n-2}, {j, 0, k}];
Array[a, 25] (* Jean-François Alcover, Nov 01 2020 *)

a(n) = 2 - n + 2 * Sum_{k=0..n-2} Sum_{j=0..k} k!/j!.
For n >= 2, a(n) = 1+n+2*Sum_{k=2..n} C(n,k)*(k-1)! = 1+n+2*n!*Sum_{k=2..n} 1/(k*(n-k)!). - N. J. A. Sloane, May 03 2017
E.g.f.: exp(x)*(1-x-2*log(1-x)). Omitting the factor exp(x), this gives (essentially) the e.g.f. for A098558 (or A052849). - N. J. A. Sloane, May 03 2017

A286282 Stage at which Ken Knowlton's elevator (version 2) reaches floor n for the first time.

Original entry on oeis.org

1, 2, 5, 18, 79, 408, 2469, 17314, 138555, 1247052, 12470593, 137176614, 1646119479, 21399553360, 299593747197, 4493906208138, 71902499330419, 1222342488617364, 22002164795112825, 418041131107143982, 8360822622142879983, 175577275065000480024, 3862700051430010560949

Offset: 1

N. J. A. Sloane, May 09 2017

nonn

Indices of records in A286281.
Theorem: Let b(n) = Sum_{k=0..n} n!/k! = A000522(n). Then a(n) = 2*b(n-1)-n+2-2*(n-1)!. - R. L. Graham, May 10 2017
This implies the following recurrence (conjectured by N. J. A. Sloane on May 09 2017): a(1)=1, and for n>=1, a(n+1) = n*a(n) + n^2 - 3*n + 3. From the asymptotic expansion of b(n) (see A000522), we have a(n) ~ 2*(e-1)*(n-1)!.

N. J. A. Sloane, Table of n, a(n) for n = 1..400

Cf. A000522, A002627, A286281, A285201.

A286282 := proc(n)
    2*A002627(n-1)-n+2 ;
end proc:
seq(A286282(n),n=1..21) ; # R. J. Mathar, May 21 2017

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]; Rest@ Map[First, Values@ PositionIndex@ FoldList[Max, 0, f@ 200000]] - 1 (* Michael De Vlieger, May 10 2017, Version 10 *)

times = {1: 1, 2: 1, 3: 1, 4: 1, 5: 1, 6: 1, 7: 1, 8: 1, 9: 1, 10: 1, 11: 1, 12: 1, 13: 1, 14: 1, 15: 1, 16: 1}
first = {1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0, 10: 0, 11: 0, 12: 0, 13: 0, 14: 0, 15: 0, 16: 0}
floor = 1
steps = 1
while floor < 17:
    if first[floor] == 0:
        first[floor] = 1
        print("First Time: ",floor,steps)
    if floor == 1:
        floor += 1
    else:
        if times[floor] < floor:
            times[floor] += 1
            floor -= 1
        else:
            times[floor] = 0
            floor += 1
    steps += 1
print(floor, steps)
# David Consiglio, Jr., May 09 2017

a(n) = 2*A002627(n-1) - (n-2). - N. J. A. Sloane, May 15 2017
Conjecture: a(n) +(-n-2)*a(n-1) +3*(n-1)*a(n-2) +(-3*n+8)*a(n-3) +(n-4)*a(n-4)=0. - R. J. Mathar, May 21 2017
Conjecture: (n+1)*a(n) +(-n^2+3*n-27)*a(n-1) +3*(-n^2+10*n-13)*a(n-2) +(n-3)*(4*n-17)*a(n-3)=0. - R. J. Mathar, May 21 2017

a(10)-a(13) from David Consiglio, Jr., May 09 2017

Further terms added by N. J. A. Sloane, May 10 2017 based on R. L. Graham's formula.

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A285200 a(n) = floor the elevator is on at the n-th stage of Ken Knowlton's elevator problem, version 1.

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A285201 Stage at which Ken Knowlton's elevator (version 1) reaches floor n for the first time.

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A286282 Stage at which Ken Knowlton's elevator (version 2) reaches floor n for the first time.

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