A285201 -id:A285201 - 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;

A285203 Local high points in A285200.

Original entry on oeis.org

2, 3, 3, 4, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 6, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 6, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 5, 3, 4, 3, 4, 3, 6, 3, 4, 3, 4, 3, 5

Offset: 1

N. J. A. Sloane, May 02 2017

nonn

Ken Knowlton, Illustration showing what floor the elevator is on for the first 49 stages

Cf. A285200, A285201.

hit:=Array(1..50,0);
hit[1]:=1; a:=[1]; dir:=1; f:=1; pk:=[];
for s from 2 to 900 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;
if s>2 and a[s-2]a[s] then pk:=[op(pk),a[s-1]]; fi;
od:
a; # A285200
pk; # A285203

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.

A285202 A285200 displayed as an irregular triangle read by rows.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, May 03 2017

This is Ken Knowlton's elevator sequence A285200, but with an extra 1 each time the elevator returns to (or starts at) the first floor.

Row lengths are A285204. The row maxima are given in A285203 (for n>0).

			Triangle begins:
1,
1, 2, 1,
1, 2, 3, 2, 1,
1, 2, 3, 2, 1,
1, 2, 3, 4, 3, 2, 1,
1, 2, 3, 2, 1,
1, 2, 3, 4, 3, 2, 1,
1, 2, 3, 2, 1,
1, 2, 3, 4, 3, 2, 1,
1, 2, 3, 2, 1,
1, 2, 3, 4, 5, 4, 3, 2, 1,
...

Cf. A285200, A285201, A285203, A285204.

A360830 Numbers that when concatenated with the natural numbers from 1 to N are divisible by the corresponding order number.

Original entry on oeis.org

1, 3, 6, 42, 84, 252, 2772, 36036, 612612, 11639628, 267711444, 803134332, 23290895628, 722017764468, 1444035528936, 53429314570632, 2190601897395912, 94195881588024216, 4427206434637138152, 30990445042459967064

Offset: 1

Rodolfo Kurchan, Feb 22 2023

There is an elevator with the numbers from 1 to N. Each number goes up in the elevator and can go up as long as the concatenation of this number with the number of the floor (1 to N) is a multiple of the floor.
Example, the number 1 reaches floor 2, because 11 is divisible by 1, 12 is divisible by 2, but it does not reach floor 3 because 13 is not divisible by 3.
The sequence shows the numbers that can go higher in the elevator than the previous number.
For example, 2 cannot go higher than 1, so it does not appear, instead number 3 can go up to floor 3, since 31 is divisible by 1, 32 is divisible by 2 and 33 is divisible by 3, instead I couldn't get to floor 4 because 34 is not divisible by 4.

			42 goes after 6, because it is the smallest number that can go more than the 6th floor that can go number 6. 421, 422, 423, 424, 425, 426 and 427 are divisible by 1, 2, 3, 4, 5, 6 and 7, but 42 cannot go to floor 8th, because 428 it is not divisible by 8.
   n |         a(n)         | Maximum Elevator floor
  ---------------------------------------------------
   1 |                    1 |          2
   2 |                    3 |          3
   3 |                    6 |          6
   4 |                   42 |          7
   5 |                   84 |          8
   6 |                  252 |         10
   7 |                 2772 |         12
   8 |                36036 |         16
   9 |               612612 |         18
  10 |             11639628 |         22
  11 |            267711444 |         26
  12 |            803134332 |         28
  13 |          23290895628 |         30
  14 |         722017764468 |         31
  15 |        1444035528936 |         36
  16 |       53429314570632 |         40
  17 |     2190601897395912 |         42
  18 |    94195881588024216 |         46
  19 |  4427206434637138152 |         48
  20 | 30990445042459967064 |         52
  ...

Jaime Poniachik, Problem El Ascensor, La Odisea del Ingenio, May 1990.

Cf. A285201, A286282.

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.

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A285203 Local high points in A285200.

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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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A285202 A285200 displayed as an irregular triangle read by rows.

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A360830 Numbers that when concatenated with the natural numbers from 1 to N are divisible by the corresponding order number.

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