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

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

Author

N. J. A. Sloane, May 09 2017

Keywords

• nonn

Comments

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)!.

Links

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

Crossrefs

Cf. A000522, A002627, A286281, A285201.

Programs

Maple

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

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

Python

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

Formula

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

Extensions

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.