A333709 a(n) is the number of iterations for x = x + 1/x needed until x + 1/x > n, starting with x = 1.

0, 1, 3, 6, 11, 16, 22, 30, 38, 48, 58, 70, 82, 95, 110, 125, 142, 159, 178, 197, 218, 239, 262, 285, 310, 335, 362, 389, 418, 447, 478, 509, 542, 575, 610, 645, 682, 719, 757, 797, 837, 879, 921, 965, 1009, 1055, 1101, 1149, 1197, 1247, 1297, 1349, 1401, 1455, 1509, 1565, 1621, 1679, 1737, 1797

Offset

1,3

Links

Peter Luschny, Table of n, a(n) for n = 1..10000

Formula

a(n) < (1/2)*n^2. - Peter Luschny, Apr 12 2020

Example

Illustrating the iterations:
n  iter   x   1/x   x+1/x > n  a(n)=iter
1   0   1.00  1.00  2.00  > 1   0
2   1   2.00  0.50  2.50  > 2   1
    2   2.50  0.40  2.90  > 3
3   3   2.90  0.34  3.24  > 3   3
    4   3.24  0.30  3.55  > 4
    5   3.55  0.28  3.83  > 4
4   6   3.83  0.26  4.09  > 4   6

Prog

(Python)
from math import ceil
x = 1
l = []
for i in range(0, 2**11):
    h = x
    x += 1/x
    if ceil(h) != ceil(x):
        l.append(i)
print(l)
(PARI)
c=x=1; for(n=0, 1e4, h=x; o=c; c=ceil(x+=1./x); if(c>o, print1(n", ")))
\\ Charles R Greathouse IV, Apr 06 2020
(Julia)
using Nemo
function A333709List(bound, prec)
    R = RealField(prec)
    L = []
    x = R(1)
    z = ZZ(1)
    for iter in 0:bound
        x += 1/x
        if z < x
            push!(L, iter)
            z += 1
        end
    end
    return L
end
# Choose higher precision for larger bounds!
A333709List(2000, 64) |> println  # Peter Luschny, Apr 11 2020

Crossrefs

Sequence in context: A111582 A366471 A024401 * A266252 A267260 A116940

Adjacent sequences: A333706 A333707 A333708 * A333710 A333711 A333712

Keyword

nonn

Author

Sebastian F. Orellana, Apr 02 2020

Status

approved