A153671 Minimal exponents m such that the fractional part of (101/100)^m obtains a maximum (when starting with m=1).

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 110, 180, 783, 859, 1803, 7591, 10763, 19105, 50172, 355146, 1101696, 1452050, 3047334, 3933030

Offset

1,2

Comments

Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (101/100)^m is greater than the fractional part of (101/100)^k for all k, 1<=k<m.

The next such number must be greater than 10^6.

a(84) > 10^7. Robert Price, Mar 21 2019

Links

Table of n, a(n) for n=1..83.

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((101/100)^m) > fract((101/100)^a(k-1))}, where fract(x) = x-floor(x).

Example

a(5)=5, since fract((101/100)^5)=0.05101005, but fract((101/100)^k)=0.01, 0.0201, 0.030301, 0.04060401 for 1<=k<=4; thus fract((101/100)^5)>fract((101/100)^k) for 1<=k<5.

Mathematica

p = 0; Select[Range[1, 20000],
If[FractionalPart[(101/100)^#] > p, p = FractionalPart[(101/100)^#];
True] &] (* Robert Price, Mar 21 2019 *)

Prog

(Python)
A153671_list, m, n, k, q = [], 1, 101, 100, 0
while m < 10**4:
    r = n % k
    if r > q:
        q = r
        A153671_list.append(m)
    m += 1
    n *= 101
    k *= 100
    q *= 100 # Chai Wah Wu, May 16 2020

Crossrefs

Cf. A153663, A154130, A153675, A153679, A153687, A153695, A091560, A153711, A153719.

Sequence in context: A090108 A323048 A090109 * A090107 A130696 A146297

Adjacent sequences: A153668 A153669 A153670 * A153672 A153673 A153674

Keyword

nonn

Author

Hieronymus Fischer, Jan 06 2009

Extensions

a(72)-a(83) from Robert Price, Mar 21 2019

Status

approved