A153695 Minimal exponents m such that the fractional part of (10/9)^m obtains a maximum (when starting with m=1).

1, 2, 3, 4, 5, 6, 13, 17, 413, 555, 2739, 3509, 3869, 5513, 12746, 31808, 76191, 126237, 430116, 477190, 1319307, 3596185

Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

• nonn
• more

Comments

Recursive definition: a(1)=1, a(n) = least number m > a(n-1) such that the fractional part of (10/9)^m is greater than the fractional part of (10/9)^k for all k, 1 <= k < m.

The next such number must be greater than 2*10^5.

a(23) > 10^7. - Robert Price, Mar 24 2019

Examples

			a(7)=13, since fract((10/9)^13) = 0.93..., but fract((10/9)^k) < 0.89 for 1 <= k <= 12; thus fract((10/9)^13) > fract((10/9)^k) for 1 <= k < 13 and 13 is the minimal exponent > 6 with this property.
		

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153699, A154130, A091560, A153711, A153719.

Programs

Mathematica

$MaxExtraPrecision = 100000;
p = 0; Select[Range[1, 20000],
If[FractionalPart[(10/9)^#] > p, p = FractionalPart[(10/9)^#];
True] &] (* Robert Price, Mar 24 2019 *)

Python

A153695_list, m, m10, m9, q = [], 1, 10, 9, 0
while m < 10**4:
    r = m10 % m9
    if r > q:
        q = r
        A153695_list.append(m)
    m += 1
    m10 *= 10
    m9 *= 9
    q *= 9 # Chai Wah Wu, May 16 2020

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((10/9)^m) > fract((10/9)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(19)-a(22) from Robert Price, Mar 24 2019