A153687 Minimal exponents m such that the fractional part of (11/10)^m obtains a maximum (when starting with m=1).

1, 2, 3, 4, 5, 6, 7, 23, 56, 77, 103, 320, 1477, 1821, 2992, 15290, 180168, 410498, 548816, 672732, 2601223

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 (11/10)^m is greater than the fractional part of (11/10)^k for all k, 1<=k

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

a(22) > 10^7. Robert Price, Mar 19 2019

Examples

			a(8)=23, since fract((11/10)^23)= 0.9543..., but fract((11/10)^k)<0.95 for 1<=k<=22;
thus fract((11/10)^23)>fract((11/10)^k) for 1<=k<23 and 23 is the minimal exponent > 7 with this property.
		

Crossrefs

Cf. A153663, A153671, A153683, A153679, A154130, A153695, A091560, A153711, A153719.

Programs

Mathematica

p = 0; Select[Range[1, 50000],
If[FractionalPart[(11/10)^#] > p, p = FractionalPart[(11/10)^#];
True] &] (* Robert Price, Mar 19 2019 *)

Python

A153687_list, m, n, k, q = [], 1, 11, 10, 0
while m < 10**4:
    r = n % k
    if r > q:
        q = r
        A153687_list.append(m)
    m += 1
    n *= 11
    k *= 10
    q *= 10 # Chai Wah Wu, May 16 2020

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((11/10)^m) > fract((11/10)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(18)-a(21) from Robert Price, Mar 19 2019

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