A153679 Minimal exponents m such that the fractional part of (1024/1000)^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, 82, 134, 1306, 2036, 6393, 34477, 145984, 2746739, 2792428, 8460321

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 (1024/1000)^m is greater than the fractional part of (1024/1000)^k for all k, 1<=k

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

a(40) > 10^7. Robert Price, Mar 16 2019

Examples

			a(30)=82, since fract((1024/1000)^82)= 0.99191990..., but fract((1024/1000)^k)<0.9893 for 1<=k<=81; thus fract((1024/1000)^82)>fract((1024/1000)^k) for 1<=k<82 and 82 is the minimal exponent >29 with this property.
		

Crossrefs

Cf. A153663, A153671, A153675, A154130, A153687, A153695, A091560, A153711, A153719.

Programs

Mathematica

$MaxExtraPrecision = 10000;
p = 0;
Select[Range[1, 50000],
If[FractionalPart[(1024/1000)^#] > p,
p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 16 2019 *)

Python

A153679_list, m, n, k, q = [], 1, 1024, 1000, 0
while m < 10**4:
    r = n % k
    if r > q:
        q = r
        A153679_list.append(m)
    m += 1
    n *= 1024
    k *= 1000
    q *= 1000 # Chai Wah Wu, May 16 2020

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((1024/1000)^m) > fract((1024/1000)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(37)-a(39) from Robert Price, Mar 16 2019

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.