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A153677 Minimal exponents m such that the fractional part of (1024/1000)^m obtains a minimum (when starting with m=1). 12
1, 68, 142, 341, 395, 490, 585, 1164, 1707, 26366, 41358, 46074, 120805, 147332, 184259, 205661, 385710, 522271, 3418770, 3675376, 9424094 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Recursive definition: a(1)=1, a(n) is the least positive integer m such that the fractional part of (1024/1000)^m is less than the fractional part of (1024/1000)^k for all k, 1 <= k < m.

a(21) >= 4.5*10^6. - David A. Corneth, Mar 15 2019

a(22) > 10^7. Robert Price, Mar 16 2019

LINKS

Table of n, a(n) for n=1..21.

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).

EXAMPLE

a(2)=68, since fract((1024/1000)^68) = 0.016456..., but fract((1024/1000)^k) >= 0.024 for 1 <= k <= 67; thus fract((1024/1000)^68) < fract((1024/1000)^k) for 1 <= k < 68.

MATHEMATICA

$MaxExtraPrecision = 10000;

p = .999;

Select[Range[1, 50000],

If[FractionalPart[(1024/1000)^#] < p,

p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 15 2019 *)

PROG

(PARI) upto(n) = my(res = List(), r = 1, p = 1); for(i=1, n, c = frac(p *= 1.024); if(c<r, r=c; print1(i", "); listput(res, i))); res \\ David A. Corneth, Mar 15 2019

CROSSREFS

Cf. A081464, A153669, A153681, A154130, A153685, A153693, A153701, A137994, A153717.

Sequence in context: A111379 A044191 A044572 * A234876 A044319 A044700

Adjacent sequences:  A153674 A153675 A153676 * A153678 A153679 A153680

KEYWORD

nonn,more

AUTHOR

Hieronymus Fischer, Jan 06 2009

EXTENSIONS

a(18) from Robert Price, Mar 15 2019

a(19)-a(20) from David A. Corneth, Mar 15 2019

a(21) from Robert Price, Mar 16 2019

STATUS

approved

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Last modified May 18 17:28 EDT 2021. Contains 343996 sequences. (Running on oeis4.)