Location via proxy:   [ UP ]  
[Report a bug]   [Manage cookies]                
A153717
Minimal exponents m such that the fractional part of (Pi-2)^m obtains a minimum (when starting with m=1).
8
1, 20, 23, 24, 523, 2811, 3465, 3776, 4567, 6145, 8507, 9353, 19790, 41136, 62097, 72506, 107346
OFFSET
1,2
COMMENTS
Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (Pi-2)^m is less than the fractional part of (Pi-2)^k for all k, 1<=k<m.
The next such number must be greater than 200000.
a(18) > 300000. - Robert Price, Mar 26 2019
FORMULA
Recursion: a(1):=1, a(k):=min{ m>1 | fract((Pi-2)^m) < fract((Pi-2)^a(k-1))}, where fract(x) = x-floor(x).
EXAMPLE
a(3)=23, since fract((Pi-2)^23)=0.0260069.., but fract((Pi-2)^k)>=0.1326... for 1<=k<=22; thus fract((Pi-2)^23)<fract((Pi-2)^k) for 1<=k<23.
MATHEMATICA
$MaxExtraPrecision = 100000;
p = 1; Select[Range[1, 10000],
If[FractionalPart[(Pi - 2)^#] < p, p = FractionalPart[(Pi - 2)^#];
True] &] (* Robert Price, Mar 26 2019 *)
KEYWORD
nonn,more
AUTHOR
Hieronymus Fischer, Jan 06 2009
STATUS
approved