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A153711 Minimal exponents m such that the fractional part of Pi^m obtains a maximum (when starting with m=1). 8
1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of Pi^m is greater than the fractional part of Pi^k for all k, 1<=k<m.
The next such number must be greater than 100000.
a(11) > 300000. - Robert Price, Mar 25 2019
LINKS
FORMULA
Recursion: a(1):=1, a(k):=min{ m>1 | fract(Pi^m) > fract(Pi^a(k-1))}, where fract(x) = x-floor(x).
EXAMPLE
a(3)=15, since fract(Pi^15)= 0.9693879984..., but fract(Pi^k)<=0.8696... for 1<=k<=14; thus fract(Pi^15)>fract(Pi^k) for 1<=k<15 and 15 is the minimal exponent > 2 with this property.
MATHEMATICA
$MaxExtraPrecision = 100000;
p = 0; Select[Range[1, 10000],
If[FractionalPart[Pi^#] > p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 25 2019 *)
CROSSREFS
Cf. A001672.
Sequence in context: A169597 A280288 A153712 * A116049 A184236 A023651
KEYWORD
nonn,more
AUTHOR
Hieronymus Fischer, Jan 06 2009
EXTENSIONS
a(9)-a(10) from Robert Price, Mar 25 2019
STATUS
approved

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Last modified March 28 09:04 EDT 2024. Contains 371240 sequences. (Running on oeis4.)