A153663 Minimal exponents m such that the fractional part of (3/2)^m reaches a maximum (when starting with m=1).

1, 5, 8, 10, 12, 14, 46, 58, 105, 157, 163, 455, 1060, 1256, 2677, 8093, 28277, 33327, 49304, 158643, 164000, 835999, 2242294, 25380333, 92600006

Offset

1,2

Comments

Recursive definition: a(1)=1, a(n) = least number m such that the fractional part of (3/2)^m is greater than the

fractional part of (3/2)^k for all k, 1<=k<m.

The fractional part of k=835999 is .999999 5 which is greater than (k-1)/k. The fractional part of k=2242294 is .999999 8 which is greater than (k-1)/k. The fractional part of k=25380333 is .999999 98 which is greater than (k-1)/k. The fractional part of k=92600006 is .999999 998 which is greater than (k-1)/k. So, all additional numbers in this sequence must be in A153664 and >3*10^8. - Robert Price, May 09 2012

Links

Table of n, a(n) for n=1..25.

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((3/2)^m) > fract((3/2)^a(k-1))}, where fract(x) = x-floor(x).

Example

a(2)=5, since fract((3/2)^5)=0.59375, but fract((3/2)^k)=0.5, 0.25, 0.375, 0.0625 for 1<=k<=4; thus
fract((3/2)^5)>fract((3/2)^k) for 1<=k<5.

Mathematica

a[1] = 1; a[n_] := a[n] = For[m = a[n-1]+1, True, m++, f = FractionalPart[(3/2)^m]; If[AllTrue[Range[m-1], f > FractionalPart[(3/2)^#]&], Print[n, " ", m]; Return[m]]];
Array[a, 21] (* Jean-François Alcover, Feb 25 2019 *)

Crossrefs

Cf. A002379, A153661, A153662, A153664, A153665, A153666, A153667, A153668.

Sequence in context: A314377 A314378 A314379 * A065528 A360398 A050936

Adjacent sequences: A153660 A153661 A153662 * A153664 A153665 A153666

Keyword

nonn,more,changed

Author

Hieronymus Fischer, Dec 31 2008

Extensions

a(22)-a(25) from Robert Price, May 09 2012

Status

approved