A153663 - OEIS

%I #14 May 12 2024 18:11:11

%S 1,5,8,10,12,14,46,58,105,157,163,455,1060,1256,2677,8093,28277,33327,

%T 49304,158643,164000,835999,2242294,25380333,92600006

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

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

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

%C 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

%F 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).

%e 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

%e fract((3/2)^5)>fract((3/2)^k) for 1<=k<5.

%t 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]]];

%t Array[a, 21] (* _Jean-François Alcover_, Feb 25 2019 *)

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

%K nonn,more

%O 1,2

%A _Hieronymus Fischer_, Dec 31 2008

%E a(22)-a(25) from _Robert Price_, May 09 2012