A153673 -id:A153673 - cp's OEIS frontend

A153669 Minimal exponents m such that the fractional part of (101/100)^m obtains a minimum (when starting with m=1).

1, 70, 209, 378, 1653, 2697, 4806, 13744, 66919, 67873, 75666, 81125, 173389, 529938, 1572706, 4751419, 7159431, 7840546, 15896994, 71074288, 119325567

Offset: 1

Hieronymus Fischer, Jan 06 2009

Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (101/100)^m is less than the fractional part of (101/100)^k for all k, 1<=k

The next term is greater than 2*10^8.

			a(2)=70, since fract((101/100)^70)=0.00676..., but fract((101/100)^k)>=0.01 for 1<=k<=69; thus fract((101/100)^70)<fract((101/100)^k) for 1<=k<70.

Cf. A081464, A154130, A153673, A153677, A153685, A153693, A153701, A137994, A153717.

p = 1; Select[Range[1, 5000],
If[FractionalPart[(101/100)^#] < p, p = FractionalPart[(101/100)^#];
True] &] (* Robert Price, Mar 21 2019 *)

Recursion: a(1):=1, a(k):=min{ m>1 | fract((101/100)^m) < fract((101/100)^a(k-1))}, where fract(x) = x-floor(x).

a(15)-a(21) from Robert Gerbicz, Nov 22 2010

A154134 Greatest number m such that the fractional part of (4/3)^A154130(n) <= 1/m.

Original entry on oeis.org

3, 6, 10, 30, 124, 238, 405, 6430, 22869, 32657, 224544, 2396968, 15229280, 28274047, 53458049, 134537968

Offset: 1

Hieronymus Fischer, Jan 11 2009

			a(3)=10 since 1/11<fract((4/3)^A154130(3))=fract((4/3)^13)=0.09238...<=1/10.

Cf. A153665, A153673, A153705, A153712, A154143, A154150.

Cf. A002379, A064628.

a(n):=floor(1/fract((4/3)^A154130(n))), where fract(x) = x-floor(x).

a(12)-a(16) from Robert Price, May 10 2012

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cp's OEIS Frontend

A153669 Minimal exponents m such that the fractional part of (101/100)^m obtains a minimum (when starting with m=1).

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