cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

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

Original entry on oeis.org

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

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Author

Hieronymus Fischer, Dec 31 2008

Keywords

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
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

Examples

			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.
		

Programs

  • 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 *)

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).

Extensions

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

A153701 Minimal exponents m such that the fractional part of e^m obtains a minimum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 9, 29, 45, 75, 135, 219, 732, 1351, 3315, 4795, 4920, 5469, 28414, 37373
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Comments

Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of e^m is less than the fractional part of e^k for all k, 1<=k
The next such number must be greater than 100000.
a(18) > 300,000. Robert Price, Mar 23 2019

Examples

			a(4)=9, since fract(e^9)=0.08392..., but fract(e^k)>=0.08553... for 1<=k<=8; thus fract(e^9)<fract(e^k) for 1<=k<9.
		

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
    p = 1; Select[Range[1, 300000],
    If[FractionalPart[E^#] < p, p = FractionalPart[E^#]; True] &] (* Robert Price, Mar 23 2019 *)

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract(e^m) < fract(e^a(k-1))}, where fract(x) = x-floor(x).
Showing 1-2 of 2 results.