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 1-8 of 8 results.

A153707 Greatest number m such that the fractional part of e^A091560(m) >= 1-(1/m).

Original entry on oeis.org

3, 23, 27, 41, 59, 261, 348, 2720, 3198, 6064, 72944, 347065
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=23, since 1-(1/24) = 0.9583...> fract(e^A091560(2)) = fract(e^8) = 0.95798.. >= 0.95652... >= 1-(1/23).
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
    A091560 = {1,8,19,76,166,178,209,1907,20926,22925,32653,119136};
    Floor[1/(1-FractionalPart[E^A091560])] (* Robert Price, Apr 18 2019 *)

Formula

a(n):=floor(1/(1-fract(e^A091560(n)))), where fract(x) = x-floor(x).

Extensions

a(12) from Robert Price, Apr 18 2019

A153711 Minimal exponents m such that the fractional part of Pi^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710
Offset: 1

Views

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 Pi^m is greater than the fractional part of Pi^k for all k, 1<=k
The next such number must be greater than 100000.
a(11) > 300000. - Robert Price, Mar 25 2019

Examples

			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.
		

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
    p = 0; Select[Range[1, 10000],
    If[FractionalPart[Pi^#] > p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 25 2019 *)

Formula

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

Extensions

a(9)-a(10) from Robert Price, Mar 25 2019

A153719 Minimal exponents m such that the fractional part of (Pi-2)^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 4, 5, 39, 56, 85, 557, 911, 2919, 2921, 4491, 11543, 15724, 98040, 110932, 126659
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 (Pi-2)^m is greater than the fractional part of (Pi-2)^k for all k, 1<=k
The next such number must be greater than 200000.
a(19) > 300000. - Robert Price, Mar 26 2019

Examples

			a(6)=39, since fract((Pi-2)^39)= 0.9586616565..., but fract((Pi-2)^k)<=0.9389018... for 1<=k<=38; thus fract((Pi-2)^39)>fract((Pi-2)^k) for 1<=k<39 and 39 is the minimal exponent > 5 with this property.
		

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
    p = 0; Select[Range[1, 10000],
    If[FractionalPart[(Pi - 2)^#] > p, p = FractionalPart[(Pi - 2)^#];
    True] &] (* Robert Price, Mar 26 2019 *)

Formula

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

A153699 Greatest number m such that the fractional part of (10/9)^A153695(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 2, 3, 8, 15, 264, 334, 465, 683, 713, 758, 8741, 15912, 18920, 38560, 409895
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(7)=15, since 1-(1/16)=0.9375>fract((10/9)^A153695(7))=fract((10/9)^13)=0.9341...>=1-(1/15).
		

Formula

a(n):=floor(1/(1-fract((10/9)^A153695(n)))), where fract(x) = x-floor(x).

A153675 Greatest number m such that the fractional part of (101/100)^A153671(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 7, 9, 11, 13, 19, 30, 76, 81, 238, 913, 1334, 4645, 6812, 17396, 351085, 552184
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(5)=1, since 1-(1/2)=0.5>fract((101/100)^A153671(5))=fract((101/100)^5)=0.0510...>=1-(1/1).
		

Formula

a(n):=floor(1/(1-fract((101/100)^A153671(n)))), where fract(x) = x-floor(x).

A153683 Greatest number m such that the fractional part of (1024/1000)^A153679(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 4, 5, 6, 9, 17, 93, 123, 1061, 1360, 4137, 66910, 571666, 1192010
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(5)=1, since 1-(1/2)=0.5>fract((1024/1000)^A153679(5))=fract((1024/1000)^5)=0.0510...>=1-(1/1).
		

Formula

a(n):=floor(1/(1-fract((1024/1000)^A153679(n)))), where fract(x) = x-floor(x).

A153691 Greatest number m such that the fractional part of (11/10)^A153687(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 19, 21, 28, 151, 200, 709, 767, 5727, 15908, 162819, 302991
Offset: 1

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(6)=4, since 1-(1/5)=0.8>fract((11/10)^A153687(6))=fract((11/10)^6)=0.771...>=1-(1/4).
		

Formula

a(n):=floor(1/(1-fract((11/10)^A153687(n)))), where fract(x) = x-floor(x).

A154136 Greatest number m such that the fractional part of (4/3)^A154132(m) >= 1-(1/m).

Original entry on oeis.org

1, 4, 88, 1228, 2253, 4562, 8183, 167378
Offset: 1

Author

Hieronymus Fischer, Jan 11 2009

Keywords

Examples

			a(3)=88, since 1-(1/89)=0.988764...>fract((4/3)^A154132(3))=fract((4/3)^8)=0.988721...>0.988636...=1-(1/88).
		

Formula

a(n):=floor(1/(1-fract((4/3)^A154132(n)))), where fract(x) = x-floor(x).
Showing 1-8 of 8 results.