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.

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A153668 Greatest number m such that the fractional part of (3/2)^A153664(n) >= 1-(1/m).

Original entry on oeis.org

2, 14, 222, 1772, 2747, 12347, 135794, 90529, 222246, 570361, 2134829, 6901329, 4600886, 3067257, 5380892, 75503109, 814558605, 543039070, 362026046, 241350697, 160900465, 107266976, 101721580, 190708740, 127139160
Offset: 1

Views

Author

Hieronymus Fischer, Dec 31 2008

Keywords

Examples

			a(2)=14, since 1-(1/15)=0.933...>fract((3/2)^A153664(2))=fract((3/2)^14)=0.929...>=1-(1/14).

Crossrefs

Cf. A002379, A081464, A153662, A153663, A153664, A153665, A153666, A153667.

Programs

  • Mathematica
    A153664 = {1, 14, 163, 1256, 2677, 8093, 49304, 49305, 158643, 164000, 835999, 2242294, 2242295, 2242296, 3965133, 25380333, 92600006, 92600007, 92600008, 92600009, 92600010, 92600011, 9267816, 125040717, 125040718};
Table[fp = FractionalPart[(3/2)^A153664[[n]]]; m = Floor[1/(1 - fp)];
While[fp >= 1 - (1/m), m++]; m - 1, {n, 1, Length[A153664]}] (* Robert Price, Mar 26 2019 *)

Formula

a(n) = floor(1/(1-fract((3/2)^A153664(n)))), where fract(x) = x-floor(x).

Extensions

a(11)-a(25) from Robert Price, May 10 2012

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

Cf. A153663, A153671, A153679, A153687, A153695, A154130, A153715, A153723.
Cf. A000149.

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.

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A153707, A153715, A154130, A153719.
Cf. A001672.

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

A153715 Greatest number m such that the fractional part of Pi^A153711(m) >= 1-(1/m).

Original entry on oeis.org

1, 7, 32, 53, 189, 2665, 10810, 26577, 128778, 483367
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(3) = 32, since 1-(1/33) = 0.9696... > fract(Pi^A153711(3)) = fract(Pi^15) = 0.96938... >= 0.96875 = 1-(1/32).

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A091560, A153711, A154130, A153723.
Cf. A001672.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
A153711 = {1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710};
Floor[1/(1-FractionalPart[Pi^A153711])] (* Robert Price, Apr 18 2019 *)

Formula

a(n) = floor(1/(1-fract(Pi^A153711(n)))), where fract(x) = x-floor(x).

Extensions

a(9)-a(10) from Robert Price, Apr 18 2019

A153723 Greatest number m such that the fractional part of (Pi-2)^A153719(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 3, 16, 24, 45, 158, 410, 946, 1182, 8786, 16159, 20188, 61392, 78800, 78959, 217556
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(5) = 16, since 1-(1/17) = 0.941176... > fract((Pi-2)^A153719(5)) = fract((Pi-2)^5) = 0.9389... >= 0.9375 = 1-(1/16).

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A091560, A153711, A153719, A154130.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
A153719 = {1, 2, 3, 4, 5, 39, 56, 85, 557, 911, 2919, 2921, 4491,
   11543, 15724, 98040, 110932, 126659};
Floor[1/(1 - FractionalPart[(Pi - 2)^A153719])] (* Robert Price, Apr 18 2019 *)

Formula

a(n) = floor(1/(1-fract((Pi-2)^A153719(n)))), where fract(x) = x-floor(x).

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

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-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.

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A153707, A153715, A153723, A154130.

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

Views

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).

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A154130, A153707, A153715, A153723.

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

Views

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).

Crossrefs

Cf. A153663, A154130, A153667, A153683, A153691, A153699, A153707, A153715, A153723.

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

Views

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).

Crossrefs

Cf. A153663, A153671, A153679, A154130, A153691, A153699, A153707, A153715, A153723.

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

Views

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).

Crossrefs

Cf. A153663, A153671, A153679, A153683, A154130, A153699, A153707, A153715, A153723.

Formula

a(n):=floor(1/(1-fract((11/10)^A153687(n)))), where fract(x) = x-floor(x).
Previous Showing 11-20 of 21 results. Next

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