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 1-6 of 6 results.

A137994 a(n) is the smallest integer > a(n-1) such that {Pi^a(n)} < {Pi^a(n-1)}, where {x} = x - floor(x), a(1)=1.

Original entry on oeis.org

1, 3, 81, 264, 281, 472, 1147, 2081, 3207, 3592, 10479, 12128, 65875, 114791, 118885
Offset: 1

Views

Author

Leroy Quet and M. F. Hasler, Mar 14 2008

Keywords

Comments

The sequence was suggested by Leroy Quet on Pi day 2008, cf. A138324.
The next such number must be greater than 100000. - Hieronymus Fischer, Jan 06 2009
a(16) > 300,000. - Robert Price, Mar 25 2019

Examples

			a(3)=81, since {Pi^81}=0.0037011283.., but {Pi^k}>=0.0062766802... for 1<=k<=80; thus {Pi^81}<{Pi^k} for 1<=k<81. - _Hieronymus Fischer_, Jan 06 2009
		

Crossrefs

Programs

  • Mathematica
    $MaxExtraPrecision = 10000;
    p = .999;
    Select[Range[1, 5000],
    If[FractionalPart[Pi^#] < p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 12 2019 *)
  • PARI
    default(realprecision,10^4); print1(a=1); for(i=1,100, f=frac(Pi^a); until( frac(Pi^a++)
    				

Extensions

a(11)-a(13) from Hieronymus Fischer, Jan 06 2009
Edited by R. J. Mathar, May 21 2010
a(14)-a(15) from Robert Price, Mar 12 2019

A153705 Greatest number m such that the fractional part of e^A153701(n) <= 1/m.

Original entry on oeis.org

1, 2, 11, 11, 23, 28, 69, 85, 115, 964, 1153, 1292, 1296, 1877, 34015, 156075, 952945
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(3)=11 since 1/12 < fract(e^A153701(3)) = fract(e^3) = 0.0855... <= 1/11.
		

Crossrefs

Programs

  • Mathematica
    A153701 = {1, 2, 3, 9, 29, 45, 75, 135, 219, 732, 1351, 3315, 4795,
       4920, 5469, 28414, 37373};
    Table[fp = FractionalPart[E^A153701[[n]]]; m = Floor[1/fp];
    While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153701]}] (* Robert Price, Mar 25 2019 *)

Formula

a(n) = floor(1/fract(e^A153701(n))), where fract(x) = x-floor(x).

A153697 Greatest number m such that the fractional part of (10/9)^A153693(n) <= 1/m.

Original entry on oeis.org

9, 11, 30, 82, 6131, 26735, 29430, 76172, 151439, 227416, 771341, 2712159, 4490404
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=11 since 1/12 < fract((10/9)^A153693(2)) = fract((10/9)^7) = 0.09075... <= 1/11.
		

Crossrefs

Programs

  • Mathematica
    A153693 = {1, 7, 50, 62, 324, 3566, 66877, 108201, 123956, 132891,
       182098, 566593, 3501843};
    Table[fp = FractionalPart[(10/9)^A153693[[n]]]; m = Floor[1/fp];
    While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153693]}] (* Robert Price, Mar 25 2019 *)

Formula

a(n) = floor(1/fract((10/9)^A153693(n))), where fract(x) = x-floor(x).

Extensions

a(12)-a(13) from Robert Price, Mar 25 2019

A153689 Greatest number m such that the fractional part of (11/10)^A153685(n) <= 1/m.

Original entry on oeis.org

10, 18, 253, 618, 6009, 6767, 21386, 697723, 4186162, 31102351
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=18 since 1/19 < fract((11/10)^A153685(2)) = fract((11/10)^17) = 0.0544... <= 1/18.
		

Crossrefs

Programs

  • Mathematica
    A153685 = {1, 17, 37, 237, 599, 615, 6638, 13885, 1063942, 9479731};
    Table[fp = FractionalPart[(11/10)^A153685[[n]]]; m = Floor[1/fp];
    While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153685]}] (* Robert Price, Mar 25 2019 *)

Formula

a(n) = floor(1/fract((11/10)^A153685(n))), where fract(x) = x-floor(x).

Extensions

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

A153673 Greatest number m such that the fractional part of (101/100)^A153669(n) <= 1/m.

Original entry on oeis.org

100, 147, 703, 932, 1172, 3389, 7089, 8767, 11155, 17457, 20810, 25355, 1129226, 1741049, 1960780, 2179637, 2859688, 11014240, 75249086, 132665447, 499298451
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=147 since 1/148<fract((101/100)^A153669(2))=fract((101/100)^70)=0.00676...<=1/147.
		

Crossrefs

Programs

  • Mathematica
    A153669 = {1, 70, 209, 378, 1653, 2697, 4806, 13744, 66919, 67873,
       75666, 81125, 173389, 529938, 1572706, 4751419, 7159431, 7840546,
       15896994, 71074288, 119325567};
    Table[fp = FractionalPart[(101/100)^A153669[[n]]]; m = Floor[1/fp];
    While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153669]}] (* Robert Price, Mar 25 2019 *)

Formula

a(n) = floor(1/fract((101/100)^A153669(n))), where fract(x) = x-floor(x).

Extensions

a(15)-a(21) from Robert Price, Mar 25 2019

A153681 Greatest number m such that the fractional part of (1024/1000)^A153677(n) <= 1/m.

Original entry on oeis.org

41, 60, 76, 116, 233, 463, 718, 1350, 12472, 13733, 17428, 27955, 32276, 41155, 62437, 69643, 111085, 811799, 2656810, 11462221, 56414953
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=60 since 1/61 < fract((1024/1000)^A153677(2)) = fract((1024/1000)^68) = 0.0164... <= 1/60.
		

Crossrefs

Programs

  • Mathematica
    A153677 = {1, 68, 142, 341, 395, 490, 585, 1164, 1707, 26366, 41358,
       46074, 120805, 147332, 184259, 205661, 385710, 522271, 3418770,
       3675376, 9424094};
    Table[fp = FractionalPart[(1024/1000)^A153677[[n]]]; m = Floor[1/fp];
    While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153677]}] (* Robert Price, Mar 25 2019 *)

Formula

a(n) = floor(1/fract((1024/1000)^A153677(n))), where fract(x) = x-floor(x).

Extensions

a(18)-a(21) from Robert Price, Mar 25 2019
Showing 1-6 of 6 results.