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.

A153721 Greatest number m such that the fractional part of (Pi-2)^A153717(n) <= 1/m.

Original entry on oeis.org

7, 7, 38, 318, 393, 396, 484, 2076, 2619, 4099, 5264, 8556, 18070, 20732, 27209, 73351, 356362
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(3)=38 since 1/39<fract((Pi-2)^A153717(3))=fract((Pi-2)^23)=0.02600...<=1/38.

Crossrefs

Cf. A081464, A153669, A153677, A153685, A153693, A153701, A137994, A153717, A154130.

Programs

  • Mathematica
    A153717 = {1, 20, 23, 24, 523, 2811, 3465, 3776, 4567, 6145, 8507, 9353, 19790, 41136, 62097, 72506, 107346};
Table[fp = FractionalPart[(Pi - 2)^A153717[[n]]]; m = Floor[1/fp];
While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A153717]}] (* Robert Price, Mar 26 2019 *)

Formula

a(n) = floor(1/fract((Pi-2)^A153717(n))), where fract(x) = x-floor(x).

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

Original entry on oeis.org

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

Examples

			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.

Crossrefs

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

Programs

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

Formula

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).

Extensions

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

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

Cf. A001203, A138324, A001672.
Cf. A081464, A153669, A153677, A153685, A153693, A153705, A153713, A154130, A153717. - Hieronymus Fischer, Jan 06 2009

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

A153677 Minimal exponents m such that the fractional part of (1024/1000)^m obtains a minimum (when starting with m=1).

Original entry on oeis.org

1, 68, 142, 341, 395, 490, 585, 1164, 1707, 26366, 41358, 46074, 120805, 147332, 184259, 205661, 385710, 522271, 3418770, 3675376, 9424094
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Comments

Recursive definition: a(1)=1, a(n) is the least positive integer m such that the fractional part of (1024/1000)^m is less than the fractional part of (1024/1000)^k for all k, 1 <= k < m.
a(21) >= 4.5*10^6. - David A. Corneth, Mar 15 2019
a(22) > 10^7. Robert Price, Mar 16 2019

Examples

			a(2)=68, since fract((1024/1000)^68) = 0.016456..., but fract((1024/1000)^k) >= 0.024 for 1 <= k <= 67; thus fract((1024/1000)^68) < fract((1024/1000)^k) for 1 <= k < 68.

Crossrefs

Cf. A081464, A153669, A153681, A154130, A153685, A153693, A153701, A137994, A153717.

Programs

  • Mathematica
    $MaxExtraPrecision = 10000;
p = .999;
Select[Range[1, 50000],
If[FractionalPart[(1024/1000)^#] < p,
p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 15 2019 *)
  • PARI
    upto(n) = my(res = List(), r = 1, p = 1); for(i=1, n, c = frac(p *= 1.024); if(cDavid A. Corneth, Mar 15 2019

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((1024/1000)^m) < fract((1024/1000)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(18) from Robert Price, Mar 15 2019
a(19)-a(20) from David A. Corneth, Mar 15 2019
a(21) from Robert Price, Mar 16 2019

A153685 Minimal exponents m such that the fractional part of (11/10)^m obtains a minimum (when starting with m=1).

Original entry on oeis.org

1, 17, 37, 237, 599, 615, 6638, 13885, 1063942, 9479731
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 (11/10)^m is less than the fractional part of (11/10)^k for all k, 1<=k
The next such number must be greater than 2*10^5.
a(11) > 10^7. Robert Price, Mar 19 2019

Examples

			a(2)=17, since fract((11/10)^17)=0.05447.., but fract((11/10)^k)>=0.1 for 1<=k<=16; thus fract((11/10)^17)<fract((11/10)^k) for 1<=k<17.

Crossrefs

Cf. A081464, A153669, A153689, A153677, A154130, A153693, A153701, A137994, A153717.

Programs

  • Mathematica
    p = 1; Select[Range[1, 50000],
 If[FractionalPart[(11/10)^#] < p, p = FractionalPart[(11/10)^#];
True] &] (* Robert Price, Mar 19 2019 *)

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((11/10)^m) < fract((11/10)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(9)-a(10) from Robert Price, Mar 19 2019

A153693 Minimal exponents m such that the fractional part of (10/9)^m obtains a minimum (when starting with m=1).

Original entry on oeis.org

1, 7, 50, 62, 324, 3566, 66877, 108201, 123956, 132891, 182098, 566593, 3501843
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 (10/9)^m is less than the fractional part of (10/9)^k for all k, 1 <= k < m.
The next such number must be greater than 2*10^5.
a(14) > 10^7. - Robert Price, Mar 24 2019

Examples

			a(2)=7, since fract((10/9)^7) = 0.09075.., but fract((10/9)^k) >= 0.11... for 1 <= k <= 6; thus fract((10/9)^7) < fract((10/9)^k) for 1 <= k < 7.

Crossrefs

Cf. A081464, A153669, A153677, A153685, A153697, A154130, A153701, A137994, A153717.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
p = 1; Select[Range[1, 10000],
If[FractionalPart[(10/9)^#] < p, p = FractionalPart[(10/9)^#];
True] &] (* Robert Price, Mar 24 2019 *)

Formula

Recursion: a(1):=1, a(k):=min{ m>1 | fract((10/9)^m) < fract((10/9)^a(k-1))}, where fract(x) = x-floor(x).

Extensions

a(12)-a(13) from Robert Price, Mar 24 2019

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

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 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.

Crossrefs

Cf. A153661, A153669, A153677, A153685, A153693, A153705, A154130, A137994, A153717, A000149.

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).

A153725 Least number m such that floor((3^n-m)/(2^n-m)) > floor(3^n/2^n).

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 7, 4, 8, 7, 12, 9, 17, 4, 8, 16, 99, 20, 39, 235, 49, 97, 194, 885, 1106, 439, 2059, 968, 4034, 5268, 3070, 1163, 2325, 4649, 9297, 18593, 16210, 4452, 8903, 67524, 68757, 49124, 98248, 39360, 288234, 17763, 35526, 567677, 1135354
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Comments

Provided A002379(n) = floor((3^n-1)/(2^n-1)) holds (which is proved only for 1 < n <= 305000), then a(n) > 1.

Examples

			a(5)=2, since floor((3^5-1)/(2^5-1)) = floor(242/31) = 7 = floor(243/32) = floor(3^5/2^5), but floor((3^5-2)/(2^5-2)) = floor(241/30) = 8 > 7.

Links

Crossrefs

Cf. A002379, A081464, A153701, A137994, A153717, A154130.

Programs

  • Mathematica
    Table[n3 = 3^n; n2 = 2^n; m = 1;
While[Floor[(n3 - m)/(n2 - m)] <= Floor[n3/n2], m++]; m, {n,1,50}] (* Robert Price, Mar 27 2019 *)
  • PARI
    a(n) = my(f = floor(3^n/2^n)); ceil(((f + 1)*(2^n) - 3^n)/f) \\ David A. Corneth, Mar 27 2019

Formula

a(n) = ceiling(((f + 1)*(2^n) - 3^n)/f) where f = floor(3^n/2^n). - David A. Corneth, Mar 27 2019
Showing 1-8 of 8 results.

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