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.

Previous Showing 51-58 of 58 results.

A153692 Greatest number m such that the fractional part of (11/10)^A153688(n) >= 1-(1/m).

Original entry on oeis.org

1, 19, 151, 200, 709, 5727, 15908, 162819, 120479, 109526, 302991
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=19, since 1-(1/20)=0.95>fract((11/10)^A153688(2))=fract((11/10)^7)=0.9487...>=1-(1/19).

Crossrefs

Cf. A153664, A153672, A153680, A153684, A154130, A153700, A153708, A153716, A153724.

Formula

a(n):=floor(1/(1-fract((11/10)^A153688(n)))), where fract(x) = x-floor(x).

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

Cf. A154130, A153669, A153665, A153681, A153689, A153697, A153705, A153713, A153721.

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

A153674 Greatest number m such that the fractional part of (101/100)^A153670(n) <= 1/m.

Original entry on oeis.org

100, 49, 33, 24, 19, 16, 13, 12, 10, 147, 703, 676, 932, 3389, 7089, 1129226, 1741049, 1356464, 1960780, 11014240, 75249086, 28657625, 132665447, 499298451
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(5) = 19 since 1/20 < fract((101/100)^A153670(5)) = fract((101/100)^5) = 0.0510... <= 1/19.

Crossrefs

Cf. A153662, A154130, A153666, A153682, A153690, A153698, A153706, A153714, A153722.

Formula

a(n) = floor(1/fract((101/100)^A153670(n))), where fract(x) = x - floor(x).

Extensions

a(18)-a(24) from Jinyuan Wang, Mar 03 2020

A153676 Greatest number m such that the fractional part of (101/100)^A153672(n) >= 1-(1/m).

Original entry on oeis.org

1, 76, 238, 913, 1334, 4645, 17396, 351085, 69587, 552184, 329808, 381654, 35874097, 5011174, 6220178, 33773592, 13149134, 105749940
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=76, since 1-(1/77)=0.9870...>fract((101/100)^A153672(2))=fract((101/100)^69)=0.98689...>=1-(1/76).

Crossrefs

Cf. A153664, A154130, A153668, A153684, A153692, A153700, A153708, A153716, A153724.

Formula

a(n):=floor(1/(1-fract((101/100)^A153672(n)))), where fract(x) = x-floor(x).

Extensions

a(13)-a(18) from Robert Price, May 10 2012

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

Cf. A081464, A153669, A153677, A154130, A153689, A153697, A153705, A153713, A153721.

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

A153690 Greatest number m such that the fractional part of (11/10)^A153686(n) <= 1/m.

Original entry on oeis.org

10, 4, 3, 18, 253, 58, 618, 484, 6009, 6767, 21386, 697723, 634293, 189959, 4186162, 31102351
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(4) = 18 since 1/19 < fract((11/10)^A153686(4)) = fract((11/10)^17) = 0.05447... <= 1/18.

Crossrefs

Cf. A153662, A153670, A153678, A153686, A154130, A153698, A153706, A153714, A153722.

Formula

a(n) = floor(1/fract((11/10)^A153686(n))), where fract(x) = x - floor(x).

Extensions

a(14)-a(16) from Jinyuan Wang, Mar 03 2020

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).

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
Previous Showing 51-58 of 58 results.

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