A153701 -id:A153701 - cp's OEIS frontend

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

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

Offset: 1

Hieronymus Fischer, Jan 06 2009

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

Cf. A081464, A153669, A153677, A153685, A153693, A153701, A154130, A153713, A153721.

Cf. A000149.

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 *)

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

A154130 Exponents m with decreasing fractional part of (4/3)^m.

Original entry on oeis.org

1, 4, 13, 17, 128, 485, 692, 1738, 12863, 77042, 109705, 289047, 720429, 4475944, 75629223, 182575231

Offset: 1

Hieronymus Fischer, Jan 11 2009

nonn

The next term is greater than 3*10^8.

			a(3)=13, since fract((4/3)^13)=0.0923.., but fract((4/3)^k)>=0.16... for 1<=k<=12; thus fract((4/3)^13)<fract((4/3)^k) for 1<=k<13.

Cf. A081464, A153669, A153701, A153708, A154138, A154146.

Cf. A002379, A064628.

Recursion: a(1):=1, a(k):=min{ m>1 | fract((4/3)^m) < fract((4/3)^a(k-1))}, where fract(x) = x-floor(x).

Extended by Charles R Greathouse IV, Nov 03 2009

a(15)-a(16) from Robert Gerbicz, Nov 21 2010

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

Hieronymus Fischer, Jan 06 2009

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.

			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.

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

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

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).

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

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

Hieronymus Fischer, Jan 06 2009

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

			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.

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

$MaxExtraPrecision = 10000;
p = .999;
Select[Range[1, 50000],
If[FractionalPart[(1024/1000)^#] < p,
p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 15 2019 *)

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

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).

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

Hieronymus Fischer, Jan 06 2009

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

			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.

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

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

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).

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

Hieronymus Fischer, Jan 06 2009

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

			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.

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

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

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).

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

A153717 Minimal exponents m such that the fractional part of (Pi-2)^m obtains a minimum (when starting with m=1).

Original entry on oeis.org

1, 20, 23, 24, 523, 2811, 3465, 3776, 4567, 6145, 8507, 9353, 19790, 41136, 62097, 72506, 107346

Offset: 1

Hieronymus Fischer, Jan 06 2009

Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (Pi-2)^m is less than the fractional part of (Pi-2)^k for all k, 1<=k

The next such number must be greater than 200000.

a(18) > 300000. - Robert Price, Mar 26 2019

			a(3)=23, since fract((Pi-2)^23)=0.0260069.., but fract((Pi-2)^k)>=0.1326... for 1<=k<=22; thus fract((Pi-2)^23)<fract((Pi-2)^k) for 1<=k<23.

Cf. A081464, A153669, A153677, A153685, A153693, A153701, A137994, A153721, A154130.

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

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).

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

Hieronymus Fischer, Jan 06 2009

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

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

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 *)

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

A153713 Greatest number m such that the fractional part of Pi^A137994(n) <= 1/m.

Original entry on oeis.org

7, 159, 270, 307, 744, 757, 796, 1079, 1226, 7804, 13876, 62099, 70718, 86902, 154755

Offset: 1

Hieronymus Fischer, Jan 06 2009

			a(2)=159 since 1/160<fract(Pi^A137994(2))=fract(Pi^3)=0.0062766...<=1/159.

Cf. A081464, A153669, A153677, A153685, A153693, A153701, A137994, A154130, A153721.

Cf. A001672.

A137994 = {1, 3, 81, 264, 281, 472, 1147, 2081, 3207, 3592, 10479, 12128, 65875, 114791, 118885};
Table[fp = FractionalPart[Pi^A137994[[n]]]; m = Floor[1/fp];
While[fp <= 1/m, m++]; m - 1, {n, 1, Length[A137994]}] (* Robert Price, Mar 26 2019 *)

a(n) = floor(1/fract(Pi^A137994(n))), where fract(x) = x-floor(x).

a(14)-a(15) from Robert Price, Mar 26 2019

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

Hieronymus Fischer, Jan 06 2009

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

			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.

David A. Corneth, Table of n, a(n) for n = 1..8005

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

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 *)

a(n) = my(f = floor(3^n/2^n)); ceil(((f + 1)*(2^n) - 3^n)/f) \\ David A. Corneth, Mar 27 2019

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

Showing 1-10 of 10 results.

cp's OEIS Frontend

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

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A154130 Exponents m with decreasing fractional part of (4/3)^m.

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

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

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

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

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A153717 Minimal exponents m such that the fractional part of (Pi-2)^m obtains a minimum (when starting with m=1).

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

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