A153663 -id:A153663 - cp's OEIS frontend

A153667 Greatest number m such that the fractional part of (3/2)^A153663(n) >= 1-(1/m).

2, 2, 2, 2, 3, 14, 31, 33, 69, 137, 222, 318, 901, 1772, 2747, 12347, 16540, 18198, 135794, 222246, 570361, 2134829, 6901329, 75503109, 814558605

Offset: 1

Hieronymus Fischer, Dec 31 2008

			a(5)=3, since 1-(1/4)=0.75>fract((3/2)^A153663(5))=fract((3/2)^12)=0.746...>=1-(1/3).

Cf. A002379, A081464, A153662, A153663, A153664, A153665, A153666, A153668.

A153663 = {1, 5, 8, 10, 12, 14, 46, 58, 105, 157, 163, 455, 1060, 1256, 2677, 8093, 28277, 33327, 49304, 158643, 164000, 835999, 2242294, 25380333, 92600006};
Table[fp = FractionalPart[(3/2)^A153663[[n]]]; m = Floor[1/(1-fp)];
While[fp >= 1 - (1/m), m++]; m - 1, {n, 1, Length[A153663]}] (* Robert Price, Mar 26 2019 *)

a(n) = floor(1/(1-fract((3/2)^A153663(n)))), where fract(x) = x-floor(x).

a(22)-a(25) from Robert Price, May 10 2012

A153662 Numbers k such that the fractional part of (3/2)^k is less than 1/k.

Original entry on oeis.org

1, 2, 4, 7, 3328, 3329, 4097, 12429, 12430, 12431, 18587, 44257, 112896, 129638, 4264691, 144941960, 144941961, 144941962

Offset: 1

Hieronymus Fischer, Dec 31 2008

Numbers k such that fract((3/2)^k) < 1/k, where fract(x) = x-floor(x).

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

			a(4) = 7 since fract((3/2)^7) = 0.0859375 < 1/7, but fract((3/2)^5)  = 0.59375 >= 1/5 and fract((3/2)^6) = 0.390625 >= 1/6.

Cf. A002379, A081464, A153663, A153664, A153665, A153666, A153667, A153668.

Select[Range[1000], FractionalPart[(3/2)^#] < (1/#) &] (* G. C. Greubel, Aug 24 2016 *)

a(15)-a(18) from Robert Gerbicz, Nov 21 2010

A153664 Numbers k such that the fractional part of (3/2)^k is greater than 1-(1/k).

Original entry on oeis.org

1, 14, 163, 1256, 2677, 8093, 49304, 49305, 158643, 164000, 835999, 2242294, 2242295, 2242296, 3965133, 25380333, 92600006, 92600007, 92600008, 92600009, 92600010, 92600011, 99267816, 125040717, 125040718

Offset: 1

Hieronymus Fischer, Dec 31 2008

Numbers k such that fract((3/2)^k) > 1-(1/k), where fract(x) = x-floor(x).

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

			a(2) = 14 since fract((3/2)^14) = 0.92926... > 0.92857... = 1 - (1/14), but fract((3/2)^k) <= 1 - (1/k) for 1<k<14.

Cf. A002379, A081464, A153662, A153663, A153665, A153666, A153667, A153668.

Select[Range[1000], FractionalPart[(3/2)^#] >= 1 - (1/#) &] (* G. C. Greubel, Aug 24 2016 *)

a(11)-a(25) from Robert Gerbicz, Nov 21 2010

A153671 Minimal exponents m such that the fractional part of (101/100)^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 110, 180, 783, 859, 1803, 7591, 10763, 19105, 50172, 355146, 1101696, 1452050, 3047334, 3933030

Offset: 1

Hieronymus Fischer, Jan 06 2009

nonn

Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (101/100)^m is greater than the fractional part of (101/100)^k for all k, 1<=k

The next such number must be greater than 10^6.

a(84) > 10^7. Robert Price, Mar 21 2019

			a(5)=5, since fract((101/100)^5)=0.05101005, but fract((101/100)^k)=0.01, 0.0201, 0.030301, 0.04060401 for 1<=k<=4; thus fract((101/100)^5)>fract((101/100)^k) for 1<=k<5.

Cf. A153663, A154130, A153675, A153679, A153687, A153695, A091560, A153711, A153719.

p = 0; Select[Range[1, 20000],
If[FractionalPart[(101/100)^#] > p, p = FractionalPart[(101/100)^#];
True] &] (* Robert Price, Mar 21 2019 *)

A153671_list, m, n, k, q = [], 1, 101, 100, 0
while m < 10**4:
    r = n % k
    if r > q:
        q = r
        A153671_list.append(m)
    m += 1
    n *= 101
    k *= 100
    q *= 100 # Chai Wah Wu, May 16 2020

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(72)-a(83) from Robert Price, Mar 21 2019

A153679 Minimal exponents m such that the fractional part of (1024/1000)^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 82, 134, 1306, 2036, 6393, 34477, 145984, 2746739, 2792428, 8460321

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 (1024/1000)^m is greater than the fractional part of (1024/1000)^k for all k, 1<=k

The next such number must be greater than 5*10^5.

a(40) > 10^7. Robert Price, Mar 16 2019

			a(30)=82, since fract((1024/1000)^82)= 0.99191990..., but fract((1024/1000)^k)<0.9893 for 1<=k<=81; thus fract((1024/1000)^82)>fract((1024/1000)^k) for 1<=k<82 and 82 is the minimal exponent >29 with this property.

Cf. A153663, A153671, A153675, A154130, A153687, A153695, A091560, A153711, A153719.

$MaxExtraPrecision = 10000;
p = 0;
Select[Range[1, 50000],
If[FractionalPart[(1024/1000)^#] > p,
p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 16 2019 *)

A153679_list, m, n, k, q = [], 1, 1024, 1000, 0
while m < 10**4:
    r = n % k
    if r > q:
        q = r
        A153679_list.append(m)
    m += 1
    n *= 1024
    k *= 1000
    q *= 1000 # Chai Wah Wu, May 16 2020

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(37)-a(39) from Robert Price, Mar 16 2019

A153687 Minimal exponents m such that the fractional part of (11/10)^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 23, 56, 77, 103, 320, 1477, 1821, 2992, 15290, 180168, 410498, 548816, 672732, 2601223

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 greater 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(22) > 10^7. Robert Price, Mar 19 2019

			a(8)=23, since fract((11/10)^23)= 0.9543..., but fract((11/10)^k)<0.95 for 1<=k<=22;
thus fract((11/10)^23)>fract((11/10)^k) for 1<=k<23 and 23 is the minimal exponent > 7 with this property.

Cf. A153663, A153671, A153683, A153679, A154130, A153695, A091560, A153711, A153719.

p = 0; Select[Range[1, 50000],
If[FractionalPart[(11/10)^#] > p, p = FractionalPart[(11/10)^#];
True] &] (* Robert Price, Mar 19 2019 *)

A153687_list, m, n, k, q = [], 1, 11, 10, 0
while m < 10**4:
    r = n % k
    if r > q:
        q = r
        A153687_list.append(m)
    m += 1
    n *= 11
    k *= 10
    q *= 10 # Chai Wah Wu, May 16 2020

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(18)-a(21) from Robert Price, Mar 19 2019

A153695 Minimal exponents m such that the fractional part of (10/9)^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 13, 17, 413, 555, 2739, 3509, 3869, 5513, 12746, 31808, 76191, 126237, 430116, 477190, 1319307, 3596185

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 greater 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(23) > 10^7. - Robert Price, Mar 24 2019

			a(7)=13, since fract((10/9)^13) = 0.93..., but fract((10/9)^k) < 0.89 for 1 <= k <= 12; thus fract((10/9)^13) > fract((10/9)^k) for 1 <= k < 13 and 13 is the minimal exponent > 6 with this property.

Cf. A153663, A153671, A153679, A153687, A153699, A154130, A091560, A153711, A153719.

$MaxExtraPrecision = 100000;
p = 0; Select[Range[1, 20000],
If[FractionalPart[(10/9)^#] > p, p = FractionalPart[(10/9)^#];
True] &] (* Robert Price, Mar 24 2019 *)

A153695_list, m, m10, m9, q = [], 1, 10, 9, 0
while m < 10**4:
    r = m10 % m9
    if r > q:
        q = r
        A153695_list.append(m)
    m += 1
    m10 *= 10
    m9 *= 9
    q *= 9 # Chai Wah Wu, May 16 2020

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(19)-a(22) from Robert Price, Mar 24 2019

A153665 Greatest number m such that the fractional part of (3/2)^A081464(n) <= 1/m.

Original entry on oeis.org

2, 4, 16, 25, 89, 91, 105, 127, 290, 668, 869, 16799, 92694, 137921, 257825, 350408, 419427, 723749, 5271294, 14223700, 18090494, 88123482, 706641581

Offset: 1

Hieronymus Fischer, Dec 31 2008

			a(4)=25 since 1/26<fract((3/2)^A081464(4))=fract((3/2)^29)=0.039...<=1/25.

Cf. A002379, A153662, A153663, A153664, A153666, A153667, A153668.

A081464 = {1, 2, 4, 29, 95, 153, 532, 613, 840, 2033, 2071, 3328, 12429, 112896, 129638, 371162, 1095666, 3890691, 4264691, 31685458, 61365215, 92432200, 144941960};
Table[fp = FractionalPart[(3/2)^A081464[[n]]]; m = Floor[1/fp];
While[fp <= 1/m, m++];  m - 1, {n, 1, Length[A081464]}] (* Robert Price, Mar 26 2019 *)

a(n):=floor(1/fract((3/2)^A081464(n))), where fract(x) = x-floor(x).

a(16)-a(23) from Robert Price, May 09 2012

A153666 Greatest number m such that the fractional part of (3/2)^A153662(n) <= 1/m.

Original entry on oeis.org

2, 4, 16, 11, 16799, 11199, 5536, 92694, 61796, 41197, 23242, 55710, 137921, 257825, 5271294, 706641581, 471094387, 314062925

Offset: 1

Hieronymus Fischer, Dec 31 2008

			a(3)=16 since 1/17<fract((3/2)^A153662(3))=fract((3/2)^4)=0.0625=1/16.

Cf. A002379, A081464, A153662, A153663, A153664, A153665, A153667, A153668.

A153662 = {1, 2, 4, 7, 3328, 3329, 4097, 12429, 12430, 12431, 18587, 44257, 112896, 129638, 4264691, 144941960, 144941961, 144941962};
Table[fp = FractionalPart[(3/2)^A153662[[n]]]; m = Floor[1/fp];
While[fp <= 1/m, m++];  m - 1, {n, 1, Length[A153662]}] (* Robert Price, Mar 26 2019 *)

a(n):=floor(1/fract((3/2)^A153662(n))), where fract(x) = x-floor(x).

a(15)-a(18) from Robert Price, May 09 2012

A091560 Fractional part of e^a(n) is the largest yet.

Original entry on oeis.org

1, 8, 19, 76, 166, 178, 209, 1907, 20926, 22925, 32653, 119136

Offset: 1

Jon Perry, Mar 04 2004

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

The next such number must be greater than 100000. [Hieronymus Fischer, Jan 06 2009]

a(13) > 300,000. Robert Price, Mar 23 2019

			a(2)=8, since fract(e^8)= 0.9579870417..., but fract(e^k)<=0.7182818... for 1<=k<=7;
thus fract(e^8)>fract(e^k) for 1<=k<8 and 8 is the minimal exponent > 1 with this property. [_Hieronymus Fischer_, Jan 06 2009]

Cf. A153663, A153671, A153679, A153687, A153695, A153707, A154130, A153711, A153719, A000149. [Hieronymus Fischer, Jan 06 2009]

a = 0; Do[b = N[ FractionalPart[ N[ E, 2^12]^n], 24]; If[b > a, Print[n]; a = b], {n, 1, 9400}] (* Robert G. Wilson v, Mar 16 2004 *)

E=exp(1); /* use sufficient precision! */
ym=0;for(i=1,1000,x=E^i;y=x-floor(x);if(y>ym,print1(","i);ym=y))

Recursion: a(1):=1, a(k):=min{ m>1 | fract(e^m) > fract(e^a(k-1))}, where fract(x) = x-floor(x). [Hieronymus Fischer, Jan 06 2009]

a(8) from Robert G. Wilson v, Mar 16 2004
a(9)-a(11) from Hieronymus Fischer, Jan 06 2009
a(12) from Robert Price, Mar 23 2019

Showing 1-10 of 21 results. Next

cp's OEIS Frontend

A153667 Greatest number m such that the fractional part of (3/2)^A153663(n) >= 1-(1/m).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A153662 Numbers k such that the fractional part of (3/2)^k is less than 1/k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A153664 Numbers k such that the fractional part of (3/2)^k is greater than 1-(1/k).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A153671 Minimal exponents m such that the fractional part of (101/100)^m obtains a maximum (when starting with m=1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A153679 Minimal exponents m such that the fractional part of (1024/1000)^m obtains a maximum (when starting with m=1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A153687 Minimal exponents m such that the fractional part of (11/10)^m obtains a maximum (when starting with m=1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A153695 Minimal exponents m such that the fractional part of (10/9)^m obtains a maximum (when starting with m=1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Formula

Extensions

A153665 Greatest number m such that the fractional part of (3/2)^A081464(n) <= 1/m.