A153711 -id:A153711 - cp's OEIS frontend

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

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

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

A153723 Greatest number m such that the fractional part of (Pi-2)^A153719(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 3, 16, 24, 45, 158, 410, 946, 1182, 8786, 16159, 20188, 61392, 78800, 78959, 217556

Offset: 1

Hieronymus Fischer, Jan 06 2009

			a(5) = 16, since 1-(1/17) = 0.941176... > fract((Pi-2)^A153719(5)) = fract((Pi-2)^5) = 0.9389... >= 0.9375 = 1-(1/16).

Cf. A153663, A153671, A153679, A153687, A153695, A091560, A153711, A153719, A154130.

$MaxExtraPrecision = 100000;
A153719 = {1, 2, 3, 4, 5, 39, 56, 85, 557, 911, 2919, 2921, 4491,
   11543, 15724, 98040, 110932, 126659};
Floor[1/(1 - FractionalPart[(Pi - 2)^A153719])] (* Robert Price, Apr 18 2019 *)

a(n) = floor(1/(1-fract((Pi-2)^A153719(n)))), where fract(x) = x-floor(x).

A154133 Numbers k such that the fractional part of (4/3)^k is greater than 1-(1/k).

Original entry on oeis.org

1, 2, 8, 39, 113, 3895, 4714, 8592, 34289, 60097, 942859, 2759790, 6649343, 7937397, 14480816, 19338413, 19338414, 19338415, 23187701, 124679421

Offset: 1

Hieronymus Fischer, Jan 11 2009

Numbers k such that fract((4/3)^k) > 1-(1/k), where fract(x) = x - floor(x).

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

			a(4) = 39, since fract((4/3)^39) = 0.9991861450... > 0.974358... = 1 - (1/39), but fract((4/3)^k) <= 1 - (1/k) for 8 < k < 39.

Cf. A153664, A153672, A153704, A153711, A154141, A154149.

Cf. A002379, A064628.

Select[Range[5000], N[FractionalPart[(4/3)^#], 100] > 1 - (1/#) &] (* G. C. Greubel, Sep 02 2016 *)

isok(n) = frac((4/3)^n) > 1 - 1/n; \\ Michel Marcus, Sep 03 2016

a(11)-a(20) from Robert Gerbicz, Nov 21 2010

Showing 1-8 of 8 results.

cp's OEIS Frontend

A153715 Greatest number m such that the fractional part of Pi^A153711(m) >= 1-(1/m).

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

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

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

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

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

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PARI

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A153723 Greatest number m such that the fractional part of (Pi-2)^A153719(m) >= 1-(1/m).

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