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 1-10 of 15 results. Next

A153675 Greatest number m such that the fractional part of (101/100)^A153671(m) >= 1-(1/m).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 6, 6, 7, 9, 11, 13, 19, 30, 76, 81, 238, 913, 1334, 4645, 6812, 17396, 351085, 552184
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(5)=1, since 1-(1/2)=0.5>fract((101/100)^A153671(5))=fract((101/100)^5)=0.0510...>=1-(1/1).

Crossrefs

Cf. A153663, A154130, A153667, A153683, A153691, A153699, A153707, A153715, A153723.

Formula

a(n):=floor(1/(1-fract((101/100)^A153671(n)))), where fract(x) = x-floor(x).

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

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 (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

Examples

			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.

Crossrefs

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

Programs

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

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(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

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 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

Examples

			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.

Crossrefs

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

Programs

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

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(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

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 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

Examples

			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.

Crossrefs

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

Programs

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

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(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

Views

Author

Jon Perry, Mar 04 2004

Keywords

Comments

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

Examples

			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]

Crossrefs

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

Programs

  • Mathematica
    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 *)
  • PARI
    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))

Formula

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]

Extensions

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

A153707 Greatest number m such that the fractional part of e^A091560(m) >= 1-(1/m).

Original entry on oeis.org

3, 23, 27, 41, 59, 261, 348, 2720, 3198, 6064, 72944, 347065
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(2)=23, since 1-(1/24) = 0.9583...> fract(e^A091560(2)) = fract(e^8) = 0.95798.. >= 0.95652... >= 1-(1/23).

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A154130, A153715, A153723.
Cf. A000149.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
A091560 = {1,8,19,76,166,178,209,1907,20926,22925,32653,119136};
Floor[1/(1-FractionalPart[E^A091560])] (* Robert Price, Apr 18 2019 *)

Formula

a(n):=floor(1/(1-fract(e^A091560(n)))), where fract(x) = x-floor(x).

Extensions

a(12) from Robert Price, Apr 18 2019

A153711 Minimal exponents m such that the fractional part of Pi^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710
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 Pi^m is greater than the fractional part of Pi^k for all k, 1<=k
The next such number must be greater than 100000.
a(11) > 300000. - Robert Price, Mar 25 2019

Examples

			a(3)=15, since fract(Pi^15)= 0.9693879984..., but fract(Pi^k)<=0.8696... for 1<=k<=14; thus fract(Pi^15)>fract(Pi^k) for 1<=k<15 and 15 is the minimal exponent > 2 with this property.

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A153707, A153715, A154130, A153719.
Cf. A001672.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
p = 0; Select[Range[1, 10000],
If[FractionalPart[Pi^#] > p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 25 2019 *)

Formula

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

Extensions

a(9)-a(10) from Robert Price, Mar 25 2019

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

Original entry on oeis.org

1, 7, 32, 53, 189, 2665, 10810, 26577, 128778, 483367
Offset: 1

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			a(3) = 32, since 1-(1/33) = 0.9696... > fract(Pi^A153711(3)) = fract(Pi^15) = 0.96938... >= 0.96875 = 1-(1/32).

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A091560, A153711, A154130, A153723.
Cf. A001672.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
A153711 = {1, 2, 15, 22, 58, 157, 1030, 5269, 145048, 151710};
Floor[1/(1-FractionalPart[Pi^A153711])] (* Robert Price, Apr 18 2019 *)

Formula

a(n) = floor(1/(1-fract(Pi^A153711(n)))), where fract(x) = x-floor(x).

Extensions

a(9)-a(10) from Robert Price, Apr 18 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

Views

Author

Hieronymus Fischer, Jan 06 2009

Keywords

Examples

			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).

Crossrefs

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

Programs

  • Mathematica
    $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 *)

Formula

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

A153719 Minimal exponents m such that the fractional part of (Pi-2)^m obtains a maximum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 4, 5, 39, 56, 85, 557, 911, 2919, 2921, 4491, 11543, 15724, 98040, 110932, 126659
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 (Pi-2)^m is greater than the fractional part of (Pi-2)^k for all k, 1<=k
The next such number must be greater than 200000.
a(19) > 300000. - Robert Price, Mar 26 2019

Examples

			a(6)=39, since fract((Pi-2)^39)= 0.9586616565..., but fract((Pi-2)^k)<=0.9389018... for 1<=k<=38; thus fract((Pi-2)^39)>fract((Pi-2)^k) for 1<=k<39 and 39 is the minimal exponent > 5 with this property.

Crossrefs

Cf. A153663, A153671, A153679, A153687, A153695, A153707, A153715, A153723, A154130.

Programs

  • Mathematica
    $MaxExtraPrecision = 100000;
p = 0; Select[Range[1, 10000],
If[FractionalPart[(Pi - 2)^#] > p, p = FractionalPart[(Pi - 2)^#];
True] &] (* Robert Price, Mar 26 2019 *)

Formula

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).
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