A137994 -id:A137994 - cp's OEIS frontend

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

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

A080052 Value of n such that for any value of n, Pi^n is closer to its nearest integer than any value of Pi^k for 1 <= k < n.

Original entry on oeis.org

1, 2, 3, 58, 81, 157, 1030, 5269, 12128, 65875, 114791, 118885, 151710

Offset: 1

Mark Hudson (mrmarkhudson(AT)hotmail.com), Jan 22 2003

nonn

Robert G. Wilson v used Mathematica with a changing number of digits to accommodate 24 digits to the right of the decimal point.
At 12128 the difference from an integer is 0.000016103224605297330719...
The sequence of rounded reciprocals of the distances, b(n) = round(1/(0.5-frac(Pi^a(n)-.5))) = round(1/abs(round(Pi^a(n))-Pi^a(n))), starts { 7, 8, 159, 190, 270, 2665, 10811, 26577, 62099, 70718, ... }. - M. F. Hasler, Apr 06 2008

			First term is 1 because this is just Pi = 3.14159....
Second term is 2 because Pi^2 = 9.869604... which is 0.13039... away from its nearest integer.
Pi^3 = 31.00627..., hence third term is 3.
Pi^58 is 0.00527... away from its nearest integer.

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 58, p. 21, Ellipses, Paris 2008.

Cf. A079490, A137994, A137995.

b := array(1..2000): Digits := 8000: c := 1: pos := 0: for n from 1 to 2000 do: exval := evalf(Pi^n): if (abs(exval-round(exval))

a = 1; Do[d = Abs[ Round[Pi^n] - N[Pi^n, Ceiling[ Log[10, Pi^n] + 24]]]; If[d < a, Print[n]; a = d], {n, 1, 25000}]
$MaxExtraPrecision = 10^9; a = 1; Do[d = Abs[ Round[Pi^n] - N[Pi^n, Ceiling[ Log[10, Pi^n] + 24]]]; If[d < a, Print[n]; a = d], {n, 1, 10^5}] (* Ryan Propper, Nov 13 2005 *)

f=0; for( i=1,99999, abs(frac(Pi^i)-.5)>f | next; f=abs(frac(Pi^i)-.5); print1(i",")) \\ M. F. Hasler, Apr 06 2008

More terms from Carlos Alves and Robert G. Wilson v, Jan 23 2003
One more term from Ryan Propper, Nov 13 2005
a(11)-a(13) from Jeremy Elson, Nov 13 2011

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

A153701 Minimal exponents m such that the fractional part of e^m obtains a minimum (when starting with m=1).

Original entry on oeis.org

1, 2, 3, 9, 29, 45, 75, 135, 219, 732, 1351, 3315, 4795, 4920, 5469, 28414, 37373

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 e^m is less than the fractional part of e^k for all k, 1<=k

The next such number must be greater than 100000.

a(18) > 300,000. Robert Price, Mar 23 2019

			a(4)=9, since fract(e^9)=0.08392..., but fract(e^k)>=0.08553... for 1<=k<=8; thus fract(e^9)<fract(e^k) for 1<=k<9.

Cf. A153661, A153669, A153677, A153685, A153693, A153705, A154130, A137994, A153717, A000149.

$MaxExtraPrecision = 100000;
p = 1; Select[Range[1, 300000],
If[FractionalPart[E^#] < p, p = FractionalPart[E^#]; True] &] (* Robert Price, Mar 23 2019 *)

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

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).

Showing 1-10 of 12 results. Next

Sequence A137994 could be defined as "least positive integer such that this one (without rounding) is increasing".

The term a(1)=7 is not surprising (3 + 1/7 = 3.14...) but it comes as a funny surprise that the next term, a(2)=159, matches the next 3 digits of Pi and a(3) just differs by 5 from the next 3 digits!

cp's OEIS Frontend

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

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A137995 Nearest integer to 1/frac(Pi^A137994(n)), where frac(x) = x - floor(x).

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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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A080052 Value of n such that for any value of n, Pi^n is closer to its nearest integer than any value of Pi^k for 1 <= k < n.

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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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A153701 Minimal exponents m such that the fractional part of e^m obtains a minimum (when starting with m=1).

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