A081464 -id:A081464 - cp's OEIS frontend

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

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

A002379 a(n) = floor(3^n / 2^n).

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 11, 17, 25, 38, 57, 86, 129, 194, 291, 437, 656, 985, 1477, 2216, 3325, 4987, 7481, 11222, 16834, 25251, 37876, 56815, 85222, 127834, 191751, 287626, 431439, 647159, 970739, 1456109, 2184164, 3276246, 4914369, 7371554, 11057332

Offset: 0

N. J. A. Sloane

It is an important unsolved problem related to Waring's problem to show that a(n) = floor((3^n-1)/(2^n-1)) holds for all n > 1. This has been checked for 10000 terms and is true for all sufficiently large n, by a theorem of Mahler. [Lichiardopol]
a(n) = floor((3^n-1)/(2^n-1)) holds true at least for 2 <= n <= 305000. - Hieronymus Fischer, Dec 31 2008
a(n) is also the curve length (rounded down) of the Sierpiński arrowhead curve after n iterations, let a(0) = 1. - Kival Ngaokrajang, May 21 2014
a(n) is composite infinitely often (Forman and Shapiro). More exactly, a(n) is divisible by at least one of 2, 5, 7 or 11 infinitely often (Dubickas and Novikas). - Tomohiro Yamada, Apr 15 2017

R. K. Guy, Unsolved Problems in Number Theory, E19.
D. H. Lehmer, Guide to Tables in the Theory of Numbers. Bulletin No. 105, National Research Council, Washington, DC, 1941, p. 82.
S. S. Pillai, On Waring's problem, J. Indian Math. Soc., 2 (1936), 16-44.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Arturas Dubickas and Aivaras Novikas, Integer parts of powers of rational numbers, Math. Z. 251 (2005), 635--648, available from the first author's page.
W. Forman and H. N. Shapiro, An arithmetic property of certain rational powers, Comm. Pure. Appl. Math. 20 (1967), 561-573.
R. K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 3-20.
R. K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 3-20. [Annotated scanned copy]
N. Lichiardopol, Problem 925 (BCC20.19), A number-theoretic problem, in Research Problems from the 20th British Combinatorial Conference, Discrete Math., 308 (2008), 621-630.
K. Mahler, On the fractional parts of the powers of a rational number, II, Mathematika 4 (1957), 122-124.
Kival Ngaokrajang, Illustration of Sierpinski arrowhead curve for n = 0..5
Eric Weisstein's World of Mathematics, Power Floors
Wikipedia, Sierpiński arrowhead curve

Cf. A094969-A094500, A000217, A081464, A153662, A153665, A153666.
Cf. A060692, A002380, A000079, A000244.
Cf. A046037, A070758, A070759, A067904 (Composites and Primes).
Cf. A064628 (an analog for 4/3).

a002379 n = 3^n `div` 2^n  -- Reinhard Zumkeller, Jul 11 2014

[Floor(3^n / 2^n): n in [0..40]]; // Vincenzo Librandi, Sep 08 2011

A002379:=n->floor(3^n/2^n); seq(A002379(k), k=0..100); # Wesley Ivan Hurt, Oct 29 2013

Table[Floor[(3/2)^n], {n, 0, 40}] (* Robert G. Wilson v, May 11 2004 *)
x[n_] := -(1/2) + (3/2)^n + ArcTan[Cot[(3/2)^n Pi]]/Pi; Array[x, 40] (* Fred Daniel Kline, Dec 21 2017 *)
x[n_]:=Round[-(1/2) + (3^n - 1)/(2^n - 1)]; Array[x, 39, 2] (* offset n+1, Fred Daniel Kline, Apr 13 2018 *)

makelist(floor(3^n/2^n), n, 0, 50); /* Martin Ettl, Oct 17 2012 */

a(n)=3^n>>n \\ Charles R Greathouse IV, Jun 10 2011

def A002379(n): return 3**n>>n # Chai Wah Wu, Sep 21 2022

a(n) = b(n) - (-2/3)^n where b(n) is defined by the recursion b(0):=2, b(1):=5/6, b(n+1):=(5/6)*b(n) + b(n-1). - Hieronymus Fischer, Dec 31 2008
a(n) = (1/2)*(b(n) + sqrt(b(n)^2 - (-4)^n)) (with b(n) as defined above). - Hieronymus Fischer, Dec 31 2008
3^n = a(n)*2^n + A002380(n). - R. J. Mathar, Oct 26 2012
a(n) = -(1/2) + (3/2)^n + arctan(cot((3/2)^n Pi)) / Pi. - Fred Daniel Kline, Apr 14 2018
a(n+1) = round( -(1/2) + (3^n-1)/(2^n-1) ). - Fred Daniel Kline, Apr 14 2018

More terms from Robert G. Wilson v, May 11 2004

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

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

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

A137994 a(n) is the smallest integer > a(n-1) such that {Pi^a(n)} < {Pi^a(n-1)}, where {x} = x - floor(x), a(1)=1.

Original entry on oeis.org

1, 3, 81, 264, 281, 472, 1147, 2081, 3207, 3592, 10479, 12128, 65875, 114791, 118885

Offset: 1

Leroy Quet and M. F. Hasler, Mar 14 2008

The sequence was suggested by Leroy Quet on Pi day 2008, cf. A138324.
The next such number must be greater than 100000. - Hieronymus Fischer, Jan 06 2009
a(16) > 300,000. - Robert Price, Mar 25 2019

			a(3)=81, since {Pi^81}=0.0037011283.., but {Pi^k}>=0.0062766802... for 1<=k<=80; thus {Pi^81}<{Pi^k} for 1<=k<81. - _Hieronymus Fischer_, Jan 06 2009

Cf. A001203, A138324, A001672.

Cf. A081464, A153669, A153677, A153685, A153693, A153705, A153713, A154130, A153717. - Hieronymus Fischer, Jan 06 2009

$MaxExtraPrecision = 10000;
p = .999;
Select[Range[1, 5000],
If[FractionalPart[Pi^#] < p, p = FractionalPart[Pi^#]; True] &] (* Robert Price, Mar 12 2019 *)

default(realprecision,10^4); print1(a=1); for(i=1,100, f=frac(Pi^a); until( frac(Pi^a++)

a(11)-a(13) from Hieronymus Fischer, Jan 06 2009
Edited by R. J. Mathar, May 21 2010
a(14)-a(15) from Robert Price, Mar 12 2019

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

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cp's OEIS Frontend

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

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A002379 a(n) = floor(3^n / 2^n).

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

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

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

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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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A137994 a(n) is the smallest integer > a(n-1) such that {Pi^a(n)} < {Pi^a(n-1)}, where {x} = x - floor(x), a(1)=1.

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