A153662 -id:A153662 - cp's OEIS frontend

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

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

A153663 Minimal exponents m such that the fractional part of (3/2)^m reaches a maximum (when starting with m=1).

Original entry on oeis.org

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

Offset: 1

Hieronymus Fischer, Dec 31 2008

Recursive definition: a(1)=1, a(n) = least number m such that the fractional part of (3/2)^m is greater than the

fractional part of (3/2)^k for all k, 1<=k

The fractional part of k=835999 is .999999 5 which is greater than (k-1)/k. The fractional part of k=2242294 is .999999 8 which is greater than (k-1)/k. The fractional part of k=25380333 is .999999 98 which is greater than (k-1)/k. The fractional part of k=92600006 is .999999 998 which is greater than (k-1)/k. So, all additional numbers in this sequence must be in A153664 and >3*10^8. - Robert Price, May 09 2012

			a(2)=5, since fract((3/2)^5)=0.59375, but fract((3/2)^k)=0.5, 0.25, 0.375, 0.0625 for 1<=k<=4; thus
fract((3/2)^5)>fract((3/2)^k) for 1<=k<5.

Cf. A002379, A153661, A153662, A153664, A153665, A153666, A153667, A153668.

a[1] = 1; a[n_] := a[n] = For[m = a[n-1]+1, True, m++, f = FractionalPart[(3/2)^m]; If[AllTrue[Range[m-1], f > FractionalPart[(3/2)^#]&], Print[n, " ", m]; Return[m]]];
Array[a, 21] (* Jean-François Alcover, Feb 25 2019 *)

Recursion: a(1):=1, a(k):=min{ m>1 | fract((3/2)^m) > fract((3/2)^a(k-1))}, where fract(x) = x-floor(x).

a(22)-a(25) from Robert Price, May 09 2012

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

A153670 Numbers k such that the fractional part of (101/100)^k is less than 1/k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 70, 209, 241, 378, 2697, 4806, 173389, 529938, 1334508, 1572706, 7840546, 15896994, 20204295, 71074288, 119325567

Offset: 1

Hieronymus Fischer, Jan 06 2009

Numbers k such that fract((101/100)^k) < 1/k, where fract(x) = x-floor(x).

The next term is greater than 2*10^8.

			a(10) = 70 since fract((101/100)^70) = 0.006... < 1/10, but fract((101/100)^k) > 0.1 >= 1/k for 10 <= k <= 69.

Cf. A153662, A154130, A153674, A153678, A153686, A153694, A153702, A153710, A153718.

Select[Range[1000], FractionalPart[(101/100)^#] < (1/#) &] (* G. C. Greubel, Aug 24 2016 *)

from itertools import count, islice
def A153670gen(): # generator of terms
    k10, k11 = 100, 101
    for k in count(1):
        if (k11 % k10)*k < k10:
            yield k
        k10 *= 100
        k11 *= 101
A153670_list = list(islice(A153670gen(),16)) # Chai Wah Wu, Dec 23 2021

a(18)-a(24) from Robert Gerbicz, Nov 29 2010

A153678 Numbers k such that the fractional part of (1024/1000)^k is less than 1/k.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 585, 1164, 1707, 522271, 3675376, 3906074, 9424094

Offset: 1

Hieronymus Fischer, Jan 06 2009

Numbers k such that fract((1024/1000)^k) < 1/k, where fract(x) = x-floor(x).
The next such number must be greater than 5*10^5.
a(14) > 10^7. - Robert Price, Mar 16 2019

			a(7) = 585 since fract((1024/1000)^585) = 0.00139... < 1/585, but fract((1024/1000)^k) >= 1/k for 7 <= k <= 584.

Cf. A153662, A153670, A153682, A154130, A153686, A153694, A153702, A153710, A153718.

Select[Range[2000], FractionalPart[(1024/1000)^#] < (1/#) &] (* G. C. Greubel, Aug 24 2016; corrected by Robert Price, Mar 16 2019 *)

isok(n) = frac((1024/1000)^n) < 1/n \\ Michel Marcus, Aug 06 2013

a(10)-a(13) from Robert Price, Mar 16 2019

A153686 Numbers k such that the fractional part of (11/10)^k is less than 1/k.

Original entry on oeis.org

1, 2, 3, 17, 37, 48, 237, 420, 599, 615, 6638, 13885, 13886, 62963, 1063942, 9479731

Offset: 1

Hieronymus Fischer, Jan 06 2009

Numbers k such that fract((11/10)^k) < 1/k, where fract(x) = x-floor(x).
The next such number must be greater than 2*10^5.
a(17) > 10^7. - Robert Price, Mar 19 2019

			a(4) = 17 since fract((11/10)^17) = 0.05447... < 1/17, but fract((11/10)^k) >= 1/k for 4 <= k <= 16.

Cf. A153662, A153670, A153682, A153678, A154130, A153694, A153702, A153710, A153718.

Select[Range[1000], FractionalPart[(11/10)^#] < (1/#) &] (* G. C. Greubel, Aug 24 2016 *)

A153686_list, k, k10, k11 =  [], 1, 10, 11
while k < 10**6:
    if (k11 % k10)*k < k10:
        A153686_list.append(k)
    k += 1
    k10 *= 10
    k11 *= 11 # Chai Wah Wu, Apr 01 2021

a(15)-a(16) from Robert Price, Mar 19 2019

A153694 Numbers k such that the fractional part of (10/9)^k is less than 1/k.

Original entry on oeis.org

1, 2, 7, 62, 324, 1647, 3566, 5464, 8655, 8817, 123956, 132891, 182098, 566593, 2189647, 2189648, 3501843, 3501844

Offset: 1

Hieronymus Fischer, Jan 06 2009

Numbers k such that fract((10/9)^k) < 1/k, where fract(x) = x-floor(x).
The next such number must be greater than 2*10^5.
a(19) > 10^7. - Robert Price, Mar 24 2019
Given a number k that is not only a term of this sequence but also has the property that the integer part of (10/9)^k is divisible by 9, we can expect that k+1 will likely also be a term of the sequence. E.g., k = 2189647 is a term because fract((10/9)^k) = 0.000000373557... < 0.000000456694... = 1/k, and since floor((10/9)^k) is divisible by 9, the integer and fractional parts of (10/9)^(k+1) will be exactly 10/9 times the integer and fractional parts of (10/9)^k, respectively, yielding a fractional part (10/9) * 0.000000373557... = 0.000000415064... < 0.000000456694... = 1/(k+1), so k+1 = 2189648 is also a term. - Jon E. Schoenfield, Mar 24 2019

			a(3) = 7 since fract((10/9)^7) = 0.09075... < 1/7, but fract((10/9)^k) >= 1/k for 3 <= k <= 6.

Cf. A153662, A153670, A153678, A153686, A153698, A154130, A153702, A153710, A153718.

Select[Range[1000], FractionalPart[(10/9)^#] < (1/#) &] (* G. C. Greubel, Aug 24 2016 *)

a(14)-a(18) from Robert Price, Mar 24 2019

A153702 Numbers k such that the fractional part of e^k is less than 1/k.

Original entry on oeis.org

1, 2, 3, 9, 732, 5469, 28414, 37373, 93638, 136986, 192897

Offset: 1

Hieronymus Fischer, Jan 06 2009

Numbers k such that fract(e^k) < 1/k, where fract(x) = x-floor(x).
The next such number must be greater than 100000.
a(12) > 300000. - Robert Price, Mar 23 2019

			a(4) = 9 since fract(e^9) = 0.08392... < 1/9, but fract(e^k) = 0.598..., 0.413..., 0.428..., 0.633..., 0.957... for 4 <= k <= 8, which are all greater than 1/k.

Cf. A153662, A153670, A153678, A153686, A153694, A153706, A154130, A153710, A153718.

Cf. A000149.

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

a(10)-a(11) from Robert Price, Mar 23 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

Showing 1-10 of 22 results. Next

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

Maxima

PARI

Python

Formula

Extensions

A153663 Minimal exponents m such that the fractional part of (3/2)^m reaches a maximum (when starting with m=1).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

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

A153670 Numbers k such that the fractional part of (101/100)^k is less than 1/k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Python

Extensions

A153678 Numbers k such that the fractional part of (1024/1000)^k is less than 1/k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

PARI

Extensions

A153686 Numbers k such that the fractional part of (11/10)^k is less than 1/k.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs