A064628 -id:A064628 - 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

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

A064629 a(n) = 4^n mod 3^n.

Original entry on oeis.org

0, 1, 7, 10, 13, 52, 451, 1075, 6487, 6265, 44743, 119923, 302545, 147298, 589192, 11922706, 33341917, 4227505, 146050183, 584200732, 1174541461, 4698165844, 18792663376, 43789593895, 175158375580, 700633502320, 1955245399837, 2737249942690, 18574597255747

Offset: 0

Labos Elemer, Oct 01 2001

(a(n+1) - 4*a(n))/3^n is always one of -3, -2, -1, 0, 1, 2. - Robert Israel, Dec 01 2016

Harry J. Smith and Seiichi Manyama, Table of n, a(n) for n = 0..2096 (first 201 terms from Harry J. Smith)

Cf. A002379, A060692, A064628.

Cf. k^n mod (k-1)^n: A002380 (k=3), this sequence (k=4), A138589 (k=5), A138649 (k=6), A139786 (k=7), A138973 (k=8), A139733 (k=9).

[4^n mod 3^n: n in [0..40]]; // Wesley Ivan Hurt, Dec 01 2016

A064629:=n->4^n mod 3^n: seq(A064629(n), n=0..40); # Wesley Ivan Hurt, Dec 01 2016

Table[PowerMod[4,n,3^n],{n,0,30}] (* Harvey P. Dale, Aug 26 2012 *)

{ f=t=1; for (n=0, 200, write("b064629.txt", n, " ", f%t); f*=4; t*=3 ) } \\ Harry J. Smith, Sep 20 2009

[power_mod(4,n,3^n)for n in range(0,27)] # Zerinvary Lajos, Nov 28 2009

a(26) from Harry J. Smith, Sep 20 2009

A064630 Number of parts if 4^n is partitioned into parts of size 3^n as far as possible into parts of size 2^n as far as possible and into parts of size 1^n.

Original entry on oeis.org

2, 5, 5, 16, 25, 15, 66, 121, 146, 771, 1220, 3641, 8093, 15843, 28359, 50236, 33366, 36709, 145250, 137776, 548024, 2186496, 1066102, 4251976, 16984368, 28678103, 13620614, 205950171, 100716646, 381399635, 1397934923, 3826001641

Offset: 1

Labos Elemer, Oct 01 2001

nonn

Corresponds to the only solution of the Diophantine equation 4^n = x*3^n + y*2^n + z*1^n with constraints 0 <= y < 3^n/2^n, 0 <= z < 2^n.

Binary order (cf. A029837) of a(n) is close to n.

			4^6 = 4096 = 729 + 729 + 729 + 729 + 729 + 64 + 64 + 64 + 64 + 64 + 64 + 64 + 1 + 1 + 1 = 5*3^6 + 7*2^6 + 3*1^6, so a(6) = 5 + 7 + 3 = 15.

Harry J. Smith, Table of n, a(n) for n = 1..250

Cf. A064628, A064629, A060692, A029837.

{for(n=1,32,a=divrem(4^n,3^n); b=divrem(a[2],2^n); print1(a[1]+b[1]+b[2],","))}

{ f=t=w=1; for (n=1, 250, f*=4; t*=3; w*=2; a=divrem(f, t); b=divrem(a[2], w); write("b064630.txt", n, " ", a[1]+b[1]+b[2]) ) } \\ Harry J. Smith, Sep 20 2009

a(n) = A064628(n) + floor(A064629(n)/2^n) + (A064629(n) mod 2^n) = floor(4^n/3^n) + floor((4^n mod 3^n)/2^n) + ((4^n mod 3^n) mod 2^n)

Edited by Klaus Brockhaus, May 24 2003

A065564 Numbers k such that floor((4/3)^(k+1))/floor((4/3)^k) = 4/3.

Original entry on oeis.org

4, 13, 18, 20, 21, 23, 24, 34, 44, 45, 49, 56, 60, 63, 65, 66, 67, 79, 81, 83, 85, 88, 94, 102, 107, 109, 119, 125, 126, 129, 131, 132, 133, 135, 138, 139, 144, 161, 162, 164, 172, 174, 175, 190, 194, 199, 204, 217, 218, 233, 234, 240, 249, 250, 253, 255, 258

Offset: 1

Benoit Cloitre, Nov 30 2001

nonn

Harry J. Smith, Table of n, a(n) for n = 1..1000

Cf. A064628, A065554.

isok(k) = { my(f=4/3); floor(f^(k + 1))/floor(f^k) == f } \\ Harry J. Smith, Oct 22 2009

Lim_{n->infinity} a(n)/n = 4?

a(55)-a(57) corrected by Harry J. Smith, Oct 22 2009

A065565 a(n) = floor((5/4)^n).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 3, 4, 5, 7, 9, 11, 14, 18, 22, 28, 35, 44, 55, 69, 86, 108, 135, 169, 211, 264, 330, 413, 516, 646, 807, 1009, 1262, 1577, 1972, 2465, 3081, 3851, 4814, 6018, 7523, 9403, 11754, 14693, 18367, 22958, 28698, 35873, 44841, 56051, 70064, 87581

Offset: 0

Benoit Cloitre, Nov 30 2001

nonn

a(n) is also the curvature (rounded down) of the circle inscribed in the n-th 3:4:5 triangle arranged in a spiral as shown in the illustration in the links section. - Kival Ngaokrajang, Aug 21 2013

By the result of Dubickas and Novikas, a(n) is divisible by at least one of 2, 3, 7, 11, 13 infinitely often, so that a(n) is composite infinitely often. - Tomohiro Yamada, Apr 23 2017

Harry J. Smith, Table of n, a(n) for n = 0..400
Arturas Dubickas, Aivaras Novikas, Integer parts of powers of rational numbers, Math. Z. 251 (2005), 635-648, available from the first author's page.
Kival Ngaokrajang, Illustration of some initial terms.

Cf. A002379, A094969-A094500.

Cf. A064628. - Tomohiro Yamada, Apr 23 2017

Table[ Floor[(5/4)^n], {n, 0, 41}] (* Robert G. Wilson v, May 26 2004 *)

a(n) = floor((5/4)^n) \\ Harry J. Smith, Oct 22 2009

Edited by N. J. A. Sloane at the suggestion of Stefan Steinerberger, Jun 20 2007

Offset changed from 1 to 0 by Harry J. Smith, Oct 22 2009

A240916 a(n) = 6a(n-1) + 22^(n-1) - 2 for n > 2, a(0) = a(1) = 0, a(2) = 3.

Original entry on oeis.org

0, 0, 3, 24, 158, 978, 5930, 35706, 214490, 1287450, 7725722, 46356378, 278142362, 1668862362, 10013190554, 60079176090, 360475122074, 2162850863514, 12977105443226, 77862633183642, 467175800150426, 2803054802999706, 16818328822192538

Offset: 0

Kival Ngaokrajang, Apr 14 2014

a(n) is the total number of holes of a triflake-like fractal (Mitsubishi logo) after n iterations. The scale factor for this case is 1/3, but for the actual triflake case, it is 1/2, i.e., Sierpiński triangle. The total number of sides is 3*A000302(n). The perimeter (rounded down) is A064628(n).

Kival Ngaokrajang, Illustration of triflake like fractal (Mitsubishi logo) for n = 0..3
Wikipedia, n-flake
Index entries for linear recurrences with constant coefficients, signature (9,-20,12).

Cf. A000302, A064628, A240523 (pentaflake), A240671 (heptaflake), A240572 (octaflake), A240733 (nonaflake), A240734 (decaflake), A240840 (hendecaflake), A240735 (dodecaflake), A240841 (tridecaflake).

Join[{0,0},LinearRecurrence[{9,-20,12},{3,24,158},30]] (* Harvey P. Dale, Jan 31 2015 *)

{a(n)=if(n<=0, 0, if(n<2, 0, if(n<3, 3, 6*a(n-1)+2*2^(n-1)-2)))}
  for(n=0,100,print1(a(n),", "))

concat([0,0], Vec(-x^2*(2*x^2-3*x+3)/((x-1)*(2*x-1)*(6*x-1)) + O(x^100))) \\ Colin Barker, Apr 15 2014

From Colin Barker, Apr 15 2014: (Start)
a(n) = (72-45*2^(1+n)+23*6^n)/180 for n>1.
a(n) = 9*a(n-1)-20*a(n-2)+12*a(n-3) for n>4.
G.f.: -x^2*(2*x^2-3*x+3) / ((x-1)*(2*x-1)*(6*x-1)). (End).

A064536 a(n) = (4^n mod 3^n) mod 2^n.

Original entry on oeis.org

1, 3, 2, 13, 20, 3, 51, 87, 121, 711, 1139, 3537, 8034, 15752, 27922, 49629, 33201, 35975, 143900, 136341, 545364, 2181456, 1060135, 4240540, 16962160, 28647197, 13597858, 205877827, 100616667, 381266393, 1397863922, 3825576990, 8216376565, 14181633879, 22366797148

Offset: 1

Labos Elemer, Oct 08 2001

nonn

A generalization of A002380. It arises also as a coefficient (=c1) of 1^n=1 in a special (greedy) decomposition of 4^n into like powers as follows: 4^n = c3*3^n + c2*2^n + c1*1^n.

Harry J. Smith, Table of n, a(n) for n = 1..200

Cf. A002380, A064853-A064855, A060692, A064628-A064631.

Table[Mod[PowerMod[4,n,3^n],2^n],{n,40}] (* Harvey P. Dale, Apr 09 2013 *)

a(n) = { (4^n % 3^n) % 2^n } \\ Harry J. Smith, Sep 17 2009

n = 7: 4^7 = 16384 = 7*2187 + 8*128 + 51*1 where a(7)=51, the last coefficient; A064630(7) = 7 + 8 + a(7) = 66.

A064631 a(n) = ceiling(log_2(A064630(n))).

Original entry on oeis.org

2, 3, 3, 5, 5, 4, 7, 7, 8, 10, 11, 12, 13, 14, 15, 16, 16, 16, 18, 18, 20, 22, 21, 23, 25, 25, 24, 28, 27, 29, 31, 32, 33, 34, 35, 36, 37, 37, 39, 40, 39, 42, 42, 44, 44, 46, 46, 46, 49, 50, 51, 51, 51, 54, 55, 55, 57, 57, 59, 60, 60, 61, 63, 64, 64, 66, 60, 62, 67, 70, 69, 72

Offset: 1

Labos Elemer, Oct 01 2001

nonn

In A064630, using a greedy algorithm we write 4^n = x*3^n+y*2^n+z*1^n and A064630(n) = x+y+z. This sequence is a measure of the "length" or complexity of those solutions.

Cf. A002379, A002380, A060692, A064628, A064629, A064630, A029837.

a(n) = A029837(A064630(n)) = ceiling(log_2(A064630(n))).

Initial terms corrected and entry revised by Sean A. Irvine, Jul 18 2023

A064855 a(n) = (((6^n mod 5^n) mod 4^n) mod 3^n) mod 2^n.

Original entry on oeis.org

1, 2, 0, 14, 16, 10, 66, 21, 321, 917, 2037, 1550, 2420, 15152, 27439, 46731, 110953, 137148, 336949, 703202, 805647, 181132, 5835407, 3343039, 21816283, 18528238, 95129681, 241918238, 311938330, 48698222, 1539688558, 3481498150, 8104918325, 13512884439, 22365723609

Offset: 1

Labos Elemer, Oct 08 2001

nonn

A generalization of A002380, A064536 and A064854. It arises also as a coefficient (=c1) of 1^n=1 in a special (greedy) decomposition of 6^n into like powers as follows: 6^n = c5*5^n + c4*4^n + c3*3^n + c2*2^n + c1*1^n.

Harry J. Smith, Table of n, a(n) for n = 1..200

Cf. A002380, A064536, A064854, A064855, A060692, A064628-A064631.

Table[Fold[Mod,6^n,Range[5,2,-1]^n],{n,40}]  (* Harvey P. Dale, Mar 14 2011 *)

a(n) = { (((6^n%5^n)%4^n)%3^n)%2^n } \\ Harry J. Smith, Sep 28 2009

n = 8: 6^8 = 1679616 = 4*390625 + 1*65536 + 7*6561 + 22*256 + 21*1 where a(8)=21, the last coefficient and here 6^8 is decomposed into 4 + 1 + 7 + 22 + 21 = 55 like (8th) powers.

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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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A064629 a(n) = 4^n mod 3^n.

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A064630 Number of parts if 4^n is partitioned into parts of size 3^n as far as possible into parts of size 2^n as far as possible and into parts of size 1^n.

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A065564 Numbers k such that floor((4/3)^(k+1))/floor((4/3)^k) = 4/3.

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A065565 a(n) = floor((5/4)^n).

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A240916 a(n) = 6*a(n-1) + 2*2^(n-1) - 2 for n > 2, a(0) = a(1) = 0, a(2) = 3.

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A240916 a(n) = 6a(n-1) + 22^(n-1) - 2 for n > 2, a(0) = a(1) = 0, a(2) = 3.