A136409 -id:A136409 - cp's OEIS frontend

A020915 Number of digits in base-3 representation of 2^n.

1, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45, 46, 47

Offset: 0

Clark Kimberling

For n > 0, first differences of A022331. - Michel Marcus, Oct 03 2013

William A. Tedeschi, Table of n, a(n) for n = 0..10000
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x+1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A007089, A020914, A076227, A081604, A000079.
Cf. A022331, A102525, A136409.
Cf. A022924 (first differences).

a020915 = a081604 . a000079  -- Reinhard Zumkeller, Mar 28 2015

[Round(1+Floor(n*(Log(2))/Log(3))): n in [0..80]]; // Vincenzo Librandi, Dec 05 2015

Table[Length[IntegerDigits[2^n,3]],{n,0,80}] (* Harvey P. Dale, May 02 2012 *)
Table[1 + Floor[n*Log[3, 2]], {n, 0, 73}] (* L. Edson Jeffery, Dec 04 2015 *)
IntegerLength[2^Range[0,80],3] (* Harvey P. Dale, Nov 17 2022 *)

a(n)=logint(2^n,3)+1 \\ Charles R Greathouse IV, Sep 02 2015

a(n) = 1 + floor(n*log_3(2)) = 1 + floor(n*A102525) = 1 + A136409(n). - R. J. Mathar, May 23 2009
a(n) = A081604(A000079(n)). - Reinhard Zumkeller, Jul 12 2011
a(A020914(n)) = n + 1. - Reinhard Zumkeller, Mar 28 2015

More terms from James Sellers

A063005 Difference between 2^n and the next smaller or equal power of 3.

Original entry on oeis.org

0, 1, 1, 5, 7, 5, 37, 47, 13, 269, 295, 1319, 1909, 1631, 9823, 13085, 6487, 72023, 84997, 347141, 517135, 502829, 2599981, 3605639, 2428309, 19205525, 24062143, 5077565, 139295293, 149450423, 686321335, 985222181, 808182895, 5103150191, 6719515981, 2978678759

Offset: 0

Jens Voß, Jul 02 2001

Sequence generalized : a(n) = A^n - B^(floor(log_B (A^n))) where A, B are integers. This sequence has A = 2, B = 3; A056577 has A = 3, B = 2. - Ctibor O. Zizka, Mar 03 2008

Alois P. Heinz, Table of n, a(n) for n = 0..3324

Cf. A056577, A063003, A063004.

Cf. A000079 (2^n), A000244 (3^n), A136409.

a:= n-> (t-> t-3^ilog[3](t))(2^n):
seq(a(n), n=0..40);  # Alois P. Heinz, Oct 11 2019

a[n_] := 2^n - 3^Floor[Log[3, 2] * n]; Array[a, 36, 0] (* Amiram Eldar, Nov 19 2021 *)

for(n=0,50,print1(2^n-3^floor(log(2^n)/log(3))","))

def a(n):
    m, p, target = 0, 1, 2**n
    while p <= target:  m += 1; p *= 3
    return target - 3**(m-1)
print([a(n) for n in range(36)]) # Michael S. Branicky, Nov 19 2021

a(n) = 2^n - 3^(floor (log_3 (2^n))).

a(n) = A000079(n) - 3^A136409(n). - Michel Marcus, Nov 19 2021

More terms from Ralf Stephan, Mar 20 2003

A156301 a(n) = ceiling( n * log_3(2) ) = ceiling(n * 0.6309297535714574371...).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 6, 6, 7, 7, 8, 9, 9, 10, 11, 11, 12, 12, 13, 14, 14, 15, 16, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 23, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 30, 30, 31, 31, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 43, 44, 45, 45, 46, 47

Offset: 0

Jonathan Vos Post, Feb 07 2009

a(n) is the unique k such that 1/2 <= 3^a(n)/2^(n+1) < 3/2. Equality occurs iff n = 0. See Gorman-Huang and Marks.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Li an-Ping, A note on the counterfeit coins problem, arXiv:0902.0841 [math.CO], 2009-2010.
A. Gorman-Huang and E. Marks, Approximating Powers of 2 Using Powers of 3 and Break Point Rounding, Girls' Angle Bulletin, Vol. 18, No. 5 (2025), 8-10.

Cf. A020915, A102525. - R. J. Mathar, Feb 19 2009

Cf. A136409.

a156301 = ceiling . (* logBase 3 2) . fromIntegral
-- Reinhard Zumkeller, Jul 03 2015

seq(ceil(n*log[3](2)),n=0..120) ; # R. J. Mathar, Mar 14 2009

With[{c=Log[3,2]},Ceiling[c*Range[0,80]]] (* Harvey P. Dale, Aug 07 2015 *)

from operator import sub
from sympy import integer_log
def A156301(n): return sub(*integer_log(1<Chai Wah Wu, Oct 09 2024

More terms from R. J. Mathar, Mar 14 2009

Edited by N. J. A. Sloane, May 23 2009 at the suggestion of Hagen von Eitzen

A265210 Irregular triangle read by rows in which row n lists the base 3 digits of 2^n in reverse order, n >= 0.

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 1, 0, 1, 1, 0, 1, 2, 2, 0, 2, 1, 1, 1, 1, 1, 0, 0, 1, 2, 2, 2, 0, 0, 2, 1, 2, 2, 1, 0, 1, 1, 2, 1, 2, 0, 1, 2, 2, 1, 0, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 2, 0, 2, 0, 1, 1, 1, 2, 0, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1

Offset: 0

L. Edson Jeffery, Dec 04 2015

The length of row n is A020915(n) = 1 + A136409(n).

Conjecture 1: The sequence in column k is periodic, with period p(k) = 2*3^(k-1) = A008776(k-1), k >= 1, and in which the numbers 0,1,2 appear with equal frequency, for each k>1.

			n
0:    1
1:    2
2:    1  1
3:    2  2
4:    1  2  1
5:    2  1  0  1
6:    1  0  1  2
7:    2  0  2  1  1
8:    1  1  1  0  0  1
9:    2  2  2  0  0  2
10:   1  2  2  1  0  1  1
11:   2  1  2  0  1  2  2
12:   1  0  2  1  2  1  2  1
13:   2  0  1  0  2  0  2  0  1
14:   1  1  2  0  1  1  1  1  2
15:   2  2  1  1  2  2  2  2  1  1

Cf. A000079 (powers of 2), A004642 (powers of 2 written in base 3), A008776 (2*3^n).
Cf. A265209 (base 3 digits of 2^n).
Cf. A264980 (row n read as ternary number).
Cf. A037096 (numbers constructed from the inverse case, base 2 digits of 3^n).

(* Replace Flatten with Grid to display the triangle: *)
Flatten[Table[Reverse[IntegerDigits[2^n, 3]], {n, 0, 15}]]

A265210_row(n)=Vecrev(digits(2^n,3)) \\ M. F. Hasler, Dec 05 2015

A022924 Number of 3^m between 2^n and 2^(n+1).

Original entry on oeis.org

0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 1

Offset: 0

Clark Kimberling

nonn

This is a Sturmian sequence; consider the straight line of equation y = x*log(3)/log(2), the sequence gives the number m of integer ordinate points between the abscissa points n and n+1. - Richard Aime Blavy, Jun 14 2020

David A. Corneth, Table of n, a(n) for n = 0..9999

Cf. A136409 (partial sums).

First differences of A020915.

a(n) = logint(2^(n+1),3)-logint(2^n,3) \\ Charles R Greathouse IV, Jan 16 2017

A265209 Irregular triangle read by rows in which row n lists the base-3 digits of 2^n, n >= 0.

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 0, 1, 2, 2, 1, 0, 1, 1, 1, 2, 0, 2, 1, 0, 0, 1, 1, 1, 2, 0, 0, 2, 2, 2, 1, 1, 0, 1, 2, 2, 1, 2, 2, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 1, 0, 2, 0, 2, 0, 1, 0, 2, 2, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 2, 2

Offset: 0

L. Edson Jeffery, Dec 04 2015

The length of row n is A020915(n) = 1 + A136409(n).

			Triangle begins:
  1
  2
  1  1
  2  2
  1  2  1
  1  0  1  2
  2  1  0  1
  1  1  2  0  2
  1  0  0  1  1  1
  2  0  0  2  2  2
  1  1  0  1  2  2  1
  2  2  1  0  2  1  2
  1  2  1  2  1  2  0  1
  1  0  2  0  2  0  1  0  2
  2  1  1  1  1  0  2  1  1
  1  1  2  2  2  2  1  1  2  2

Cf. A000079 (powers of 2), A003137, A004642 (powers of 2 written in base 3).

Cf. A265210 (base 3 digits of 2^n in reverse order).

(* Replace Flatten with Grid to display the triangle: *)
Flatten[Table[IntegerDigits[2^n, 3], {n, 0, 15}]]

for(n=0,15,for(k=1,#digits(2^n,3),print1(digits(2^n,3)[k],", "))) \\ Derek Orr, Dec 24 2015

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A020915 Number of digits in base-3 representation of 2^n.

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A063005 Difference between 2^n and the next smaller or equal power of 3.

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A156301 a(n) = ceiling( n * log_3(2) ) = ceiling(n * 0.6309297535714574371...).

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A265210 Irregular triangle read by rows in which row n lists the base 3 digits of 2^n in reverse order, n >= 0.

Views

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Mathematica

PARI

A022924 Number of 3^m between 2^n and 2^(n+1).

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PARI

A265209 Irregular triangle read by rows in which row n lists the base-3 digits of 2^n, n >= 0.

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Author

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Comments

Examples

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Programs

Mathematica

PARI