A076227 -id:A076227 - cp's OEIS frontend

A056576 Highest k with 2^k <= 3^n.

0, 1, 3, 4, 6, 7, 9, 11, 12, 14, 15, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 38, 39, 41, 42, 44, 45, 47, 49, 50, 52, 53, 55, 57, 58, 60, 61, 63, 64, 66, 68, 69, 71, 72, 74, 76, 77, 79, 80, 82, 84, 85, 87, 88, 90, 91, 93, 95, 96, 98, 99, 101, 103, 104, 106, 107

Offset: 0

Henry Bottomley, Jun 29 2000

nonn

			a(3)=4 because 3^3=27 and 2^4=16 is power of 2 immediately below 27.

Michael De Vlieger, 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. A000079 (powers of 2), A000244 (powers of 3), A020914, A022921.

Cf. A056850, A117630 (complement), A020857 (decimal expansion of log_2(3)), A076227, A100982.

a056576 = subtract 1 . a020914  -- Reinhard Zumkeller, May 17 2015

seq(ilog2(3^n), n= 0 .. 1000); # Robert Israel, Dec 11 2014

Table[Floor[Log[2, 3^n]], {n, 0, 69}] (* Robert G. Wilson v, Apr 06 2006 *)
Table[Floor[n*Log[2, 3]], {n, 0, 68}] (* L. Edson Jeffery, Dec 11 2014 *)

{a(n) = if( n<0, 0, logint(3^n, 2))}; /* Michael Somos, Dec 13 2014 */

def A056576(n): return (3**n).bit_length()-1 # Chai Wah Wu, Oct 09 2024

a(n) = floor(log_2(3^n)) = log_2(A000244(n)-A056576(n)) = a(n-1)+A022921(n-1).

a(n) = A020914(n) - 1. - L. Edson Jeffery, Dec 12 2014

A100982 Number of admissible sequences of order j; related to 3x+1 problem and Wagon's constant.

Original entry on oeis.org

1, 1, 2, 3, 7, 12, 30, 85, 173, 476, 961, 2652, 8045, 17637, 51033, 108950, 312455, 663535, 1900470, 5936673, 13472296, 39993895, 87986917, 257978502, 820236724, 1899474678, 5723030586, 12809477536, 38036848410, 84141805077, 248369601964

Offset: 1

Steven Finch, Jan 13 2005

Eric Roosendaal counted all admissible sequences up to order j=1000 (2005). Note: there is a typo in both Wagon and Chamberland in the definition of Wagon's constant 9.477955... The expression floor(1 + 2*i + i*log_2(3)) should be replaced by floor(1 + i + i*log_2(3)).
The length of all admissible sequences of order j is A020914(j). - T. D. Noe, Sep 11 2006
Conjecture: a(n) is given for each n > 3 by a formula using a(2)..a(n-1). This allows us to create an iterative algorithm which generates a(n) for each n > 6. This has been proved for each n <= 53. For higher values of n the algorithm must be slightly modified. - Mike Winkler, Jan 03 2018
Theorem 1: a(k) is given for each k > 1 by a formula using a(1)..a(k-1). Namely, a(1)=1 and a(k+1) = Sum_{m=1..k} (-1)^(m-1)*binomial(floor((k-m+1)*(log(3)/log(2))) + m - 1, m)*a(k-m+1) for k >= 1. - Vladimir M. Zarubin, Sep 25 2015
Theorem 2: a(n) can be generated for each n > 2 algorithmically in a Pascal's triangle-like manner from the two starting values 0 and 1. This result is based on the fact that the Collatz residues (mod 2^k) can be evolved according to a binary tree. There is a direct connection with A076227, A056576 and A022921. - Mike Winkler, Sep 12 2017
A177789 shows another theorem and algorithm for generating a(n). - Mike Winkler, Sep 12 2017

			The unique admissible sequence of order 1 is 3/2, 1/2.
The unique admissible sequence of order 2 is 3/2, 3/2, 1/2, 1/2.
The two admissible sequences of order 3 are 3/2, 3/2, 3/2, 1/2, 1/2 and 3/2, 3/2, 1/2, 3/2, 1/2.
a(13) = 8045 = binomial(floor(5*(13-2)/3), 13-2)
- Sum_{i=2..6} binomial(floor((3*(13-i)+0)/2), 13-i)*a(i)
- Sum_{i=7..11} binomial(floor((3*(13-i)-1)/2), 13-i)*a(i)
- Sum_{i=12..12} binomial(floor((3*(13-i)-2)/2), 13-i)*a(i)
= 31824 - 4368*1 - 3003*2 - 715*3 - 495*7 - 120*12 - 28*30 - 21*85 - 5*173 - 4*476 - 1*961 - 0*2652. (Conjecture)
From _Mike Winkler_, Sep 12 2017: (Start)
The next table shows how Theorem 2 works. No entry is equal to zero.
n =       3  4  5   6   7   8   9  10  11   12 .. |A076227(k)=
--------------------------------------------------|
k =  2 |  1                                       |     1
k =  3 |  1  1                                    |     2
k =  4 |     2  1                                 |     3
k =  5 |        3   1                             |     4
k =  6 |        3   4   1                         |     8
k =  7 |            7   5   1                     |    13
k =  8 |               12   6   1                 |    19
k =  9 |               12  18   7   1             |    38
k = 10 |                   30  25   8   1         |    64
k = 11 |                   30  55  33   9    1    |   128
:      |                        :   :   :    : .. |    :
--------------------------------------------------|---------
a(n) =    2  3  7  12  30  85 173 476 961 2652 .. |
The entries (k,n) in this table are generated by the rule (k+1,n) = (k,n) + (k,n-1). The last value of (k+1,n) is given by k+1 = A056576(n-1), or the highest value in column n is given twice only if A022921(n-2) = 2. Then a(n) is equal to the sum of the entries in column n. For n = 7 there is 1 = 0 + 1, 5 = 1 + 4, 12 = 5 + 7, 12 = 12 + 0. Therefore a(7) = 1 + 5 + 12 + 12 = 30. The sum of row k is equal to A076227(k). (End)
From _Ruud H.G. van Tol_, Dec 04 2023: (Start)
A tree view.
n-tree--A098294--ids-----paths-----------------a(n)
0 ._          0  0       0                       -
1 |_          1  1       10                      1
2 |_._        2  2       1100                    1
3 |_|_        2  3-4     11010     -   11100     2
4 |_|_._      3  5-7     1101100   -  1111000    3
5 |_|_|_      3  8-14    11011010  - 11111000    7
6 |_|_|_._    4  15-26   1101101100-1111110000  12
7 |_|_|_|_._  5  27-56   ...                    30
8 |_|_|_|_|_  5  57-141  ...                    85
...
For n>=1, the endpoints are at A098294(n) to the right.
(End)

T. D. Noe, Table of n, a(n) for n = 1..500
M. Chamberland, Una actualizacio del problema 3x+1, Butl. Soc. Catalana Mat. 18 (2003) 19-45.
M. Chamberland, English translation
Ruud H.G. van Tol, Sequence as counts of lattice walks
Ruud H.G. van Tol, A100982 Musings
Stan Wagon, The Collatz problem, Math. Intelligencer 7 (1985) 72-76.
Mike Winkler, On a stopping time algorithm of the 3n + 1 function, 2011.
Mike Winkler, On the structure and the behaviour of Collatz 3n + 1 sequences - Finite subsequences and the role of the Fibonacci sequence, arXiv:1412.0519 [math.GM], 2014.
Mike Winkler, New results on the stopping time behaviour of the Collatz 3x + 1 function, arXiv:1504.00212 [math.GM], 2015.
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A122790 (Wagon's constant), A076227, A056576, A022921, A098294, A177789.

Cf. A060941, A293946.

(* based on Eric Roosendaal's algorithm *) nn=100; Clear[x,y]; Do[x[i]=0, {i,0,nn+1}]; x[1]=1; t=Table[Do[y[cnt]=x[cnt]+x[cnt-1], {cnt,p+1}]; Do[x[cnt]=y[cnt], {cnt,p+1}]; admis=0; Do[If[(p+1-cnt)*Log[3]T. D. Noe, Sep 11 2006 *)

/* translation of the above code from T. D. Noe */
{limit=100; n=1; x=y=vector(limit+1); x[1]=1; for(b=2, limit, for(c=2, b+1, y[c]=x[c]+x[c-1]); for(c=2, b+1, x[c]=y[c]); a_n=0; for(c=1, b+1, if((b+1-c)*log(3)Mike Winkler, Feb 28 2015

/* algorithm for the Conjecture */
{limit=53; zn=vector(limit); zn[2]=1; zn[3]=2; zn[4]=3; zn[5]=7; zn[6]=12; f=1; e1=-1; e2=-2; for(n=7, limit, m=floor((n-1)*log(3)/log(2))-(n-1); j=(m+n-2)!/(m!*(n-2)!); if(n>6*f, if(frac(n/2)==0, e=e1, e=e2)); if(frac((n-6 )/12)==0, f++; e1=e1+2); if(frac((n-12)/12)==0, f++; e2=e2+2); Sum=a=b=0; c=1; d=5; until(c>=n-1, for(i=2+a*5+b, 1+d+a*5, if(i>11 && frac((i+2)/6)==0, b++); delta=e-a; Sum=Sum+binomial(floor((3*(n-i)+delta)/2),n-i)*zn[i]; c++); a++; for(k=3, 50, if(n>=k*6 && a==k-1, d=k+3))); zn[n]=j-Sum; print(n" "zn[n]))} \\ Mike Winkler, Jan 03 2018

/* cf. code for Theorem 2 */
{limit=100; /*or limit>100*/ p=q=vector(limit); c=2; w=log(3)/log(2); for(n=3, limit, p[1]=Sum=1; for(i=2, c, p[i]=p[i-1]+q[i]; Sum=Sum+p[i]); a_n=Sum; print(n" "a_n); for(i=1, c, q[i]=p[i]); d=floor(n*w)-floor((n-1)*w); if(d==2, c++)); } \\ Mike Winkler, Apr 14 2015

/* algorithm for Theorem 1 */
n=20; a=vector(n); log32=log(3)/log(2);
{a[1]=1; for ( k=1, n-1, a[k+1]=sum( m=1,k,(-1)^(m-1)*binomial( floor( (k-m+1)*log32)+m-1,m)*a[k-m+1] ); print(k" "a[k]) );
} \\ Vladimir M. Zarubin, Sep 25 2015

/* algorithm for Theorem 2 */
{limit=30; /*or limit>30*/ R=matrix(limit,limit); R[2,1]=0; R[2,2]=1; for(n=2, limit, print; print1("For n="n" in column n: "); Kappa_n=floor(n*log(3)/log(2)); a_n=0; for(k=n, Kappa_n, R[k+1,n]=R[k,n]+R[k,n-1]; print1(R[k+1,n]", "); a_n=a_n+R[k+1,n]); print; print(" and the sum is a(n)="a_n))} \\ Mike Winkler, Sep 12 2017

A sequence s(k), where k=1, 2, ..., n, is *admissible* if it satisfies s(k)=3/2 exactly j times, s(k)=1/2 exactly n-j times, s(1)*s(2)*...*s(n) < 1 but s(1)*s(2)*...*s(m) > 1 for all 1 < m < n.
a(n) = (m+n-2)!/(m!*(n-2)!) - Sum_{i=2..n-1} binomial(floor((3*(n-i)+b)/2), n-i)*a(i), where m = floor((n-1)*log_2(3))-(n-1) and b assumes different integer values within the sum at intervals of 5 or 6 terms. (Conjecture)
a(n) = Sum_{k=n-1..A056576(n-1)} (k,n). (Theorem 2, cf. example)
a(k) = 2*A076227(A020914(k)-1) - A076227(A020914(k)), for k > 0. - Vladimir M. Zarubin, Sep 29 2019
a(1)=1, a(n) = Sum_{k=0..A020914(n-1)-n-2} A325904(k)*binomial(A020914(n-1)-k-2, n-2) for n>1. - Benjamin Lombardo, Oct 18 2019

Two more terms from Jules Renucci (jules.renucci(AT)wanadoo.fr), Nov 02 2005

More terms from T. D. Noe, Sep 11 2006

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

Original entry on oeis.org

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

A022921 Number of integers m such that 3^n < 2^m < 3^(n+1).

Original entry on oeis.org

1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 2, 1, 2, 1, 2, 1, 2, 2, 1, 2, 1

Offset: 0

Clark Kimberling

Represents increments between successive terms of allowable dropping times in the Collatz (3x+1) problem. That is, a(n) = A020914(n+1) - A020914(n). - K. Spage, Oct 23 2009

			From _Amiram Eldar_, Mar 01 2024: (Start)
a(0) = 1 because 3^0 = 1 < 2^1 = 2 < 3^1 = 3.
a(1) = 2 because 3^1 = 3 < 2^2 = 4 < 2^3 = 8 < 3^2 = 9.
a(2) = 1 because 3^2 = 9 < 2^4 = 16 < 3^3 = 27. (End)

T. D. Noe, Table of n, a(n) for n = 0..1000
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A020914, A056576, A076227, A100982, A122437.

See also A020857 (decimal expansion of log_2(3)).

Digits := 100: c1 := log(3.)/log(2.): A022921 := n->floor((n+1)*c1)-floor(n*c1);
seq(ilog2(3^(n+1)) - ilog2(3^n), n=0 .. 1000); # Robert Israel, Dec 11 2014

i2 = 1; Table[p = i2; While[i2++; 2^i2 < 3^(n + 1)]; i2 - p, {n, 0, 98}] (* T. D. Noe, Feb 28 2014 *)
f[n_] := Floor[ Log2[ 3^n] + 1]; Differences@ Array[f, 106, 0] (* Robert G. Wilson v, May 25 2014 *)

a(n) = logint(3^(n+1),2) - logint(3^n,2) \\ Ruud H.G. van Tol, Dec 28 2022

Vec(matreduce([logint(2^i,3)|i<-[1..158]])[,2])[1..-2] \\ Ruud H.G. van Tol, Dec 29 2022

a(n) = floor((n+1)*log_2(3)) - floor(n*log_2(3)).
a(n) = A122437(n+2) - A122437(n+1) - 1. - K. Spage, Oct 23 2009
First differences of A020914. - Robert G. Wilson v, May 25 2014
First differences of A056576. - L. Edson Jeffery, Dec 12 2014
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = log_2(3) (A020857). - Amiram Eldar, Mar 01 2024

A243115 Starting values of the reduced Collatz function (A014682) where 2 to the power of the "dropping time" is greater than the starting value.

Original entry on oeis.org

3, 7, 11, 15, 23, 27, 31, 39, 47, 59, 63, 71, 79, 91, 95, 103, 111, 123, 127, 155, 159, 167, 175, 191, 199, 207, 219, 223, 231, 239, 251, 255, 283, 287, 303, 319, 327, 347, 359, 367, 383, 411, 415, 423, 447, 463, 479, 487, 495, 507, 511, 539, 543, 559, 575

Offset: 1

K. Spage, Aug 20 2014

nonn

a(n) is the lowest positive starting value of the reduced Collatz function such that all starting values (>1) that are congruent to a(n) (mod 2^d) have the same dropping time (d). The dropping time here counts the (3x+1)/2 and the x/2 steps as listed in A126241. A number is included in this sequence if 2^A126241(a(n)) > a(n).
Starting values that produce new record dropping times as listed in A060412 are necessarily a subset of this sequence.
If at least one iteration is carried out before checking that the absolute iterated value has become less than or equal to the absolute starting value, then a(n) is the lowest positive starting value such that all starting values (positive, zero or negative) that are congruent to a(n) (mod 2^d) have the same dropping time (d). Defined like this, the sequence would start with 0, 1, 3, 7.
For k>0, A076227(k) is the number of terms between 2^k and 2^(k+1)-1. - Ruud H.G. van Tol, Dec 18 2022
All terms are congruent to 3 (mod 4) since any 1 (mod 4) has dropping time A126241(4k+1) = 2, for k>=1. - Ruud H.G. van Tol, Jan 11 2023

			3 is in this sequence because the dropping time starting with 3 is A126241(3) = 4 and 2^4 > 3.

Cf. A014682, A060412, A076227, A126241, A177789, A239126.

is(t)= if(t<3||3!=t%4,0,my(x=t, d=0); until(x<=t, if(x%2, x=(x*3+1)/2, x/=2); d++); 2^d>t); \\ updated by Ruud H.G. van Tol, Jan 10 2023

Offset 1 from Ruud H.G. van Tol, Jan 10 2023

A325904 Generator sequence for A100982.

Original entry on oeis.org

1, 0, -3, -8, 15, -91, -54, 2531, -17021, 43035, -66258, 1958757, -24572453, 146991979, -287482322, -3148566077, 35506973089, -198639977241, 1006345648929, -8250266425561, 76832268802555, -517564939540551, 1890772860334557, 3323588929061820, -104547561696315008, 907385094824827328, -6313246535826877248

Offset: 0

Benjamin Lombardo, Sep 08 2019

sign

The name of this sequence is derived from its main purpose as a formula for A100982 (see link). Both formulas below stem from Mike Winkler's 2017 paper on the 3x+1 problem (see below), in which a recursive definition of A100982 and A076227 is created in 2-D space. These formulas redefine the sequences in terms of this 1-D recursive sequence.

Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A020914, A076227, A100982.

import math
numberOfTerms = 20
L6 = [1,0]
def c(n):
    return math.floor(n/(math.log2(3)-1))
def p(a,b):
    return math.factorial(a)/(math.factorial(a-b)*math.factorial(b))
def anotherTerm(newTermCount):
    global L6
    for a in range(newTermCount+1-len(L6)):
        y = len(L6)
        newElement = 0
        for k in range(y):
            newElement -= int(L6[k]*p(c(y)+y-k-2, c(y)-2))
        L6.append(newElement)
anotherTerm(numberOfTerms)
print("A325904")
for a in range(numberOfTerms+1):
    print(a, "|", L6[a])

@cached_function
def a(n):
    if n < 2: return 0^n
    A = floor(n/(log(3, 2) - 1)) - 2
    return -sum(a(k)*binomial(A + n - k, A) for k in (0..n-1))
[a(n) for n in range(100)] # Peter Luschny, Sep 10 2019

a(0)=1, a(1)=0, a(n) = -Sum_{k=0..n-1} a(k)*binomial(A325913(n)+n-k-2, A325913(n)-2) for n>1.

A182137 Size of the set of b for numbers of the form 2^n*x + b that cannot be the smallest element of a set giving a duration of infinite flight in the Collatz problem.

Original entry on oeis.org

1, 3, 6, 13, 28, 56, 115, 237, 474, 960, 1920, 3870, 7825, 15650, 31473, 63422, 126844, 254649, 509298, 1021248, 2050541, 4101082, 8219801, 16490635, 32981270, 66071490, 132455435, 264910870, 530485275, 1060970550, 2123841570, 4253619813, 8507239626, 17027951548, 34095896991, 68191793982, 136471574881, 272943149762, 546144278026, 1093108792776, 2186217585552

Offset: 1

Jérôme STORTI, Apr 14 2012

nonn

In the Collatz Problem A014682, it is possible to apply the algorithm to first degree polynomials like 2^n*x+b, where n is an integer and 0 <= b < 2^n. The iteration terminates by two cases:
1) a*x+b where a < 2^n: the polynomial is "minimized"
2) a*x+b where a is odd and a > 2^n, parity cannot be found. The polynomial cannot be minimized.
The sequence counts how many first degree polynomials end like first case for each n > 0.
The interest of this sequence is that every number that can be described by a minimized polynomial cannot be the smallest element of a set of value of T(n) = infinity.

			Example with 4x+b (0 <= b < 4):
4x is even, thus gives 2x, 2 < 4 (first case).
4x+1, is odd thus 3(4x+1)+1 = 12x+4 is even, thus (12x+4)/2/2=3x+1 3 < 4, first case.
4x+2 is even, (4x+2)/2=2x+1, 2 < 4, first case.
4x+3 with same way gives 9x+8. 9 is odd and 9 > 4, second case.
That explains why the second (n=2) term in sequence is 3.

Cf. A020914, A074473, A076227, A100982, A186109, A243115.

a[n_] := Module[{b, p0, p1, minimized = 0}, For[b = 1, b <= 2^n, b++, {p0, p1} = {b, 2^n}; While[Mod[p1, 2] == 0 && p1 >= 2^n, {p0, p1} = If[Mod[p0, 2] == 0, {p0/2, p1/2}, {3*p0+1, 3*p1}]; If[p1<2^n, minimized += 1]]]; minimized]; Table[Print[an = a[n]]; an, {n, 1, 40}] (* Jean-François Alcover, Feb 12 2014, translated from D. S. McNeil's Sage code *)

upto(P=18)= my(r=Vec([1, 1], P)); forstep(x=3,2^P,4, my(s=x, p=0); until(s<=x, s= if(s%2, 3*s+1, s)/2; if(p++ > P, next(2))); if((2^p>x), r[p]++)); for(i=2, #r, r[i]+= 2*r[i-1]); print(r); \\ Ruud H.G. van Tol, Mar 13 2023

def A182137(n):
    minimized = 0
    for b in range(2**n):
        p = [b, 2**n]
        while p[1] % 2 == 0 and p[1] >= 2**n:
            p[0],p[1] = [p[0]/2, p[1]/2] if p[0] % 2 == 0 else [3*p[0]+1, 3*p[1]]
        if p[1] < 2**n: minimized += 1
    return minimized # D. S. McNeil, Apr 14 2012

a(n) = 2^n - A076227(n) for n >= 2. - Ruud H.G. van Tol, Mar 13 2023

For n not in A020914, a(n) = 2*a(n-1). - Ruud H.G. van Tol, Apr 12 2023

More terms from D. S. McNeil, Apr 14 2012
a(31) from Jérôme STORTI, Apr 22 2012
a(32)-a(38) from Jérôme STORTI, Jul 21 2012
a(39) from Jérôme STORTI, Jul 26 2012
a(40) from Jérôme STORTI, Feb 08 2014
a(37) and a(39) corrected by Jérôme STORTI, Dec 29 2021

cp's OEIS Frontend

A056576 Highest k with 2^k <= 3^n.

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A100982 Number of admissible sequences of order j; related to 3x+1 problem and Wagon's constant.

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PARI

PARI

PARI

PARI

PARI

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

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Magma

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A022921 Number of integers m such that 3^n < 2^m < 3^(n+1).

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A243115 Starting values of the reduced Collatz function (A014682) where 2 to the power of the "dropping time" is greater than the starting value.

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A325904 Generator sequence for A100982.

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SageMath