A022921 -id:A022921 - cp's OEIS frontend

A020914 Number of digits in the base-2 representation of 3^n.

1, 2, 4, 5, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 23, 24, 26, 27, 29, 31, 32, 34, 35, 37, 39, 40, 42, 43, 45, 46, 48, 50, 51, 53, 54, 56, 58, 59, 61, 62, 64, 65, 67, 69, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 105, 107

Offset: 0

Clark Kimberling

Also, numbers k such that the first digit in the ternary expansion of 2^k is 1. - Mohammed Bouayoun (Mohammed.bouayoun(AT)sanef.com), Apr 24 2006
a(n) is the smallest integer such that n/a(n) < log_2(3). - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 31 2008
This sequence represents allowable values of the "dropping time" in the Collatz (3x+1) problem when iterated according to the function f(n) := n/2 if n is even, (3n+1)/2 otherwise, as tabulated in A126241. There is one exception, A126241(1), which has been set to zero by convention. - K. Spage, Oct 22 2009
An integer k is a term of A020914 if and only if floor(k*(1 + log(2)/log(3))) - abs(k-1)*(1 + log(2)/log(3)) - 1 >= 0. - K. Spage, Oct 22 2009
Also smallest k such that ceiling(2^k / 3^n) = 2. - Michel Lagneau, Jan 31 2012
For n > 0, first differences of A022330. - Michel Marcus, Oct 03 2013
Also the number of powers of two less than or equal to 3^n. - Robert G. Wilson v, May 25 2014
Except for 1, A020914 is the complement of A054414 and therefore these two form a pair of Beatty sequences. - Robert G. Wilson v, May 25 2014

T. D. Noe, R. J. Mathar, Table of n, a(n) for n = 0..20000
Mike Winkler, On the structure and the behaviour of Collatz 3n + 1 sequences, 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. A056576, A054414, A070939, A000244, A227048, A022330, A022921 (first differences), A126241.
Cf. A020857 (decimal expansion of log_2(3)).
Cf. A020915.
Cf. A204399 (essentially the same).

a020914 = a070939 . a000244  -- Reinhard Zumkeller, Jun 30 2013

A020914 :=n->nops(convert(3^n,base,2)):
seq(A020914(n),n=0..70); # Emeric Deutsch, Apr 30 2006
seq(ilog2(3^n)+1, n=0 .. 100); # Robert Israel, Dec 12 2014

Table[Length[IntegerDigits[3^n, 2]], {n, 0, 100}] (* Stefan Steinerberger, Apr 19 2006 *)
a[n_] := Floor[ Log2[3^n] + 1]; Array[a, 105, 0] (* Robert G. Wilson v, May 25 2014 *)

for(n=0,100,print1(floor(1+n*log(3)/log(2)),",")) \\ K. Spage, Oct 22 2009

a(n)=exponent(3^n)+1 \\ Charles R Greathouse IV, Nov 03 2022

def A020914(n): return (3**n).bit_length() # Chai Wah Wu, Oct 09 2024

a(n) = floor(1 + n*log(3)/log(2)). - K. Spage, Oct 22 2009
a(0) = 1, a(n+1) = a(n) + A022921(n). - K. Spage, Oct 23 2009
a(n) = A122437(n-1) - n. - K. Spage, Oct 23 2009
A098294(n) = a(n) + n for n > 0. - Mike Winkler, Dec 31 2010
a(n) = A070939(A000244(n)) = length of n-th row in triangle A227048. - Reinhard Zumkeller, Jun 30 2013
a(n) = 1 + floor(n*log_2(3)) = 1 + A056576(n) = 1 + floor(n*A020857). - L. Edson Jeffery, Dec 12 2014
A020915(a(n)) = n + 1. - Reinhard Zumkeller, Mar 28 2015

More terms from Stefan Steinerberger, Apr 19 2006

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

Original entry on oeis.org

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

A080763 Exchange 1's and 2's in the eta-sequence A006337.

Original entry on oeis.org

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, 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

Offset: 1

N. J. A. Sloane, Mar 25 2003

T. D. Noe, Table of n, a(n) for n=1..10000
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)

Different from A022921.

The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A003151 as the parent: A003151, A001951, A001952, A003152, A006337, A080763, A082844 (conjectured), A097509, A159684, A188037, A245219 (conjectured), A276862. - N. J. A. Sloane, Mar 09 2021

Flatten[ Table[ Nest[ Flatten[ # /. {1 -> {2, 2, 1}, 2 -> {2, 1}}] &, {2}, n], {n, 5}]] (* Robert G. Wilson v, May 06 2005 *)

from math import isqrt
def A080763(n): return 3+isqrt(m:=n*n<<1)-isqrt(m+(n<<2)+2) # Chai Wah Wu, Aug 03 2022

Digits := 100: c2 := sqrt(2.): A080763 := n->3-floor((n+1)*c2)+floor(n*c2);

A122437 Allowable values of the "dropping time" of the Collatz (3x+1) iteration.

Original entry on oeis.org

1, 3, 6, 8, 11, 13, 16, 19, 21, 24, 26, 29, 32, 34, 37, 39, 42, 44, 47, 50, 52, 55, 57, 60, 63, 65, 68, 70, 73, 75, 78, 81, 83, 86, 88, 91, 94, 96, 99, 101, 104, 106, 109, 112, 114, 117, 119, 122, 125, 127, 130, 132, 135, 138, 140, 143, 145, 148, 150, 153, 156, 158, 161

Offset: 1

T. D. Noe, Sep 06 2006

Only these numbers appear in A060445, which tabulates the "dropping time" of odd numbers. Note that all even numbers have a "dropping time" of 1.
a(n) is also the number of binary digits of 6^(n-1); for example, a(4)=8 since 6^(4-1)=216 in binary is 11011000, an 8-digit number. - Julio Cesar de la Yncera, Mar 28 2009
A positive integer (x) is an allowable value if and only if (x-1)/(1+log(2)/log(3)) - floor(x/(1+log(2)/log(3))) is not negative. - K. Spage, Oct 22 2009
Here the word "allowable" means that it is necessary for a sequence of iterates starting from odd value m to arrive at a value x = f^{floor(1+n+n*log(3)/log(2))}(m) < m, where n gives the number of odds in such a sequence including m, to have undergone precisely floor(1+n+n*log(3)/log(2)) iterations of f, where f(2*m)=m, f(2*m+1)=6*m+4. However, the formula for a(n+1) does not fully account for the order of odds and evens in such a sequence because it does not account for the effects of the "+1". Thus it is unknown whether it maximizes the value x for all values m. For example, fix m = 1 and the "+1" is enough to give the trivial cycle. So it is possible that for some m we have f^{floor(1+n+n*log(3)/log(2))}(m) >= m. - Jeffrey R. Goodwin, Aug 24 2011
The indices of the powers of 3 in A006899. - Ruud H.G. van Tol, Nov 02 2022

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A022921 (number of 2^m between 3^n and 3^(n+1)), A122442 (least k having dropping time a(n)).

Cf. A006899.

Floor[1+Range[0,100]*(1+Log[2,3])] (* T. D. Noe, Sep 08 2006 *)
Map[Length[RealDigits[ #, 2][[1]]] &, Table[10^i, {i, 0, 50}]] (* Julio Cesar de la Yncera, Mar 28 2009 *)

a(n)=logint(3^(n-1),2)+n \\ Ruud H.G. van Tol, Nov 04 2022

a(1) = 1, a(n+1) = a(n) + A022921(n-1) + 1.
a(n+1) = floor(1 + n + n*log(3)/log(2)). - T. D. Noe, Sep 08 2006
a(n) = floor((1 + log(2)/log(3))*A020914(n-1)). - K. Spage, Oct 22 2009
a(n) = A020914(n-1) + n - 1. - K. Spage, Oct 23 2009 [corrected by Ruud H.G. van Tol, Nov 03 2022]
a(n) = a(n-1)+2 if 3^(n-1) < 2^(a(n-1)+2-(n-1)); a(n) = a(n-1)+3 otherwise. - V. Barbera, Aug 12 2025

Comment corrected and edited by Jon E. Schoenfield, Feb 27 2014

A076227 Number of surviving Collatz residues mod 2^n.

Original entry on oeis.org

1, 1, 1, 2, 3, 4, 8, 13, 19, 38, 64, 128, 226, 367, 734, 1295, 2114, 4228, 7495, 14990, 27328, 46611, 93222, 168807, 286581, 573162, 1037374, 1762293, 3524586, 6385637, 12771274, 23642078, 41347483, 82694966, 151917636, 263841377, 527682754, 967378591, 1934757182, 3611535862

Offset: 0

Labos Elemer, Oct 01 2002

nonn

Number of residue classes in which A074473(m) is not constant.
The ratio of numbers of inhomogenous r-classes versus uniform-classes enumerated here increases with n and tends to 0. For n large enough ratio < a(16)/65536 = 2114/65536 ~ 3.23%.
Theorem: 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 connectedness to A100982, A056576, A022921, A020915. - Mike Winkler, Sep 12 2017
Brown's criterion ensures that the sequence is complete (see formulae). - Vladimir M. Zarubin, Aug 11 2019

			n=6: Modulo 64, eight residue classes were counted: r=7, 15, 27, 31, 39, 47, 59, 63. See A075476-A075483. For other 64-8=56 r-classes u(q)=A074473(64k+q) is constant: in 32 class u(q)=2, in 16 classes u(q)=4, in 4 classes u(q)=7 and in 4 cases u(q)=9. E.g., for r=11, 23, 43, 55 A074473(64k+r)=9 independently of k.
From _Mike Winkler_, Sep 12 2017: (Start)
The next table shows how the theorem works. No entry is equal to zero.
  k =        3  4  5   6   7   8   9  10  11   12 .. | a(n)=
-----------------------------------------------------|
  n =  2  |  1                                       |    1
  n =  3  |  1  1                                    |    2
  n =  4  |     2  1                                 |    3
  n =  5  |        3   1                             |    4
  n =  6  |        3   4   1                         |    8
  n =  7  |            7   5   1                     |   13
  n =  8  |               12   6   1                 |   19
  n =  9  |               12  18   7   1             |   38
  n = 10  |                   30  25   8   1         |   64
  n = 11  |                   30  55  33   9    1    |  128
  :       |                        :   :   :    : .. |   :
-----------------------------------------------------|------
A100982(k) = 2  3  7  12  30  85 173 476 961 2652 .. |
The entries (n,k) in this table are generated by the rule (n+1,k) = (n,k) + (n,k-1). The last value of (n+1,k) is given by n+1 = A056576(k-1), or the highest value in column n is given twice only if A022921(k-2) = 2. Then a(n) is equal to the sum of the entries in row n. For k = 7 there is: 1 = 0 + 1, 5 = 1 + 4, 12 = 5 + 7, 12 = 12 + 0. It is a(9) = 12 + 18 + 7 + 1 = 38. The sum of column k is equal to A100982(k). (End)

Isaac DeJager, Madeleine Naquin, and Frank Seidl, Colored Motzkin Paths of Higher Order, VERUM 2019.
Andreas-Stephan Elsenhans, Numerical verification of the Collatz conjecture for billion digit random numbers, arXiv:2502.16743 [math.NT], 2025.
Tomás Oliveira e Silva, Computational verification of the 3x+1 conjecture.
Eric Weisstein's World of Mathematics, Brown's Criterion
M. Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A006370, A074473, A075476-A075483, A100982, A056576, A022921, A020915, A243115.

/* call as follows: uint64_t s=survives(0,1,1,0,bits); */
uint64_t survives(uint64_t r, uint64_t m, uint64_t lm, int p2, int fp2)
{
    while(!(m&1) && (m>=lm)) {
        if(r&1) { r+=(r+1)>>1; m+=m>>1; }
        else { r>>=1; m>>=1; }
    }
    if(mPhil Carmody, Sep 08 2011 */

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

a(n) = Sum_{k=A020915(n+2)..n+1} (n,k). (Theorem, cf. example) - Mike Winkler, Sep 12 2017
From Vladimir M. Zarubin, Aug 11 2019: (Start)
a(0) = 1, a(1) = 1, and for k > 0,
a(A020914(k)) = 2*a(A020914(k)-1) - A100982(k),
a(A054414(k)) = 2*a(A054414(k)-1). (End)
a(n) = 2^n - 2^n*Sum_{k=0..A156301(n)-1} A186009(k+1)/2^A020914(k). - Benjamin Lombardo, Sep 08 2019

New terms to n=39 by Phil Carmody, Sep 08 2011

cp's OEIS Frontend

A020914 Number of digits in the base-2 representation of 3^n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Formula

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

PARI

PARI

Formula

Extensions

A080763 Exchange 1's and 2's in the eta-sequence A006337.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Python

Formula

A122437 Allowable values of the "dropping time" of the Collatz (3x+1) iteration.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A076227 Number of surviving Collatz residues mod 2^n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C