A008908 -id:A008908 - cp's OEIS frontend

A275968 Smaller of two consecutive primes p and q such that c(p) = c(q), where c(n) = A008908(n) is the length of x, f(x), f(f(x)), ... , 1 in the Collatz conjecture.

173, 409, 419, 421, 439, 487, 521, 557, 571, 617, 761, 887, 919, 1009, 1039, 1117, 1153, 1171, 1217, 1327, 1373, 1549, 1559, 1571, 1657, 1693, 1709, 1721, 1733, 1783, 1831, 1861, 1901, 1993, 1997, 2053, 2089, 2339, 2393, 2521, 2539, 2647, 2657, 2677, 2693, 2777

Offset: 1

Abhiram R Devesh, Aug 15 2016

If x is even f(x) = x/2 else f(x) = 3x + 1.

			a(1) = p = 173; q = 179
c(p) = c(q) = 32

Abhiram R Devesh, Table of n, a(n) for n = 1..10000

Cf. A006577 (Collatz trajectory lengths), A078417, A008908.

t = Table[Length@ NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # != 1 &] - 1, {n, 10^4}]; Prime@ Flatten@ Position[#, k_ /; Length@ k == 1] &@ Map[Union@ Part[t, #] &, #] &@ Partition[#, 2, 1] &@ Prime@ Range@ 410 (* Michael De Vlieger, Sep 01 2016 *)

A008908(n)=my(c=1); while(n>1, n=if(n%2, 3*n+1, n/2); c++); c
t=A008908(p=2); forprime(q=3,1e4, tt=A008908(q); if(t==tt, print1(p", ")); p=q; t=tt) \\ Charles R Greathouse IV, Sep 01 2016

import sympy
def lcs(n):
    a=1
    while n>1:
        if n%2==0:
            n=n//2
        else:
            n=(3*n)+1
        a=a+1
    return(a)
m=2
while m>0:
    n=sympy.nextprime(m)
    if lcs(m)==lcs(n):
        print(m,)
    m=n
# Abhiram R Devesh, Sep 02 2016

A224399 Numbers k such that A224166(k) = A008908(k).

Original entry on oeis.org

1, 2, 4, 8, 13, 16, 26, 32, 35, 52, 64, 70, 81, 93, 104, 128, 140, 162, 181, 186, 208, 241, 256, 280, 324, 362, 372, 416, 455, 482, 483, 512, 543, 560, 607, 643, 645, 648, 724, 744, 809, 815, 832, 903, 910, 914, 915, 964, 966, 967, 1024, 1079, 1081, 1086, 1087

Offset: 1

Michel Lagneau, Apr 05 2013

nonn

Numbers k for which the number of iterations starting with -k to reach the last number of the cycle equals the number of iterations starting with k to reach 1 in Collatz (3x+1) trajectory of +/-k.

			a(6) = 26 because A224166(26) = A008908(26) = 11.
The trajectory of - 26 is :
-26 -> - 13 -> -38 -> -19 -> -56 -> -28 -> -14 -> -7 -> -20 -> -10 -> -5 with 11 iterations (the first value is counted);
The trajectory of 26 is :
26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 with 11 iterations (the first value is counted).

Cf. A008908, A224166.

z:={1}:
  for m from -1 by -1 to -1500 do:
   lst:={m}:a:=0: x:=m: lst:=lst union {x}:
        for i from 1 to 100 do:
         lst:=lst union {x}:
         if irem(abs(x), 2)=1
         then
         x:=3*x+1: lst:=lst union {x}:
         else
         x:=x/2: lst:=lst union {x}:
         fi:
         od:
         n0:=nops(lst):
         if lst intersect z = {1}
         then
         n1:=n0-2:
         else
         n1:=n0-1:
      fi:
       a:=0:y:=-m:
          for it from 1 to 100 while (y>1) do:
             if irem(y,2)=0
             then
             y := y/2:a:=a+1:
             else
              y := 3*y+1: a := a+1:
             fi:
          od:
          if n1=a
          then
          printf(`%d, `,-m):
          else
          fi:
od:

A247717 Numbers n such that A033493(n)/A008908(n) is an integer.

Original entry on oeis.org

1, 5, 18, 58, 99, 153, 176, 228, 238, 240, 282, 341, 345, 421, 475, 585, 629, 712, 739, 779, 802, 815, 974, 995, 1078, 1088, 1237, 1257, 1262, 1290, 1346, 1398, 1424, 1459, 1673, 1694, 1724, 1731, 1802, 1811, 1916, 1928, 1988, 2170, 2222, 2260, 2272, 2275, 2317, 2365, 2397, 2410

Offset: 1

Derek Orr, Sep 22 2014

nonn

Equivalently, these are also numbers n such that the average of the numbers in the Collatz (3x+1) iteration of n is an integer.

Cf. A008908, A033493.

Tavg(n)=c=0;s=n;while(n!=1,if(n==Mod(0,2),n=n/2;c++;s+=n);if(n==Mod(1,2)&&n!=1,n=3*n+1;s+=n;c++));s/(c+1)
n=1;while(n<10^4,if(floor(Tavg(n))==Tavg(n),print1(n,", "));n++)

A260801 Primes p such that A008908(p) is also prime.

Original entry on oeis.org

2, 7, 17, 29, 31, 71, 89, 107, 113, 127, 131, 157, 181, 223, 239, 263, 271, 277, 281, 283, 313, 337, 379, 409, 419, 421, 431, 503, 547, 571, 577, 691, 701, 727, 757, 809, 821, 857, 883, 947, 953, 971, 1031, 1109, 1129, 1153, 1163, 1231, 1283, 1291, 1327, 1361, 1447, 1487, 1531, 1559, 1567, 1583

Offset: 1

Ivan N. Ianakiev, Jul 31 2015

			7 is prime and A008908(7) is 17, which is also prime. Therefore, 7 is in the sequence.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A008908.

collatz[n_]:=If[EvenQ[n],n/2,3*n+1];
seq[n_]:=Length[NestWhileList[collatz,n,#!= 1&]]
Select[Flatten[Position[seq/@Range[7!],_?PrimeQ]],PrimeQ]

T(n)=c=0;while(n!=1,if(n%2,n=3*n+1;c++);if(!(n%2),n=n/2;c++));c+1
forprime(p=1,10^3,if(isprime(T(p)),print1(p,", "))) \\ Derek Orr, Aug 27 2015

A339614 Inputs n that yield a record-breaking value of A008908(n)/(log_2(n)+1) for the Collatz conjecture.

Original entry on oeis.org

1, 3, 7, 9, 27, 26623, 35655, 52527, 142587, 156159, 230631, 626331, 837799, 1723519, 3542887, 3732423, 5649499, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231

Offset: 1

Matthew Russell Downey, Dec 10 2020

The metric A008908(n)/(log_2(n)+1) is always equal to 1 for any power of 2 (where 1 is the smallest possible value).

			a(1) = 1, which is trivial, because the first element in any sequence is record setting.
a(5) = 27, because A008908(n)/(log_2(n)+1) yields a maximum value at n=27 among the first 27 elements, and there are 4 record-breaking elements beforehand.

Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A008908, A006877, A033492.

import math
oeis_A006877 = [1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, 10971, 13255, 17647, 23529, 26623, 34239, 35655, 52527, 77031, 10623, 142587, 156159, 216367, 230631, 410011, 511935, 626331, 837799, 1117065, 1501353, 1723519, 2298025, 3064033, 3542887, 3732423, 5649499, 6649279, 8400511, 11200681, 14934241, 15733191, 31466382, 36791535, 63728127]
def stopping_time(n):
    time = 1
    while n>1:
        n = 3*n + 1 if n & 1 else n//2
        time += 1
    return time
def stopping_time_metric(n):
    time = stopping_time(n)
    logarithmic_distance = (math.log(n, 2)+1)
    return float(time/logarithmic_distance)
result = []
record_input = oeis_A006877[0]
record_stopping_time_metric = stopping_time_metric(record_input)
result.append(record_input)
for n in range(1, len(oeis_A006877)):
    current_input = oeis_A006877[n]
    current_stopping_time_metric = stopping_time_metric(current_input)
    if current_stopping_time_metric > record_stopping_time_metric:
        record_input = current_input
        record_stopping_time_metric = current_stopping_time_metric
        result.append(record_input)
for n in range(len(result)):
    print(result[n], end=", ")

A006577 Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.

Original entry on oeis.org

0, 1, 7, 2, 5, 8, 16, 3, 19, 6, 14, 9, 9, 17, 17, 4, 12, 20, 20, 7, 7, 15, 15, 10, 23, 10, 111, 18, 18, 18, 106, 5, 26, 13, 13, 21, 21, 21, 34, 8, 109, 8, 29, 16, 16, 16, 104, 11, 24, 24, 24, 11, 11, 112, 112, 19, 32, 19, 32, 19, 19, 107, 107, 6, 27, 27, 27, 14, 14, 14, 102, 22

Offset: 1

N. J. A. Sloane, Bill Gosper

The 3x+1 or Collatz problem is as follows: start with any number n. If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach 1? This is a famous unsolved problem. It is conjectured that the answer is yes.
It seems that about half of the terms satisfy a(i) = a(i+1). For example, up to 10000000, 4964705 terms satisfy this condition.
n is an element of row a(n) in triangle A127824. - Reinhard Zumkeller, Oct 03 2012
The number of terms that satisfy a(i) = a(i+1) for i being a power of ten from 10^1 through 10^10 are: 0, 31, 365, 4161, 45022, 477245, 4964705, 51242281, 526051204, 5378743993. - John Mason, Mar 02 2018
5 seems to be the only number whose value matches its total number of steps (checked to n <= 10^9). - Peter Woodward, Feb 15 2021

			a(5)=5 because the trajectory of 5 is (5,16,8,4,2,1).

R. K. Guy, Unsolved Problems in Number Theory, E16.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
David Eisenbud and Brady Haran, UNCRACKABLE? The Collatz Conjecture, Numberphile video, 2016.
Geometry.net, Links on Collatz Problem
Christian Hercher, There are no Collatz m-Cycles with m <= 91, J. Int. Seq. (2023) Vol. 26, Article 23.3.5.
Jason Holt, Log-log plot of first billion terms
Jason Holt, Plot of 1 billion values of the number of steps to drop below n (A060445), log scale on x axis
Jason Holt, Plot of 10 billion values of the number of steps to drop below n (A060445), log scale on x axis
A. Krowne, Collatz problem, PlanetMath.org.
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
J. C. Lagarias, How random are 3x+1 function iterates?, in The Mathemagician and the Pied Puzzler - A Collection in Tribute to Martin Gardner, Ed. E. R. Berlekamp and T. Rogers, A. K. Peters, 1999, pp. 253-266.
J. C. Lagarias, The 3x+1 Problem: an annotated bibliography, II (2000-2009), arXiv:0608208 [math.NT], 2006-2012.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
M. Le Brun, Email to N. J. A. Sloane, Jul 1991
Mathematical BBS, Biblography on Collatz Sequence
P. Picart, Algorithme de Collatz et conjecture de Syracuse
E. Roosendaal, On the 3x+1 problem
J. L. Simons, On the nonexistence of 2-cycles for the 3x+1 problem, Math. Comp. 75 (2005), 1565-1572.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 8.
G. Villemin's Almanach of Numbers, Cycle of Syracuse
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz Conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

See A070165 for triangle giving trajectories of n = 1, 2, 3, ....
Cf. A006370, A125731, A127885, A127886, A008908, A112695, A135282, A208981, A025586.
See also A008884, A161021, A161022, A161023.

import Data.List (findIndex)
import Data.Maybe (fromJust)
a006577 n = fromJust $ findIndex (n `elem`) a127824_tabf
-- Reinhard Zumkeller, Oct 04 2012, Aug 30 2012

A006577 := proc(n)
        local a,traj ;
        a := 0 ;
        traj := n ;
        while traj > 1 do
                if type(traj,'even') then
                        traj := traj/2 ;
                else
                        traj := 3*traj+1 ;
                end if;
                a := a+1 ;
        end do:
        return a;
end proc: # R. J. Mathar, Jul 08 2012

f[n_] := Module[{a=n,k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Table[f[n],{n,4!}] (* Vladimir Joseph Stephan Orlovsky, Jan 08 2011 *)
Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1,{n,80}] (* Harvey P. Dale, May 21 2012 *)

a(n)=if(n<0,0,s=n; c=0; while(s>1,s=if(s%2,3*s+1,s/2); c++); c)

step(n)=if(n%2,3*n+1,n/2);
A006577(n)=if(n==1,0,A006577(step(n))+1); \\ Michael B. Porter, Jun 05 2010

def a(n):
    if n==1: return 0
    x=0
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        x+=1
        if n<2: break
    return x
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 05 2017

def A006577(n):
    ct = 0
    while n != 1: n = A006370(n); ct += 1
    return ct # Ya-Ping Lu, Feb 22 2024

collatz<-function(n) ifelse(n==1,0,1+ifelse(n%%2==0,collatz(n/2),collatz(3*n+1))); sapply(1:72, collatz) # Christian N. K. Anderson, Oct 09 2024

a(n) = A006666(n) + A006667(n).
a(n) = A112695(n) + 2 for n > 2. - Reinhard Zumkeller, Apr 18 2008
a(n) = A008908(n) - 1. - L. Edson Jeffery, Jul 21 2014
a(n) = A135282(n) + A208981(n) (after Alonso del Arte's comment in A208981), if 1 is reached, otherwise a(n) = -1. - Omar E. Pol, Apr 10 2022
a(n) = 2*A007814(n + 1) + a(A085062(n)) + 1 for n > 1. - Wing-Yin Tang, Jan 06 2025

More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001

"Escape clause" added to definition by N. J. A. Sloane, Jun 06 2017

A070165 Irregular triangle read by rows giving trajectory of n in Collatz problem.

Original entry on oeis.org

1, 2, 1, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 5, 16, 8, 4, 2, 1, 6, 3, 10, 5, 16, 8, 4, 2, 1, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 8, 4, 2, 1, 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 10, 5, 16, 8, 4, 2, 1, 11, 34, 17, 52, 26, 13

Offset: 1

Eric W. Weisstein, Apr 23 2002

n-th row has A008908(n) entries (unless some n never reaches 1, in which case the triangle ends with an infinite row). [Escape clause added by N. J. A. Sloane, Jun 06 2017]
A216059(n) is the smallest number not occurring in n-th row; see also A216022.
Comment on the mp3 file from Gordon Charlton (the recording artist Beat Frequency). The piece uses the first 3242 terms (i.e. the first 100 hailstone sequences), with pitch modulus 36, duration modulus 2. Its musicality stems from the many repetitions and symmetries within the sequence, and in particular the infrequency of multiples of 3. This means that when the pitch modulus is a multiple of 12 the notes are predominantly in the symmetric octatonic scale, known to modern classical composers as the second of Messiaen's modes of limited transposition, and to jazz musicians as half-whole diminished. - N. J. A. Sloane, Jan 30 2019

			The irregular array a(n,k) starts:
n\k   0  1  2  3  4   5  6   7  8  9 10 11 12 13 14 15 16 17 18 19
1:    1
2:    2  1
3:    3 10  5 16  8   4  2   1
4:    4  2  1
5:    5 16  8  4  2   1
6:    6  3 10  5 16   8  4   2  1
7:    7 22 11 34 17  52 26  13 40 20 10  5 16  8  4  2  1
8:    8  4  2  1
9:    9 28 14  7 22  11 34  17 52 26 13 40 20 10  5 16  8  4  2  1
10:  10  5 16  8  4   2  1
11:  11 34 17 52 26  13 40  20 10  5 16  8  4  2  1
12:  12  6  3 10  5  16  8   4  2  1
13:  13 40 20 10  5  16  8   4  2  1
14:  14  7 22 11 34  17 52  26 13 40 20 10  5 16  8  4  2  1
15:  15 46 23 70 35 106 53 160 80 40 20 10  5 16  8  4  2  1
... Reformatted and extended by _Wolfdieter Lang_, Mar 20 2014

T. D. Noe, Rows n = 1..100 of triangle, flattened
Gordon Charlton ("Beat Frequency"), Hailstone Trajectory (mp3 file)
David Eisenbud and Brady Haran, UNCRACKABLE? The Collatz Conjecture, Numberphile Video, 2016.
David Rabahy, Hailstone Sequence presented as a spreadsheet
Anatoly E. Voevudko, File of first 10K Collatz sequences
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006370 (step), A008908 (row lengths), A033493 (row sums).
Cf. A220237 (sorted rows), A347270 (array), A192719.
Cf. A070168 (Terras triangle), A256598 (reduced triangle).
Cf. A254311, A257480 (and crossrefs therein).
Cf. A280408 (primes).

a070165 n k = a070165_tabf !! (n-1) !! (k-1)
a070165_tabf = map a070165_row [1..]
a070165_row n = (takeWhile (/= 1) $ iterate a006370 n) ++ [1]
a070165_list = concat a070165_tabf
-- Reinhard Zumkeller, Oct 07 2011

T:= proc(n) option remember; `if`(n=1, 1,
      [n, T(`if`(n::even, n/2, 3*n+1))][])
    end:
seq(T(n), n=1..15);  # Alois P. Heinz, Jan 29 2021

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Flatten[Table[Collatz[n], {n, 10}]] (* T. D. Noe, Dec 03 2012 *)

row(n, lim=0)={if (n==1, return([1])); my(c=n, e=0, L=List(n)); if(lim==0, e=1; lim=n*10^6); for(i=1, lim, if(c%2==0, c=c/2, c=3*c+1); listput(L, c); if(e&&c==1, break)); return(Vec(L)); } \\ Anatoly E. Voevudko, Mar 26 2016; edited by Michel Marcus, Aug 10 2021

def a(n):
    if n==1: return [1]
    l=[n, ]
    while True:
        if n%2==0: n/=2
        else: n = 3*n + 1
        if n not in l:
            l+=[n, ]
            if n<2: break
        else: break
    return l
for n in range(1, 101): print(a(n)) # Indranil Ghosh, Apr 14 2017

T(n,k) = T^{(k)}(n) with the k-th iterate of the Collatz map T with T(n) = 3*n+1 if n is odd and T(n) = n/2 if n is even, n >= 1. T^{(0)}(n) = n. k = 0, 1, ..., A008908(n) - 1. - Wolfdieter Lang, Mar 20 2014

Name specified and row length A-number corrected by Wolfdieter Lang, Mar 20 2014

A033496 Numbers m that are the largest number in their Collatz (3x+1) trajectory.

Original entry on oeis.org

1, 2, 4, 8, 16, 20, 24, 32, 40, 48, 52, 56, 64, 68, 72, 80, 84, 88, 96, 100, 104, 112, 116, 128, 132, 136, 144, 148, 152, 160, 168, 176, 180, 184, 192, 196, 200, 208, 212, 224, 228, 232, 240, 244, 256, 260, 264, 272, 276, 280, 288, 296, 304, 308, 312, 320, 324

Offset: 1

Jeff Burch

Or, possible peak values in 3x+1 trajectories: 1,2 and m=16k+4,16k+8,16k but not for all k; those 4k numbers [like m=16k+12 and others] which cannot be such peaks are listed in A087252.

Possible values of A025586(m) in increasing order. See A275109 (number of times each value of a(n) occurs in A025586). - Jaroslav Krizek, Jul 17 2016

			These peak values occur in 1, 3, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 21, 22, 27, 30, 39, 44, 71, 75, 1579 [3x+1]-iteration trajectories started with different initial values. This list most probably is incomplete.
From _Hartmut F. W. Hoft_, Jun 24 2016: (Start)
Let n be the maximum in some Collatz trajectory and let F(n), the initial fan of n, be the set of all initial values less than or equal to n whose Collatz trajectories lead to n as their maximum. Then the size of F(n) never equals 2, 4, 5, 7 or 10 (see the link).
Conjecture: Every number k > 10 occurs as the size of F(n) for some n.
Fans F(n) of size k, for all 10 < k < 355, exist for 4 <= n <= 50,000,000. The largest fan in this range, F(41163712), has size 7450.
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 2000 terms from T. D. Noe)
Hartmut F. W. Hoft, initial Collatz fans
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A025586, A087251-A087256, A015999, A275109.
Cf. A006370, A008908, A070165.
Cf. A095384 (contains a definition of Collatz[]).
Cf. A105730, A233293.

a033496 n = a033496_list !! (n-1)
a033496_list = 1 : filter f [2, 4 ..] where
   f x = x == maximum (takeWhile (/= 1) $ iterate a006370 x)
-- Reinhard Zumkeller, Oct 22 2015

Set(Sort([Max([k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]): n in [1..2^10] | Max([k eq 1 select n else IsOdd(Self(k-1)) and not IsOne(Self(k-1)) select 3*Self(k-1)+1 else Self(k-1) div 2: k in [1..5*n]]) le 2^10])) // Jaroslav Krizek, Jul 17 2016

Collatz[a0_Integer, maxits_:1000] := NestWhileList[If[EvenQ[ # ], #/2, 3# + 1] &, a0, Unequal[ #, 1, -1, -10, -34] &, 1, maxits]; (* Collatz[n] function definition by Eric Weisstein *)
Select[Range[324], Max[Collatz[#]] == # &] (* T. D. Noe, Feb 28 2013 *)

def a(n):
    if n<2: return [1]
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        if n not in l:
            l.append(n)
            if n<2: break
        else: break
    return l
print([n for n in range(1, 501) if max(a(n)) == n]) # Indranil Ghosh, Apr 14 2017

A008908(a(n)) = A159999(a(n)). - Reinhard Zumkeller, May 04 2009

Max(A070165(a(n),k): k=1..A008908(a(n))) = A070165(a(n),1) = a(n). - Reinhard Zumkeller, Oct 22 2015

A089357 a(n) = 2^(6*n).

Original entry on oeis.org

1, 64, 4096, 262144, 16777216, 1073741824, 68719476736, 4398046511104, 281474976710656, 18014398509481984, 1152921504606846976, 73786976294838206464, 4722366482869645213696, 302231454903657293676544, 19342813113834066795298816

Offset: 0

Douglas Winston (douglas.winston(AT)srupc.com), Dec 26 2003

For n > 0, numbers M such that a(n) is the highest power of 2 in the Collatz (3x+1) iteration are given by 2^k*(a(n)-1)/3 for any k >= 0. Example: For n = 1, the numbers such that 64 is the highest power of 2 in the Collatz (3x+1) iteration are given by 2^k*(64-1)/3 = 21*2^k for any k >= 0. See A008908 for more information on the Collatz (3x+1) iteration. - Derek Orr, Sep 22 2014

Powers of 64. - Alexander Fraebel, Aug 29 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (64).

Cf. A000079, A000302, A001018, A001025, A008588, A008908, A009976.

[2^(6*n): n in [0..20]]; // Vincenzo Librandi, Jun 07 2011

seq(2^(6*n), n=0..14); # Nathaniel Johnston, Jun 26 2011

2^(6 Range[0,20]) (* or *) NestList[64#&,1,20] (* Harvey P. Dale, Sep 28 2011 *)

makelist(2^(6*n),n,0,20); /* Martin Ettl, Nov 12 2012 */

a(n)=64^n \\ Charles R Greathouse IV, Jul 02 2016

G.f.: 1/(1-64*x). - Philippe Deléham, Nov 24 2008
a(n) = 63*a(n-1) + 64^(n-1), a(0)=1. - Vincenzo Librandi, Jun 07 2011
E.g.f.: exp(64*x). - Ilya Gutkovskiy, Jul 02 2016
a(n) = A000079(A008588(n)). - Wesley Ivan Hurt, Jul 02 2016
a(n) = 64*a(n-1). - Miquel Cerda, Oct 27 2016

cp's OEIS Frontend

A186701 Partial sums of Collatz sequence (A008908).

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A275968 Smaller of two consecutive primes p and q such that c(p) = c(q), where c(n) = A008908(n) is the length of x, f(x), f(f(x)), ... , 1 in the Collatz conjecture.

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A224399 Numbers k such that A224166(k) = A008908(k).

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A247717 Numbers n such that A033493(n)/A008908(n) is an integer.

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A260801 Primes p such that A008908(p) is also prime.

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A339614 Inputs n that yield a record-breaking value of A008908(n)/(log_2(n)+1) for the Collatz conjecture.

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A006577 Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.

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A070165 Irregular triangle read by rows giving trajectory of n in Collatz problem.

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