A006666 -id:A006666 - cp's OEIS frontend

A211981 Numbers n such that floor(2^A006666(n)/3^A006667(n)) = n.

1, 2, 3, 4, 5, 8, 10, 16, 21, 32, 42, 64, 75, 85, 113, 128, 151, 170, 227, 256, 341, 512, 682, 1024, 1365, 2048, 2730, 4096, 5461, 7281, 8192, 10922, 14563, 16384, 21845, 32768, 43690, 65536, 87381, 131072, 174762, 262144, 349525, 466033, 524288, 699050, 932067

Offset: 1

Michel Lagneau, Feb 13 2013

nonn

A006666 and A006667 give the number of halving and tripling steps to reach 1 in 3x+1 problem.
Properties of this sequence:
A006667(a(n)) <= 3, and if a(n) is even then a(n)/2 is in the sequence.
The sequence A000079(n) (power of 2) is included in this sequence.
{a(n)} = E1 union E2 where E1 = {A000079(n)} union {5, 10, 21, 85, 170, 227, 341, 682, 1365, 2730, 5461, ...} and E2 = {75, 113, 151, 7281, ...}. If an element k of E1 generates the Collatz sequence of iterates k -> T_1(k) -> T_2(k) -> T_3(k) -> ... then any T_i(k) is an element of  E1 of the form [2^a /3^b] where a = A006666(n), or A006666(n)-1, or ... and b = A006667(n), or A006667(n)-1, or ... But if k is an element of E2, there exists at least an element T_i(k) that is not in the sequence a(n). For example 75 -> 226 ->113 -> 340 -> ... and 226 is not in the sequence because, if [x] = [2^a /3^b] = [ x. x0 x1 x2 ...], the rational number 0.x0 x1 x2 ... > 0.666666.... => [2^a /3^(b-1)] of the form [(3x+2).y0 y1 y2 ...], and this integer is different from T_(i+1)(k) = [(3x+1).y0 y1 y2 ...] = 3x+1.
Example: T_2(75) = floor(2^10 /3^2) = 113 => floor(2^10/3^1) = 341 instead T_3(75) = 340.

			227 is in the sequence because A006666(227) = 11, A006667(227) = 2 => floor(2^11/3^2) = 227.
The Collatz trajectory of 227 is 227 -> 682 -> 341 -> 1024 -> 512 -> ... -> 2 -> 1, and 227 is in the subset E1 implies the following Collatz iterates:
227 = floor(2^11/3^2);
682 = floor(2^11/3^1);
341 = floor(2^10/3^1);
1024 = floor(2^10/3^0);
512 = floor(2^9/3^0);
256 = floor(2^8/3^0);
128 = floor(2^7/3^0);
...
2 = floor(2^1/3^0);
1 = floor(2^0/3^0);
With the numbers of E1, we obtain another formulation of the Collatz problem.

Cf. A006666, A006667.

A:= proc(n) if type(n, 'even') then n/2; else 3*n+1 ; end if; end proc:
B:= proc(n) a := 0 ; x := n ; while x > 1 do x := A(x) ; a := a+1 ; end do; a ; end proc:
C:= proc(n) a := 0 ; x := n ; while x > 1 do if type(x, 'even') then x := x/2 ; else x := 3*x+1 ; a := a+1 ; end if; end do; a ; end proc:
D:= proc(n) C(n) ; end proc:
A006666:= proc(n) B(n)- C(n) ; end:
A006667:= proc(n) C(n)- D(n) ; end:
G:= proc(n) floor(2^ A006666 (n)/3^ A006667 (n)) ; end:
for i from 1 to 1000000 do: if G(i) =i then printf(`%d, `,i):else fi:od:

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 30; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; If[Floor[2^ev/3^od] == n, AppendTo[t, n]]]; t (* T. D. Noe, Feb 13 2013 *)

A225089 a(n) = floor(2^A006666(m)/3^A006667(m)) - m, where m = 2n + 1.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 0, 2, 0, 1, 2, 1, 2, 1, 0, 1, 4, 3, 1, 0, 3, 2, 4, 4, 2, 5, 5, 4, 3, 5, 0, 6, 3, 2, 8, 7, 6, 8, 2, 7, 0, 10, 6, 5, 4, 7, 8, 10, 9, 8, 4, 3, 10, 9, 11, 14, 9, 12, 7, 6, 10, 9, 0, 14, 13, 12, 7, 6, 5, 10, 17, 13, 15, 0, 13, 12, 16, 15, 5, 8

Offset: 1

Michel Lagneau, Apr 27 2013

nonn

A006666 and A006667 are the number of halving and tripling steps to reach 1 in 3x+1 problem.
Properties of this sequence:
a(m) = 0 for m =  A211981(m).

			a(9) = 3 because floor(2^A006666(19)/3^A006667(19)) - 19 = floor(2^14 /3^6) - 19 = floor(22.474622) - 19 = 22 - 19 = 3.

Michel Lagneau, Table of n, a(n) for n = 1..10000

Cf. A006666, A006667, A211981, A075680.

A:= proc(n) if type(n, 'even') then n/2; else 3*n+1 ; end if; end proc:
B:= proc(n) a := 0 ; x := n ; while x > 1 do x := A(x) ; a := a+1 ; end do; a ; end proc:
C:= proc(n) a := 0 ; x := n ; while x > 1 do if type(x, 'even') then x := x/2 ; else x := 3*x+1 ; a := a+1 ; end if; end do; a ; end proc:
D:= proc(n) C(n) ; end proc:
A006666:= proc(n) B(n)- C(n) ; end:
A006667:= proc(n) C(n)- D(n) ; end:
G:= proc(n) floor(2^A006666 (n)/3^A006667 (n)) ; end:
for i from 1 to 100 do: printf(`%d, `, G(i)-i):od:

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; nn = 100; t = {}; n = 0; While[Length[t] < nn, n++; c = Collatz[n]; ev = Length[Select[c, EvenQ]]; od = Length[c] - ev - 1; AppendTo[t, Floor[2^ev/3^od]-n]]; t

A076182 a(n) = A006666(n) mod 2.

Original entry on oeis.org

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

Offset: 1

Benoit Cloitre, Nov 01 2002

nonn

Cf. A014682, A006513, A006666, A076092.

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

(define (A076182 n) (modulo (A006666 n) 2))
(definec (A006666 n) (if (= 1 n) 0 (+ 1 (A006666 (A014682 n))))) ;; With memoization-macro definec
(define (A014682 n) (if (even? n) (/ n 2) (/ (+ n n n 1) 2)))
;; Antti Karttunen, Aug 13 2017

a(n) = A006513(n) - 1.

Apparently also the antiparity of A064433. - Ralf Stephan, Nov 16 2004

A304715 For any n > 0, if A006666(n) >= 0, then a(n) = Sum_{i = 0..A006666(n)-1} 2^i * [T^i(n) == 0 (mod 2)] (where [] is an Iverson bracket and T^i denotes the i-th iterate of the Collatz function A014682); otherwise a(n) = -1.

Original entry on oeis.org

0, 1, 28, 3, 14, 57, 1896, 7, 7586, 29, 948, 115, 118, 3793, 3824, 15, 474, 15173, 15180, 59, 62, 1897, 1912, 231, 60722, 237, 1102691417057682138372, 7587, 7590, 7649, 137836427132210267296, 31, 242890, 949, 956, 30347, 30350, 30361, 7772616, 119

Offset: 1

Rémy Sigrist, May 17 2018

In other words, when a(n) >= 0, the binary representation of a(n) encodes the tripling and halvings steps of the Collatz compressed trajectory of n up to the first occurrence of the number 1 (where zeros and ones respectively denote tripling and halving steps).

			The first terms, alongside the binary representation of a(n) and the Collatz compressed trajectory of a(n) up to the first 1 in reverse order, are:
  n    a(n)       bin(a(n))  rev(traj(n))
  --   ----       ---------  ------------
   1      0               0  (1)
   2      1               1  (1, 2)
   3     28           11100  (1, 2, 4, 8, 5, 3)
   4      3              11  (1, 2, 4)
   5     14            1110  (1, 2, 4, 8, 5)
   6     57          111001  (1, 2, 4, 8, 5, 3, 6)
   7   1896     11101101000  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7)
   8      7             111  (1, 2, 4, 8)
   9   7586   1110110100010  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7, 14, 9)
  10     29           11101  (1, 2, 4, 8, 5, 10)
  11    948      1110110100  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11)
  12    115         1110011  (1, 2, 4, 8, 5, 3, 6, 12)
  13    118         1110110  (1, 2, 4, 8, 5, 10, 20, 13)
  14   3793    111011010001  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7, 14)
  15   3824    111011110000  (1, 2, 4, 8, 5, 10, 20, 40, 80, 53, 35, 23, 15)
  16     15            1111  (1, 2, 4, 8, 16)
  17    474       111011010  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17)
  18  15173  11101101000101  (1, 2, 4, 8, 5, 10, 20, 13, 26, 17, 11, 7, 14, 9, 18)

Cf. A000120, A006666, A014682, A029837, A220071, A248573.

a(n) = my (v=0); for (k=0, oo, if (n==1, return (v), n%2, n = (3*n+1)/2, n = n/2; v += 2^k))

a(2^k) = 2^k - 1 for any k >= 0.
a(2*n) = 2*a(n) + 1.
A029837(a(n)+1) = A006666(n).
A000120(a(n)) = A220071(n).
a(A248573(n)) < a(A248573(n+1)) for any n >= 0. - Rémy Sigrist, Nov 09 2018

A232462 Number of iterations of the map n -> f(f(f(...f(n)...))) to reach the end of the cycle, where f(n) = A006666(n), the initial number n is not counted.

Original entry on oeis.org

0, 1, 4, 2, 3, 0, 6, 5, 8, 4, 5, 7, 7, 8, 8, 3, 9, 9, 9, 1, 1, 6, 6, 6, 4, 6, 7, 8, 8, 8, 11, 4, 10, 5, 5, 9, 9, 9, 7, 7, 7, 7, 2, 8, 8, 8, 11, 9, 10, 10, 10, 9, 9, 12, 12, 9, 7, 9, 7, 9, 9, 7, 7, 1, 10, 10, 10, 6, 6, 6, 11, 4, 0, 4, 6, 4, 4, 7, 7, 6, 4, 7, 7, 6, 6, 2, 2, 8, 2, 8, 8, 8, 8, 11, 11, 5, 7, 10, 10, 10, 10, 10, 10, 5, 7, 5, 2, 5, 5, 5, 9, 9, 5, 7, 7, 9, 9, 7, 7, 9

Offset: 1

Christoph Neubauer, Nov 24 2013

nonn

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. As multiplying by 3 and adding 1 always gives an even number, the next step is always a division by 2. So there already exists a shortened version of Collatz problem: If n is even, multiply it by three, add one and divide the result by two, else divide it by two. A006666(n), which is the Number of halving steps to reach 1 in '3x+1' problem, also is the number of all steps for the shortened form to reach 1. If you map this function onto itself, you'll get a sequence that ends with 1 or 6 (for 6, 20, 21, 43, 64, 86, ...) or 73 (for 73, 216, 218, ...).

			For n = 3, the mapping of the shortened Collatz sequence is [3, 5, 4, 2, 1], its number of steps is 4. For n=6 the sequence is [6], the number of steps is 0.

Christoph Neubauer, Table of n, a(n) for n = 1..10000

#!/usr/bin/env python
# output to
outputStr = "b232462.txt"
# max index
maxIndex = 10000
# goodies
sep=" "
eol="\n"
# subroutine for collatz sequence; classical and shortened
def collatz (n, k):
    # put starting number to list
    list = [n]
    # collatz assumption used as end criterion
    while (n > 1):
        # collatz formula
        if (n%2 == 0):
            n = n // 2
        else:
            n = (3*n + 1) // k  # if k==2 --> shortened
        # put new number to list
        list.append(n);
    # return complete list
    return list
# subroutine for collatz composition; classical and shortened
# composition:  collatz (collatz (collatz (...)))
def composition (n, k):
    # put starting number to list
    list = [n]
    # calculate collatz(starting number)
    l = len (collatz(n, k))-1
    # while we do not have a cycle
    while ((l >= 1) and (not l in list)):
        # put new number to list
        list.append(l)
        # get next collatz number
        l = len (collatz(l, k))-1
    # return complete list
    return list
# open output
output = open (outputStr, 'w')
# for index 1 to maxIndex
for i in range (1, maxIndex+1):
    # compute complete composition sequence
    list = composition(i, 2)
    # get number of steps
    l = len(list)-1
    # write to file
    output.write (str(i)+sep+str(l)+eol)
# close output
output.close ()

A265099 Least k such that floor(2^A006666(k)/3^A006667(k)) - k = n.

Original entry on oeis.org

1, 6, 9, 19, 18, 27, 33, 37, 36, 50, 43, 56, 59, 66, 57, 74, 78, 72, 97, 87, 86, 98, 112, 119, 118, 134, 123, 115, 114, 130, 149, 148, 157, 135, 179, 144, 153, 187, 220, 174, 173, 172, 197, 196, 255, 224, 238, 219, 236, 203, 249, 268, 247, 246, 230, 229, 228

Offset: 0

Michel Lagneau, Dec 01 2015

nonn

A006666 and A006667 are the number of halving and tripling steps to reach 1 in 3x+1 problem.
Conjecture: k exists for all n.
In other words, given an integer n, there always exists at least an integer k and a pair of integers (a, b) such that n + k = 2^a/3^b where a is the number of halving steps to reach 1, and b is the number of tripling steps to reach 1, in the 3x+1 problem.

			a(0) = 1 because A006666(1) = 0 and A006667(1) = 0 => floor(2^0/3^0) - 1 = 1 - 1 = 0;
a(1) = 6 because A006666(6) = 6 and A006667(6) = 2 => floor(2^6/3^2) - 6 = floor(64/9) - 6 = 7 - 6 = 1.

Cf. A006666, A006667, A075680, A211981, A225089.

lst={};Do[Collatz[k_]:=NestWhileList[If[EvenQ[#],#/2,3 #+1]&,k,#>1&];nn=500;t={};k=0;While[Length[t]

A277367 a(n) = gcd(A006666(n), A006667(n)) where A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

Original entry on oeis.org

0, 1, 1, 2, 1, 2, 1, 3, 1, 1, 2, 1, 1, 1, 1, 4, 3, 2, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 5, 2, 1, 1, 3, 3, 3, 1, 1, 1, 1, 1, 4, 4, 4, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 2, 10, 10, 1, 1, 3, 3

Offset: 1

Michel Lagneau, Oct 11 2016

nonn

			a(17) = 3 because gcd(A006666(17), A006667(17)) = gcd(9, 3) = 3.

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

Cf. A006666, A006667, A277068.

nn:=100:
for n from 1 to nn do:
m:=n:it0:=0:it1:=0:
   for j from 1 to 1000 while(m<>1) do:
    if irem(m,2)=0
     then
     m:=m/2:it0:=it0+1:
     else
     m:=3*m+1:it1:=it1+1:
    fi:
   od:
   q:=gcd(it0,it1):printf(`%d, `,q):
  od:

Table[GCD[Count[NestWhileList[If[OddQ@ #, 3 # + 1, #/2] &, n, # > 1 &], ?EvenQ], Count[Differences[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # > 1 &]], ?Positive]], {n, 87}] (* Michael De Vlieger, Oct 13 2016, after Harvey P. Dale at A006666 and A006667 *)

a(n) = {my(se = 0); my(so = 0); while (n!=1, if (n % 2, so++; n = 3*n+1, se++; n = n/2);); gcd(se, so);} \\ Michel Marcus, Oct 13 2016

a(2^m) = m.

A281938 a(n) is the least k such that gcd(A006666(k), A006667(k)) = n.

Original entry on oeis.org

2, 4, 8, 16, 32, 64, 128, 256, 512, 82, 129, 4096, 327, 16384, 32768, 1249, 35655, 159, 4926, 283, 377, 502, 603, 799, 1063, 1417, 1889, 2518, 3356, 4472, 5960, 7944, 10594, 14124, 18833, 25110, 33481, 44641, 59521, 79361, 105814, 141084, 188113, 250817, 334422

Offset: 1

Michel Lagneau, Feb 02 2017

nonn

A006666: Number of halving steps to reach 1 in '3x+1' problem.
A006667: number of tripling steps to reach 1 in '3x+1' problem.
a(n) = 2^n for n = 1, 2, 3, 3, 4, 5, 6, 7, 8, 9, 12, 14, 15.
The primes in the sequence are 2, 283, 1063, 1249, 1889, 44641, ...

			a(10) = 82 because gcd(A006666(82), A006667(82)) = gcd(70, 40) = 10, and there is no k < 82 such that gcd(A006666(k), A006667(k)) = 10.

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

Cf. A006577, A006666, A006667.

for n from 1 to 45 do:
ii:=0:
for k from 2 to 10^7 while(ii=0) do:
  m:=k:s1:=0:s2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     s2:=s2+1:m:=m/2:
     else
     s1:=s1+1:m:=3*m+1:
    fi:
   od:
    if gcd(s1,s2)=n
     then
     ii:=1:printf(`%d %d \n`,n,k):
     else
    fi:
od:
od:

Function[w, First /@ Lookup[w, Function[k, If[k == {}, #, Take[#, First@ k]]]@ Complement[Range@ Max@ #, #]] &@ Keys@ w]@ KeySort@ PositionIndex@ Table[GCD[Count[NestWhileList[If[OddQ[#], 3 # + 1, #/2] &, n, # > 1 &], ?(EvenQ[#] &)], Count[Differences[NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]], ?Positive]], {n, 2^16}] (* Michael De Vlieger, Feb 02 2017, Version 10, after Harvey P. Dale at A006666 and A006667 *)

A304119 Numerators of record low values of the ratio n*3^A006667(n)/2^A006666(n).

Original entry on oeis.org

1, 27, 1701, 6561, 1760826122505, 115093142840908791, 460166680231540515, 1840049047529878113

Offset: 1

Michel Lagneau, May 03 2018

This has been verified for n up to 10^7.
Conjecture: Consider A006666 and A006667, the sequences giving the number of halving and tripling steps to reach 1 in 3x+1 problem. There exists a rational constant c such that c <= n*3^A006667(n)/2^A006666(n) <= 1 where c = 1840049047529878113/2305843009213693952 is the last term in the sequence of the ratios.
Note that n*3^A006667(n)/2^A006666(n) = 1 if n is a power of 2 (A000079).
It seems that n*3^A006667(n)/2^A006666(n) = c for n = 993*2^k, k = 0, 1, 2, ... In this case, c = 993*2^k*3^32/2^(61+k), where 32 = A006667(993*2^k) and 61+k = A006666(993*2^k). For example, c = 993*3^32/2^61 = 1986*3^32/2^62 = 3972*3^32/2^63 = 7944*3^32/2^64 = ...

			For n=1 to 10 the ratios are: 1, 1, 27/32, 1, 15/16, 27/32, 1701/2048, 1, 6561/8192, 15/16, so the low records are 1, 27/32, 1701/2048, 6561/8192, ...

Cf. A006666, A006667, A127789 (for the indices where these records occur).

q=1; Collatz[n_]:=NestWhileList[If[EvenQ[#], #/2, 3 #+1]&, n, #>1&]; nn=5000; t={}; n=0; While[Length[t]

ht(n) = my(h, t); while(n>1, if(n%2, n=3*n+1; t++, n>>=1; h++)); return([h, t]);
lista(nn) = {m = 2; for (n=1, nn, v = ht(n); newm = n*3^v[2]/2^v[1]; if (newm < m, print1(numerator(newm), ", "); m = newm));} \\ Michel Marcus, May 06 2018

A123481 Record n for the sequence A006666(n)/log(n), for n > 1; i.e., the n at which the number of steps to 1 for the halving-Collatz iteration divided by log(n) sets a new maximum.

Original entry on oeis.org

2, 3, 7, 9, 27, 230631, 626331, 837799, 1723519, 3732423, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231

Offset: 1

Diego Dominici (dominicd(AT)newpaltz.edu), Apr 16 2007

nonn

These numbers (for n > 2, at least) are called "Abby-Normal" numbers by Dominici in the referenced paper (definition 21); 2 is included here to agree with the RECORDS transform convention. - D. S. McNeil, Mar 26 2009

Three further terms are known (see Roosendaal) but they may not be the next three terms. - N. J. A. Sloane, Oct 21 2012

D. Dominici, Working with 2s and 3s, arXiv:0704.1057
Eric Roosendaal, 3x+1 Completeness and Gamma Records - From _N. J. A. Sloane_, Oct 21 2012

Definition edited by D. S. McNeil, Mar 26 2009
Extended by D. S. McNeil, Mar 26 2009
a(13) from Donovan Johnson, Feb 16 2011
a(14)-a(16) from Roosendaal's table added by N. J. A. Sloane, Oct 21 2012

cp's OEIS Frontend

A211981 Numbers n such that floor(2^A006666(n)/3^A006667(n)) = n.

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Mathematica

A225089 a(n) = floor(2^A006666(m)/3^A006667(m)) - m, where m = 2n + 1.

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A076182 a(n) = A006666(n) mod 2.

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A304715 For any n > 0, if A006666(n) >= 0, then a(n) = Sum_{i = 0..A006666(n)-1} 2^i * [T^i(n) == 0 (mod 2)] (where [] is an Iverson bracket and T^i denotes the i-th iterate of the Collatz function A014682); otherwise a(n) = -1.

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A232462 Number of iterations of the map n -> f(f(f(...f(n)...))) to reach the end of the cycle, where f(n) = A006666(n), the initial number n is not counted.

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A265099 Least k such that floor(2^A006666(k)/3^A006667(k)) - k = n.

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A277367 a(n) = gcd(A006666(n), A006667(n)) where A006666 and A006667 are respectively the number of halving and tripling steps in the '3x+1' problem.

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A281938 a(n) is the least k such that gcd(A006666(k), A006667(k)) = n.

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