A006666 -id:A006666 - cp's OEIS frontend

A182078 Odd numbers n such that the reduced Collatz map n -> (3n+1)/2^k gives a trajectory of decreasing odd numbers.

5, 13, 17, 21, 45, 53, 69, 85, 113, 141, 181, 213, 241, 277, 301, 321, 341, 369, 401, 453, 565, 725, 753, 853, 909, 965, 1069, 1109, 1137, 1205, 1285, 1365, 1425, 1477, 1605, 1713, 1813, 1933, 1969, 2261, 2417, 2573, 2577, 2625, 2901, 2957, 3013, 3213, 3413

Offset: 1

Michel Lagneau, Apr 10 2012

nonn

			45 is in the sequence because 45 generates the trajectory of odd numbers : 45 -> 17 -> -> 13 -> 5 -> 1.

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

Cf. A006666, A075677 (reduced Collatz map), A256598 (trajectory).

for n from 3 by 2 to 5000 do:i:=0:x:=n:n0:=n: u0:=0:for it from 1 to 1000 while(n0<>1 and u0=0) do: for a from 1 to 100 while(x mod 2 = 0 ) do: i:=i+1:x:=x/2: od:if x > n0 then u0:=1:else i:=i+1:n0:=x :x:=3*n0+1: fi:od: if u0=0 then printf(`%d, `,n):else fi:od:

A281665 Numbers m such that A006667(m)/A006577(m) = 1/3.

Original entry on oeis.org

159, 283, 377, 502, 503, 603, 615, 668, 669, 670, 799, 807, 888, 890, 892, 893, 1063, 1065, 1095, 1186, 1187, 1188, 1189, 1190, 1417, 1435, 1580, 1581, 1582, 1585, 1586, 1587, 1889, 1913, 1947, 1959, 1963, 2104, 2106, 2108, 2109, 2113, 2114, 2115, 2119, 2518

Offset: 1

Michel Lagneau, Jan 31 2017

nonn

A006667: number of tripling steps to reach 1 in '3x+1' problem.
A006577: number of halving and tripling steps to reach 1 in '3x+1' problem.
The corresponding number of iterations A006577(a(n)) is given by the sequence 54, 60, 63, 66, 66, 69, 69, 69, 69, 69, 72, 72, 72, 72, 72, 72, 75, 75, ... and the set of the distinct values of this sequence is {b(n)} = {54, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, ...}. We observe that {b(k)} = {54} union {60 + 3*k} for k = 1, 2, ...

			159 is in the sequence because A006667(159)/A006577(159) = 18/54 = 1/3.

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

Cf. A006577, A006666, A006667.

nn:=10000:
for n from 2 to 3000 do:
  m:=n: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:
   s:=s1/(s1+s2):
    if s=1/3
     then
     printf(`%d, `,n):
     else
    fi:
od:

A282408 Numbers k for which the number of odd members and the number of even members in the Collatz (3x+1) trajectory are both a perfect square.

Original entry on oeis.org

1, 2, 16, 17, 322, 323, 512, 1595, 1598, 1599, 1609, 1614, 1615, 1627, 1641, 1643, 1657, 1663, 1776, 1780, 1781, 1784, 1786, 2176, 2208, 2216, 2218, 2240, 2256, 2260, 2261, 2274, 2275, 2400, 2408, 2410, 2416, 2417, 2420, 2421, 3844, 3845, 3846, 3848, 3850, 3852

Offset: 1

Michel Lagneau, Feb 14 2017

nonn

Or numbers m such that A078719(m) and A006666(m) are both a perfect square.
For k < 5*10^6, the fourteen distinct pairs of squares in the order of appearance are: (1, 0), (1, 1), (1, 4), (4, 9), (36, 64), (1, 9), (25, 49), (4, 16), (16, 36), (9, 25), (1, 16), (81, 144), (4, 25) and (64, 121).
The numbers 2^(m^2) = A002416(m) are in the sequence, and the corresponding pairs of squares are (1, m^2).
Number of terms <= 10^h: 2, 4, 7, 202, 203, 474, 20888, etc. Robert G. Wilson v, Feb 14 2017

			17 is in the sequence because the Collatz trajectory is 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 => the number of odd members is 4 = 2^2 and number of even members is 9 = 3^2.

Robert G. Wilson v, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A000079, A002416, A078719, A006666, A006667.

nn:=10^6:
for n from 1 to 10000 do:
  m:=n:i1:=1:i2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     m:=m/2:i2:=i2+1:
     else
     m:=3*m+1:i1:=i1+1:
    fi:
   od:
    if sqrt(i1)=floor(sqrt(i1)) and sqrt(i2)=floor(sqrt(i2))
     then
     printf(`%d, `,n):
     else
    fi:
od:

fQ[n_] := Block[{m = n, e = 0, o = 1}, While[m > 1, If[OddQ@ m, m = 3 m + 1; o++, m /= 2; e++]]; IntegerQ@ Sqrt@ e && IntegerQ@ Sqrt@ o]; Select[ Range@ 3855, fQ] (* Robert G. Wilson v, Feb 14 2017 *)
memsqQ[n_]:=Module[{col=NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]}, AllTrue[ {Sqrt[ Count[ col, ?(EvenQ[#]&)]],Sqrt[Count[col,?(OddQ[ #]&)]]},IntegerQ]]; Select[Range[4000],memsqQ] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Sep 04 2018 *)

A282409 Numbers n for which the number of odd members and the number of even members in the Collatz (3x+1) trajectory are both prime.

Original entry on oeis.org

3, 9, 10, 12, 13, 22, 23, 40, 42, 73, 88, 90, 92, 93, 114, 115, 118, 119, 144, 148, 149, 152, 154, 162, 163, 164, 165, 166, 192, 208, 212, 213, 226, 227, 251, 295, 318, 319, 350, 351, 576, 592, 596, 597, 608, 616, 618, 625, 640, 642, 643, 648, 650, 652, 653

Offset: 1

Michel Lagneau, Feb 14 2017

nonn

Or numbers m such that A078719(m) and A006666(m) are both prime.

The distinct pairs of primes in the order of appearance are: (3, 5), (7, 13), (2, 5), (3, 7), (5, 11), (2, 7), (43, 73), (5, 13), (11, 23), (7, 17), (41, 71), (3, 11), (23, 43), (19, 37),...

			13 is in the sequence because its Collatz trajectory is 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 which contains 3 odd members and 7 even members.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A078719, A006666, A006667.

nn:=10^6:
for n from 1 to 500 do:
  m:=n:i1:=1:i2:=0:
   for i from 1 to nn while(m<>1) do:
    if irem(m,2)=0
     then
     m:=m/2:i2:=i2+1:
     else
     m:=3*m+1:i1:=i1+1:
    fi:
   od:
     if isprime(i1) and isprime(i2)
     then
     printf(`%d, `,n):
     else
    fi:od:

is(n)=my(e,o=1); while(n>1, n=if(n%2, o++; 3*n+1, e++; n/2)); isprime(e) && isprime(o) \\ Charles R Greathouse IV, Feb 14 2017

from sympy import isprime
def a(n):
    l=[n]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        l.append(n)
        if n<2: break
    o=list(filter(lambda i: i%2==1, l))
    e=list(filter(lambda i: i%2==0, l))
    return [o, e]
print([n for n in range(2, 1001) if isprime(len(a(n)[0])) and isprime(len(a(n)[1]))]) # Indranil Ghosh, Apr 14 2017

A316783 Seed values of hailstone sequences for which, by the time they first reach 1, the ratio of odd terms to even terms is exactly 1/2. (The seed value itself is included in the ratio calculation.)

Original entry on oeis.org

4, 5, 6, 11, 14, 15, 18, 19, 25, 33, 43, 57, 59, 78, 79, 105, 135, 139, 185, 187, 191, 246, 247, 249, 254, 255, 329, 338, 339, 359, 427, 438, 439, 443, 450, 451, 478, 479, 569, 585, 590, 591, 601, 636, 637, 638, 758, 759, 767, 779, 786, 787, 801, 849, 850, 851

Offset: 1

Matt Enlow, Jul 13 2018

nonn

1/2 is the most common odd/even ratio for these sequences. For seed values from 2 to 2^17, it is the odd/even ratio for 4454 of those sequences. The next most common ratio is 5/11, with 1834 sequences. (This is all empirical observation, using Mathematica.)

			The hailstone sequence starting with 11 (and ending at 1) is 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Five of those terms are odd, and the other ten are even, giving an odd/even ratio of 1/2. Therefore 11 is a term in this sequence.

Cf. A006370, A006666, A078719.

b:= proc(n) option remember; `if`(n=1, [1, 0],
      `if`(n::odd, [1, 0]+b(3*n+1), [0, 1]+b(n/2)))
    end:
a:= proc(n) option remember; local k; for k from a(n-1)
      +1 while (l-> 2*l[1]<>l[2])(b(k)) do od; k
    end: a(0):=0:
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 02 2018

collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1]&, n, # != 1 &];
A316783 = Select[Range[2, 1000], With[{c = Mod[collatz[#], 2]}, 2 Total[c] == Total[1 - c]] &]

is(n) = my(x=n, even=0, odd=0); while(1, if(x%2==0, x=x/2; even++, x=3*x+1; odd++); if(x==1, odd++; break)); odd/even==1/2 \\ Felix Fröhlich, Jul 13 2018

{ k : A006666(k)/A078719(k) = 2 }. - Alois P. Heinz, Aug 02 2018

A348094 If the Collatz trajectory of n reaches 1, say after k steps, and there is an integer m > n such that T^i(m) and T^i(n) have the same parity for i = 0..k (where T^i denotes the i-th iterate of the Collatz map A006370), then a(n) is the least such m, otherwise a(n) is -1.

Original entry on oeis.org

2, 4, 35, 8, 21, 70, 2055, 16, 8201, 42, 1035, 140, 141, 4110, 4111, 32, 529, 16402, 16403, 84, 85, 2070, 2071, 280, 65561, 282, 1180591620717411303451, 8220, 8221, 8222, 147573952589676412959, 64, 262177, 1058, 1059, 32804, 32805, 32806, 8388647, 168

Offset: 1

Rémy Sigrist, Sep 29 2021

nonn

When a(n) > 0, the binary expansion of A125711(n) is a prefix of that of A125711(a(n)).

			The first terms, alongside the binary representations of A125711(n) and of A125711(a(n)), are:
  n  a(n)  h(n)               h(a(n))
  -  ----  -----------------  --------------------------------------
  1     2                  1                                      11
  2     4                 11                                     111
  3    35           10101111                          10101111101111
  4     8                111                                    1111
  5    21             101111                                10111111
  6    70          110101111                         110101111101111
  7  2055  10101011011101111  10101011011101111110111010101111101111
  8    16               1111                                   11111

Cf. A006370, A006666, A070165, A125711.

A348094[n_] := n+2^(Length[NestWhileList[If[OddQ[#], 3#+1, #]/2 &, n, #>1 &]]-1);
Array[A348094, 50] (* Paolo Xausa, Apr 05 2024 *)

a(n) = { my (h=0, r=n); while (r>1, if (r%2, r=3*r+1, r=r/2; h++)); n+2^h }

a(2^k) = 2^(k+1) for any k >= 0.

a(n) = n + 2^A006666(n) when A006666(n) >= 0.

A365502 In the Collatz problem, total stopping times for iteration of the 3x+1 function corresponding to the starting points given by A248037.

Original entry on oeis.org

1, 5, 11, 13, 70, 278, 319, 329, 349, 374, 384, 416, 429, 592, 966, 1134, 1404

Offset: 1

Paolo Xausa, Sep 06 2023

The 3x+1 function (A014682), denoted by T(x) in the literature, is defined as T(x) = (3x+1)/2 if x is odd, T(x) = x/2 if x is even.
The total stopping time is the number of iterations of the T(x) map required to reach 1.
Except for a(1), these values are listed in the third column of Table 1 in Kontorovich and Lagarias (2009, 2010).
See A365503 for corresponding number of odd iterates before reaching 1.

Alex V. Kontorovich and Jeffrey C. Lagarias, Stochastic Models for the 3x+1 and 5x+1 Problems, arXiv:0910.1944 [math.NT], 2009, pp. 9-11, and in Jeffrey C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, American Mathematical Society, 2010, pp. 138-140.
Index entries for sequences related to 3x+1 (or Collatz) problem

Subsequence of A006666.

Cf. A014682, A248037, A365503.

a(n) = A006666(A248037(n)).

A370350 Number of steps to go from n to 1 in a variant of the Collatz iteration x -> (x / 5 if 5 divides x, x + (x+(5-(x mod 5)))/5 otherwise), or -1 if 1 is never reached.

Original entry on oeis.org

0, 4, 3, 2, 1, 7, 17, 6, 16, 5, 15, 5, 5, 14, 4, 4, 13, 11, 11, 3, 12, 10, 10, 20, 2, 11, 9, 9, 19, 8, 69, 10, 8, 8, 18, 17, 18, 68, 9, 7, 7, 8, 77, 16, 17, 67, 8, 17, 8, 6, 7, 76, 15, 7, 16, 66, 7, 16, 7, 6, 75, 6, 75, 14, 6, 6, 16, 65, 6, 15, 6, 15, 24, 74

Offset: 1

Giuseppe Ciacco, Feb 16 2024

Because ceiling(a/b) = (a+(b-(a mod b)))/b, this sequence equals the total number of steps to reach 1 in a variant of the Collatz problem using the iteration f(x) := x/k if k divides x, x+ceiling(x/k) otherwise.
Specifically, here we set k=5 because it is the next integer (after 2) that is not in A368136; i.e. the iteration used here has no loops for starting values up to 5*5=25, apart from the loop containing 1.
It is not known if there exists n such that a(n) = -1 (either by iteration reaching a non-elementary loop which implies n>5*5, or by iteration growing without bound).

			For n = 11, the following trajectory is obtained:
  11, 14, 17, 21, 26, 32, 39, 47, 57, 69, 83, 100, 20, 4, 5, 1
which requires 15 steps to reach 1, therefore a(11) = 15.

Giuseppe Ciacco, Table of n, a(n) for n = 1..10000
Walter Carnielli, Some natural generalizations of the Collatz Problem, Applied Mathematics E-Notes 15 (2015): 207-215.
Wikipedia, Collatz Conjecture.
OEIS Wiki, 3x+1 problem.

Cf. A006666, A368136.

s={};Do[c=0;a=n;While[a>1,If[Divisible[a,5],a=a/5,a=a+Ceiling[a/5]];c++];AppendTo[s,c],{n,74}];s (* James C. McMahon, Feb 28 2024 *)

def a(n, C = 5):
    s = 0
    while n > 1:
        d, r = divmod(n, C)
        n = n + 1 + d if r else d
        s += 1
    return s
print([a(n) for n in range(1, 75)])
# Giuseppe Ciacco and Robert Munafo, Mar 25 2024

A174539 Starting numbers n such that the number of halving and tripling steps to reach 1 under the Collatz 3x+1 map is a perfect square.

Original entry on oeis.org

1, 2, 7, 12, 13, 16, 44, 45, 46, 80, 84, 85, 98, 99, 100, 101, 102, 107, 129, 153, 156, 157, 158, 169, 272, 276, 277, 280, 282, 300, 301, 302, 350, 351, 512, 576, 592, 596, 597, 608, 616, 618, 625, 642, 643, 644, 645, 646, 648, 649, 650, 651, 652, 653, 654, 655, 662, 663

Offset: 1

Michel Lagneau, Mar 21 2010

nonn

Numbers n such that A006577(n) is a perfect square.

			44, 45 and 46 are in the sequence because the number of steps as counted in A006577 for each of them is 16 = 4^2, a perfect square.

Index to sequences related to the 3x+1 problem

Cf. A006577, A008908, A064433, A006666, A006667, A033491

with(numtheory):for x from 1 to 200 do traj:=0: n1:=x: x1:=x: for p from 1 to 20 while(irem(x1,2)=0)do p1:=2^p: xx1:=x1: x1:=floor(n1/p1): traj:=traj+1:od:
n:=x1: for q from 1 to 100 while(n<>1)do n1:=3*n+1: traj:=traj+1: x0:=irem(n1,2): for p from 1 to 20 while(x0=0)do p1:=2^p: xx1:=x1: x1:=floor(n1/p1): x0:=n1-p1*x1: traj:=traj+1: od: traj:=traj-1: n:=xx1:od:
if(sqrt(traj))=floor(sqrt(traj)) then print(x):else fi:od:

htsQ[n_]:=With[{len=Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1},IntegerQ[Sqrt[len]]]; Select[Range[700],htsQ] (* Harvey P. Dale, Jan 01 2023 *)

{n: A006577(n) in A000290}.

Unspecific references removed - R. J. Mathar, Mar 31 2010

Corrected and extended by Harvey P. Dale, Jan 01 2023

A268338 Numbers that cycle under the following transformation: if m is even, divide by 2, if m is congruent to 1 mod 4, multiply by 3 and add 1; if m is congruent to 3 mod 4, multiply by 7 and add 1.

Original entry on oeis.org

1, 2, 4, 19, 23, 31, 38, 41, 46

Offset: 1

David Consiglio, Jr., Feb 01 2016

Some numbers appear to grow indefinitely under these rules, but it is possible that they may eventually cycle at some point. All numbers up to 50 either cycle or transform to another number that cycles (typically 1). 51 is the first open case: it may eventually cycle or may continue to grow indefinitely.

			23 is a member of this sequence. 23 is congruent to 3 mod 4.  As a result, 23 transforms to 23*7+1 = 162.  From there 162 -> 81 -> 244 -> 122 -> 61 -> 184 -> 92 -> 46 -> 23.  23 is the least member of this cycle.
49 is not a member of this sequence because it eventually reduces to 19, which cycles.

Cf. A267703, A267970, A267969, A006666, A006370, A005186, A267970, A232711.

a = 1
b = 1
prev = []
keep = []
count = 0
while b < 51:
    keep.append(a)
    flag1 = False
    flag2 = False
    if a % 2 == 0:
        a /= 2
    elif a % 4 == 1:
        a = a*3+1
    else:
        a = a*7+1
    if count > 50:
        b += 1
        a = b
        count = 0
        keep = []
    if keep.count(a) == 2 and a not in prev and a <= 50:
        prev.append(a)
        count = 0
        keep = []
        b += 1
        a = b
    count += 1
print(sorted(prev))
# David Consiglio, Jr., Feb 01 2016

Corrected and edited by David Consiglio, Jr., Apr 20 2016

cp's OEIS Frontend

A182078 Odd numbers n such that the reduced Collatz map n -> (3n+1)/2^k gives a trajectory of decreasing odd numbers.

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A281665 Numbers m such that A006667(m)/A006577(m) = 1/3.

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A282408 Numbers k for which the number of odd members and the number of even members in the Collatz (3x+1) trajectory are both a perfect square.

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Mathematica

A282409 Numbers n for which the number of odd members and the number of even members in the Collatz (3x+1) trajectory are both prime.

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A316783 Seed values of hailstone sequences for which, by the time they first reach 1, the ratio of odd terms to even terms is exactly 1/2. (The seed value itself is included in the ratio calculation.)

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A348094 If the Collatz trajectory of n reaches 1, say after k steps, and there is an integer m > n such that T^i(m) and T^i(n) have the same parity for i = 0..k (where T^i denotes the i-th iterate of the Collatz map A006370), then a(n) is the least such m, otherwise a(n) is -1.

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Mathematica

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A365502 In the Collatz problem, total stopping times for iteration of the 3x+1 function corresponding to the starting points given by A248037.

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A370350 Number of steps to go from n to 1 in a variant of the Collatz iteration x -> (x / 5 if 5 divides x, x + (x+(5-(x mod 5)))/5 otherwise), or -1 if 1 is never reached.

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