A070165 -id:A070165 - cp's OEIS frontend

A225761 Numerators of the sums of reciprocals of the Collatz (3x+1) sequence beginning with n and stopping at 1.

1, 3, 617, 7, 171, 219, 766329, 15, 1061683, 179, 102151, 677, 497, 785777, 10380059, 31, 8861, 360377, 60226515, 183, 2731, 103919, 3339321, 229, 1548244271, 505, 129481899470258402665619129356105706380861444925035330406812603986229803685477, 113643

Offset: 1

Nico Brown, May 14 2013

nonn

If the sum of the reciprocals of a Collatz sequence is bounded, there are no cycles other than 4,2,1.

a(n) = numerator of sum (1/A070165(n,k): k = 1..A006577(n)). - Reinhard Zumkeller, May 16 2013

			For n=9 the Collatz sequence is {9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 4, 2, 1}.  So the sum of the reciprocals is 1/9 + 1/28 + 1/14 + 1/7 + 1/22 + 1/11 + ... + 1/4 + 1/2 + 1/1 = 1061683/350064, whose numerator is 1061683.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A087226, A225784 (denominators).

Cf. A225843.

import Data.Ratio (numerator)
a225761 = numerator . sum . map (recip . fromIntegral) . a070165_row
-- Reinhard Zumkeller, May 16 2013

Table[Numerator[Total[1/Collatz[n]]], {n, 40}] (* T. D. Noe, May 15 2013 *)

Extended by T. D. Noe, May 15 2013

A235801 Length of n-th horizontal line segment in a diagram of a two-dimensional version of the 3x+1 (or Collatz) problem.

Original entry on oeis.org

0, 1, 2, 3, 7, 5, 6, 7, 8, 9, 17, 11, 12, 13, 14, 15, 27, 17, 18, 19, 20, 21, 37, 23, 24, 25, 26, 27, 47, 29, 30, 31, 32, 33, 57, 35, 36, 37, 38, 39, 67, 41, 42, 43, 44, 45, 77, 47, 48, 49, 50, 51, 87, 53, 54, 55, 56, 57, 97, 59, 60, 61, 62, 63, 107, 65, 66

Offset: 0

Omar E. Pol, Jan 15 2014

nonn

In the diagram every cycle is represented by a directed graph.
After (3x + 1) the next step is (3y + 1).
After (x/2) the next step is (y/2).
A235800(n) gives the length of n-th vertical line segment, from left to right, in the same diagram.

			The first part of the diagram in the first quadrant:
. . . . . . . . . . . . . . . . . . . . . . . .
.              _ _|_ _|_ _|_ _|_ _|_ _|_ _|_ _.
.             |   |   |   |   |   |   |   |_|_.
.             |   |   |   |   |   |   |  _ _|_.
.             |   |   |   |   |   |   |_|_ _|_.
.             |   |   |   |   |   |  _ _|_ _|_.
.             |   |   |   |   |   |_|_ _|_ _|_.
.          _ _|_ _|_ _|_ _|_ _|_ _ _|_ _|_ _|_.
.         |   |   |   |   |   |_|_ _|_ _|_ _|_.
.         |   |   |   |   |  _ _|_ _|_ _|_ _|_.
.         |   |   |   |   |_|_ _|_ _|_ _|_ _|_.
.         |   |   |   |  _ _|_ _|_ _|_ _|_ _|_.
.         |   |   |   |_|_ _|_ _|_ _|_ _|_ _| . 11
.      _ _|_ _|_ _|_ _ _|_ _|_ _|_ _|_ _|     . 17
.     |   |   |   |_|_ _|_ _|_ _|_ _|         .  9
.     |   |   |  _ _|_ _|_ _|_ _|             .  8
.     |   |   |_|_ _|_ _|_ _|                 .  7
.     |   |  _ _|_ _|_ _|                     .  6
.     |   |_|_ _|_ _|                         .  5
.  _ _|_ _ _|_ _|                             .  7
. |   |_|_ _|                                 .  3
. |  _ _|                                     .  2
. |_|                                         .  1
. . . . . . . . . . . . . . . . . . . . . . . .  0
.                                              a(n)
.
For an explanation of this diagram as the skeleton of a piping model see A235800. - _Omar E. Pol_, Dec 30 2021

Cf. A347270 (all 3x+1 sequences).
Companion of A235800.
Cf. A000027, A004767, A006370, A014682, A016957, A070165, A235795.

from _future_ import division
A235801_list = [n if n % 6 != 4 else 10*(n//6)+7 for n in range(10**4)] # Chai Wah Wu, Sep 26 2016

a(n) = 10*k - 3, if n is of the form (6*k-2), k>=1, otherwise a(n) = n.
From Chai Wah Wu, Sep 26 2016: (Start)
a(n) = 2*a(n-6) - a(n-12) for n > 11.
G.f.: x*(x^2 + 1)*(x^3 + 2*x^2 + 1)*(x^5 + x^4 + 2*x + 1)/(x^12 - 2*x^6 + 1). (End)

A258145 Row lengths of the irregular array in A256598.

Original entry on oeis.org

1, 3, 2, 6, 7, 5, 3, 6, 4, 7, 2, 5, 8, 42, 6, 40, 9, 4, 7, 12, 41, 10, 5, 39, 8, 8, 3, 42, 11, 11, 6, 40, 9, 9, 4, 38, 43, 4, 7, 12, 7, 41, 2, 10, 10, 34, 5, 39, 44, 8, 8, 32, 13, 37, 42, 25, 3, 11, 6, 11, 35

Offset: 0

Wolfdieter Lang, May 27 2015

a(n) is the number of odd numbers in row A070165(2*n+1) and also in row A070168(2*n+1), for n >= 0 (if it turns out that a row 2*n+1 has infinite row length one puts a(n) = 0).

Cf. A256598, A070165, A070168.

A258772 Number of fixed points in the Collatz (3x+1) trajectory of n.

Original entry on oeis.org

1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0

Offset: 1

Derek Orr, Jun 09 2015

nonn

This sequence uses the definition in A006370: if n is odd, n -> 3n+1 and if n is even, n -> n/2.

The number 3 appears first at a(187561). Do all nonnegative numbers appear? See A258821.

			For n = 5, the trajectory is T(5) = [5, 16, 8, 4, 2, 1]. Since the fourth term in this sequence is 4, this is a fixed point. Since there is only one fixed point, a(5) = 1.
For n = 6, the trajectory is T(6) = [6, 3, 10, 5, 16, 8, 4, 2, 1]. Here, the k-th term in this trajectory does not equal k for any possible k. So a(6) = 0.

Cf. A006370, A070165.

A258772[n_]:=Count[MapIndexed[{#1}==#2&,NestWhileList[If[OddQ[#],3#+1,#/2]&,n,#>1&]],True];Array[A258772,100] (* Paolo Xausa, Nov 06 2023 *)

Tvect(n)=v=[n];while(n!=1,if(n%2,k=(3*n+1);v=concat(v,k);n=k);if(!(n%2),k=n/2;v=concat(v,k);n=k));v
for(n=1,200,d=Tvect(n);c=0;for(i=1,#d,if(d[i]==i,c++));print1(c,", "))

A263716 Irregular triangle read by rows: numbers in the Collatz conjecture in the order of their first appearance.

Original entry on oeis.org

1, 2, 3, 10, 5, 16, 8, 4, 6, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 9, 28, 14, 12, 15, 46, 23, 70, 35, 106, 53, 160, 80, 18, 19, 58, 29, 88, 44, 21, 64, 32, 24, 25, 76, 38, 27, 82, 41, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182

Offset: 0

Daniel Suteu, Oct 24 2015

This is the irregular triangle read by rows giving trajectory of n in the Collatz problem, flattened and with all the repeated terms deleted.

This sequence goes to infinity as n gets larger. On the Collatz conjecture this sequence is a permutation of the positive integers. [Corrected by Charles R Greathouse IV, Jul 29 2016]

			Triangle begins:
1;
2;
3, 10, 5, 16, 8, 4;
...
The Collatz trajectories for the first five positive integers are {1}, {2, 1}, {3, 10, 5, 16, 8, 4, 2, 1}, {4, 2, 1}, {5, 16, 8, 4, 2, 1}.
From {2, 1} we delete 1 because it has already occurred. From {3, 10, 5, ..., 4, 2, 1} we delete {2, 1} because both numbers have already occurred. We completely get rid of {4, 2, 1} because it has already occurred as the tail end of {3, 10, 5, ...}, and we also completely get rid of {5, 16, 8, ...} for the same reason.
This leaves us with {1}, {2}, {3, 10, 5, 16, 8, 4}, thus accounting for the first eight terms of this sequence.

Cf. A006577, A070165, A222118 (row lengths).

Cf. A347265 (essentially the same).

collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; DeleteDuplicates[Flatten[Table[collatz[n], {n, 20}]]] (* Alonso del Arte, Oct 24 2015 *)

func collatz(n) is cached {  # automatically memoized function
    say n;                   # prints the first unseen numbers
    n.is_one ? 0
             : (n.is_even ? collatz(n/2)
                          : collatz(3*n + 1));
}
range(1, Math.inf).each { |i| collatz(i) }

row(n) = {
   if seen[n]: stop
   else: write(n) and do:
   | n is one: stop
   | n is odd: n <- 3*n+1
   | n is even: n <- n/2
}

A280408 Irregular triangle read by rows listing the prime numbers that appear from the trajectory of n in Collatz Problem.

Original entry on oeis.org

2, 2, 3, 5, 2, 2, 5, 2, 3, 5, 2, 7, 11, 17, 13, 5, 2, 2, 7, 11, 17, 13, 5, 2, 5, 2, 11, 17, 13, 5, 2, 3, 5, 2, 13, 5, 2, 7, 11, 17, 13, 5, 2, 23, 53, 5, 2, 2, 17, 13, 5, 2, 7, 11, 17, 13, 5, 2, 19, 29, 11, 17, 13, 5, 2, 5, 2, 2, 11, 17, 13, 5, 2, 23, 53, 5, 2, 3, 5, 2, 19, 29, 11, 17, 13, 5, 2

Offset: 1

Matthew Campbell, Jan 02 2017

			The irregular array a(n,k) starts:
n\k   1   2   3   4   5   6
...
1:    2
2:    2
3:    3   5   2
4:    2
5:    5   2
6:    3   5   2
7:    7  11  17  13   5   2
8:    2
9:    7  11  17  13   5   2
10:   5  2
11:  11  17  13   5   2
12:   3   5   2
13:  13   5   2
14:   7  11  17  13   5   2
15:  23  53   5   2

David Radcliffe, Table of n, a(n) for n = 1..10000

Cf. A070165, A280409.

Table[Select[NestWhileList[If[EvenQ@ #, #/2, 3 # + 1] &, n, # > 1 &], PrimeQ], {n, 2, 30}] // Flatten (* Michael De Vlieger, Jan 02 2017 *)

from sympy import isprime
def a(n):
    if n==1: return [2]
    l=[n, ]
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        l+=[n, ]
        if n<2: break
    return list(filter(lambda i: isprime(i), l))
for n in range(1, 21): print(a(n)) # Indranil Ghosh, Apr 14 2017

A304030 a(n) is the number of steps at which the Collatz ('3x+1') trajectory of n crosses its initial value, or -1 if the number of crossings is infinite.

Original entry on oeis.org

0, 0, 1, 0, 1, 4, 3, 0, 5, 2, 3, 2, 3, 8, 3, 0, 3, 10, 7, 0, 1, 6, 1, 0, 9, 2, 3, 6, 7, 4, 3, 0, 13, 4, 1, 4, 5, 8, 9, 0, 7, 2, 5, 2, 3, 4, 5, 0, 5, 8, 5, 0, 1, 14, 9, 0, 7, 2, 3, 6, 7, 12, 9, 0, 5, 6, 3, 0, 1, 4, 7, 0, 13, 2, 1, 2, 3, 8, 3, 0, 7, 14, 11, 0, 1, 8, 3, 0, 3, 2, 7, 4, 5, 12, 9, 0, 19, 4, 1, 0

Offset: 1

Jon E. Schoenfield, May 04 2018

nonn

Some treatments of the Collatz conjecture view trajectories as starting to cycle when they reach 1, continuing with 4, 2, 1, 4, 2, 1, ..., while others view trajectories as terminating as soon as 1 is reached; this sequence treats trajectories as terminating at 1.
If the Collatz conjecture is true, then for n > 1, a(n) == n (mod 2).
If there exists any number n whose Collatz trajectory enters a cycle that includes values above and below n, then the number of crossings would be infinite. If the Collatz conjecture is true, then there exists no such value of n.
If a(k) = 0, then a(2^j * k) = 0, for j>0. Therefore the primitives are 1, 20, 24, 52, 56, 68, 72, 84, 88, 100, 116, ..., . - Robert G. Wilson v, May 19 2018

			The Collatz trajectory of 6 crosses its initial value (6) a total of 4 times, so a(6) = 4:
.
                           16
                           / \
                          /   \
              10         /     \
              / \       /       \
             /   \     /         8
  6---------*-----*---*-----------*--------
   \       /       \ /             \
    \     /         5               \
     \   /                           \
      \ /                             4
       3                               \
                                        ...
(Each "*" represents a crossing.)

Cf. A006370, A070165, A349325.

Collatz[n_] := NestWhileList[ If[ OddQ@#, 3# +1, #/2] &, n, # > 1 &]; f[n_] := Block[{x = Length[ SplitBy[ Collatz@ n, # < n +1 &]] - 1}, If[ OddQ@ n && n > 1, x - 1, x]]; Array[f, 100] (* Robert G. Wilson v, May 05 2018 *)

def A304030(n):
    prevc = c = n
    h = 0
    while c > 1:
        if c % 2:
            c = 3*c+1
            if prevc < n and c > n: h += 1
        else:
            c //= 2
            if prevc > n and c < n: h += 1
        prevc = c
    return h
print([A304030(n) for n in range(1, 100)]) # Paolo Xausa, Feb 22 2022

A333860 The maximum Hamming (binary) weight of the elements of the Collatz orbit of n, or -1 if 1 is never reached.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 3, 2, 3, 2, 3, 3, 4, 1, 3, 3, 4, 2, 3, 3, 4, 2, 4, 3, 8, 3, 4, 4, 8, 1, 4, 3, 4, 3, 3, 4, 5, 2, 8, 3, 4, 3, 4, 4, 8, 2, 3, 4, 4, 3, 4, 8, 8, 3, 4, 4, 5, 4, 5, 8, 8, 1, 3, 4, 4, 3, 3, 4, 8, 3, 8, 3, 4, 4, 4, 5, 6, 2, 5, 8, 8, 3, 4, 4, 5

Offset: 1

Markus Sigg, Apr 08 2020

			The Collatz orbit of 3 is 3,10,5,16,8,4,2,1. The Hamming weights are 2,2,2,1,1,1,1,1. The maximum is a(3) = 2.

Markus Sigg, Table of n, a(n) for n = 1..10000

Cf. A000120, A006370, A008908, A070165, A333861, A352895, A352897.

a[n_] := Max[DigitCount[#, 2, 1] & /@ NestWhileList[If[OddQ[#], 3*# + 1, #/2] &, n, # > 1 &]]; Array[a, 100] (* Amiram Eldar, Jul 29 2023 *)

a(n) = {
my(c = hammingweight(n));
while(n>1, n = if(n%2 == 0, n/2, 3*n+1); c = max(c, hammingweight(n)));
c;
}

a(n) = max(A000120(n), A352895(n)) = max(A000120(n), a(A006370(n))). - Antti Karttunen, Apr 10 2022

Escape clause added to the definition by Antti Karttunen, Apr 10 2022

A025587 '3x+1' record-setters (blowup factor).

Original entry on oeis.org

1, 3, 7, 15, 27, 703, 1819, 4255, 4591, 9663, 26623, 60975, 77671, 113383, 159487, 1212415, 2684647, 3041127, 3873535, 4637979, 5656191, 6416623, 6631675, 19638399, 80049391, 210964383, 319804831, 1410123943, 70141259775, 77566362559

Offset: 0

David W. Wilson

This sequence uses the highest even number reached, which will always be a power of 2 larger than A295163. - Howard A. Landman, Nov 20 2017
A proper subsequence of A006884. - Robert G. Wilson v, Dec 23 2017
Let m be the maximum value in row n of A070165. This sequence is the record transform of the sequence m/n for n >= 1. - Michael De Vlieger, Mar 13 2018

Howard A. Landman, Table of n, a(n) for n = 0..33 (a(0)-a(23) from _David W. Wilson_, a(24)-a(26) from Larry Reeves, a(27) from _Jud McCranie_)
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A295163 for maximum odd number reached, and A061523 for blowup factors.

Cf. A006884, A070165.

// First column is this sequence.
// Second column is the maximum (even) N reached.
// Third column is A061523, the ratio of those.
// NOTE: This could be made faster by special-casing 1,
// starting at 3, and incrementing by 4, since all terms except 1
// are congruent to 3 (mod 4).
#include    
long long    i=1, n, max_n;
long double    max_ratio=1.0, ratio;
int main()
{
    while(1)
    {
        n = i;
        max_n = n;
        while (n > i) // Can stop as soon as we drop below start.
        {
            n = 3*n + 1;
            max_n = (n > max_n) ? n : max_n;
            while (!(n&1))
            {
                n >>= 1;
            }
         }
        ratio = (double) max_n / (double) i;
        if (ratio > max_ratio)
        {
            max_ratio = ratio;
            printf("%lld\t%lld\t%Lf\n", i, max_n, max_ratio);
        }
        i += 2;
    }
}
// Howard A. Landman, Nov 14 2017

With[{s = Array[Max@ NestWhileList[If[EvenQ@#, #/2, 3 # + 1] &, #, # > 1 &]/# &, 2^18]}, Map[FirstPosition[s, #][[1]] &, Union@ FoldList[Max, s]]] (* Michael De Vlieger, Mar 13 2018 *)

More terms from Larry Reeves (larryr(AT)acm.org), May 03 2001
a(27) from Jud McCranie, Apr 23 2012
a(26) corrected (was missing least significant digit) by Howard A. Landman, Nov 14 2017

A101229 Perfect inverse "3x+1 conjecture" (See comments for rules).

Original entry on oeis.org

1, 2, 4, 1, 2, 4, 8, 16, 5, 10, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144, 12288, 24576, 49152, 98304, 196608, 393216, 786432, 1572864, 3145728, 6291456, 12582912, 25165824, 50331648, 100663296, 201326592, 402653184, 805306368

Offset: 1

Alexandre Wajnberg, Jan 22 2005

Perfect inverse "3x+1 conjecture": rule 1: multiply n by 2 to give n' = 2n. rule 2: when n'=(3x+1), do n"= (n'-1)/3 (n" integer) Additional rule: rule 2 is applied once for any number n' (otherwise, the sequence beginning with 1 would be the cycle "1 2 4 1 2 4 1 2 4 1..."); then apply rule 1.
This gives a particular sequence of hailstone numbers which may be considered as a central axis for all the hailstone number sequences. The perfect inverse "3x+1 conjecture" falls rapidly into the sequence 3 6 12 24 48 96... which will never give a number to which apply the 2nd rule.
a(n) for n >= 11 written in base 2: 11, 110, 11000, 110000, ..., i.e.: 2 times 1, (n-11) times 0 (see A003953(n-10)). - Jaroslav Krizek, Aug 17 2009

			The first 4 is followed by 1 because 4 = 3*1 + 1, so rule 2: (4-1)/3 = 1;
the second 4 is followed by 8 because the 2nd rule has already been applied, so rule 1: 4*2 = 8.

R. K. Guy, Collatz's Sequence, Section E16 in Unsolved Problems in Number Theory, 2nd ed. New York: Springer-Verlag, pp. 215-218, 1994.

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Collatz Problem
Index entries for linear recurrences with constant coefficients, signature (2).

Cf. A070165, A006577, A006667, A006666, A070167.

R:=PowerSeriesRing(Integers(), 45); Coefficients(R!( x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1) )); // G. C. Greubel, Mar 20 2019

Rest[CoefficientList[Series[x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1), {x, 0, 45}], x]] (* G. C. Greubel, Mar 20 2019 *)
LinearRecurrence[{2},{1,2,4,1,2,4,8,16,5,10,3},40] (* Harvey P. Dale, May 06 2023 *)

my(x='x+O('x^45)); Vec(x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1)) \\ G. C. Greubel, Mar 20 2019

a=(x*(17*x^10+27*x^8+7*x^3-1)/(2*x-1)).series(x, 45).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 20 2019

a(n) = 3*2^(n-11) = 2^(n-11) + 2^(n-10) for n >= 11. - Jaroslav Krizek, Aug 17 2009
From Colin Barker, Apr 28 2013: (Start)
a(n) = 2*a(n-1) for n>11.
G.f.: x*(17*x^10+27*x^8+7*x^3-1) / (2*x-1). (End)

More terms from Joshua Zucker, May 18 2006

Edited by G. C. Greubel, Mar 20 2019

cp's OEIS Frontend

A225761 Numerators of the sums of reciprocals of the Collatz (3x+1) sequence beginning with n and stopping at 1.

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A235801 Length of n-th horizontal line segment in a diagram of a two-dimensional version of the 3x+1 (or Collatz) problem.

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A258145 Row lengths of the irregular array in A256598.

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A258772 Number of fixed points in the Collatz (3x+1) trajectory of n.

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A263716 Irregular triangle read by rows: numbers in the Collatz conjecture in the order of their first appearance.

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Sidef

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A280408 Irregular triangle read by rows listing the prime numbers that appear from the trajectory of n in Collatz Problem.

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A304030 a(n) is the number of steps at which the Collatz ('3x+1') trajectory of n crosses its initial value, or -1 if the number of crossings is infinite.

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A333860 The maximum Hamming (binary) weight of the elements of the Collatz orbit of n, or -1 if 1 is never reached.

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