A006666 -id:A006666 - cp's OEIS frontend

A064433 Number of iterations of A064455 to reach 2 (or 1 in the case of 1).

1, 1, 2, 6, 3, 5, 7, 12, 4, 14, 6, 11, 8, 8, 13, 13, 5, 10, 15, 15, 7, 7, 12, 12, 9, 17, 9, 71, 14, 14, 14, 68, 6, 19, 11, 11, 16, 16, 16, 24, 8, 70, 8, 21, 13, 13, 13, 67, 10, 18, 18, 18, 10, 10, 72, 72, 15, 23, 15, 23, 15, 15, 69, 69, 7, 20, 20, 20, 12, 12, 12, 66, 17, 74, 17

Offset: 1

Jonathan Ayres (Jonathan.ayres(AT)btinternet.com), Oct 01 2001

Similar to 3x+1 series (A008908). Does this sequence converge to 2 for all values of n (true for all values of n up to 100000)? The inverse sequence using next n = n-int(n/2) for n even and n+int(n/2) for n odd leads to 3 (?) possible end sequences (1), (5, 7, 10) and (17, 25, 37, 55, 82, 41, 61, 91, 136, 68, 34)
Starting with a number n, the next value generated is n+int(n/2) if n is even, n-int(n/2) if n is odd; a(n) is the number of iteration for the initial value n to reach the limit of 1 to 2
Collatz's 3N+1 function as isometry over the dyadics is N->N/2 if even, but N->(3N+1)/2 if odd, including the (necessary) halving into each tripling step. Counting steps until reaching 1 in this way leads to this sequence instead of A008908. - Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009
The value at each step of a trajectory starting with n (n>1) is equal to the value plus one at the same step of the row starting with (n-1) of the irregular triangle of the abbreviated (Terras-modified) Collatz sequence (A070168). - K. Spage, Aug 07 2014

			a(4) = 6. Starting with 4, 4 is even so the next number is 4+int(4/2) = 6, 6 is even so next number is 6+int(6/2) = 9, 9 is odd so next number is 9-int(9/2) = 5, 5 is odd so next number is 5-int(5/2) = 3, 3 is odd so next number is 3-int(3/2)=2, so giving a sequence of 4,6,9,5,3,2: 6 numbers.
a(5) = 3. Starting with 5, A064455(5) = 3, A064455(3) = 2, so giving a trajectory of 5,3,2: 3 numbers. - _K. Spage_, Aug 07 2014

Antti Karttunen, Table of n, a(n) for n = 1..19683
M. del P. Canales Chacon and M. J. Vielhaber, Structural and Computational Complexity of Isometries and Their Shift Commutators, Electr. Colloq. on Computational Cpx., ECCC TR04-057, 2004. [From Michael Vielhaber (vielhaber(AT)gmail.com), Nov 18 2009]

Cf. A006666, A008908, A064455, A070168.

Table[Length@ NestWhileList[If[EvenQ@ #, 3 #/2, (# + 1)/2] &, n, # != 1 + Boole[n > 1] &], {n, 75}] (* Michael De Vlieger, Sep 24 2016 *)

A064455(n) = {if(n%2, (n + 1)/2, 3*n/2)}
A064433(n) = {my(c=1); if(n==1, 1, while(n!=2, n=A064455(n); c++); c)} \\ K. Spage, Aug 07 2014

a(n) = A006666(n-1) + 1. - K. Spage, Aug 04 2014

A092892 Smallest starting value in a Collatz '3x+1' sequence such that the sequence contains exactly n halving steps.

Original entry on oeis.org

1, 2, 4, 8, 5, 3, 6, 12, 24, 17, 11, 7, 14, 9, 18, 36, 25, 49, 33, 65, 43, 86, 57, 39, 78, 153, 105, 203, 135, 270, 185, 123, 246, 169, 329, 219, 159, 295, 569, 379, 283, 505, 377, 251, 167, 111, 222, 444, 297, 593, 395, 263, 175, 350, 233, 155, 103, 206, 137, 91, 182

Offset: 0

Hugo Pfoertner, Mar 11 2004

First occurrence of n in A006666.

The graph of this sequence has features similar to those of A092893, but with the x-axis scaled by log(3)/log(2). - T. D. Noe, Apr 09 2007

			a(5)=3 because the Collatz sequence 3,10,5,16,8,4,2,1 is the first sequence containing 5 halving steps.

T. D. Noe, Table of n, a(n) for n=0..500
Eric Weisstein's World of Mathematics, Collatz Problem.
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006666, A092893.

import Data.List (elemIndex); import Data.Maybe (fromJust)
a092892 = (+ 1) . fromJust . (`elemIndex` a006666_list)
-- Reinhard Zumkeller, Mar 14 2014

coll[n_]:= NestWhileList[If[EvenQ[#], #/2, 3#+1] &, n, #>1 &]; Table[i = 1; While[Count[coll[i], ?EvenQ] != n, i++]; i, {n, 0, 60}] (* _Jayanta Basu, Jun 05 2013 *)

A127789 Record indices of 2^h(n)/(3^t(n)*n), where h and t are the number of halving resp. tripling steps in the '3x + 1' problem.

Original entry on oeis.org

1, 3, 7, 9, 505, 559, 745, 993

Offset: 1

Paul Boddington, Apr 04 2007

Original name: In the '3x + 1' problem we would expect a positive integer n to be approximately equal to 2^j / 3^k where j is the number of halving steps and k is the number of '3x + 1' steps required to reach 1. The numbers in this sequence are those for which 2^j / (n*3^k) sets a new record.
Eric Roosendaal calls 2^j / (n*3^k) the residue of n and conjectures that 993 yields the highest residue. - T. D. Noe, Apr 08 2007
It has been verified that the next value, if it exists, is larger than 2^32 ~ 4.3e9. We do not need an "escape clause" (as for A006577, A006666, A006667, ...) in this sequence since the unlikely case of a possibly undefined ratio is irrelevant for the list of records. - M. F. Hasler, May 07 2018

Eric Roosendaal, On the 3x+1 Problem

Cf. A006370 (Collatz map), A014682 (condensed version using (3n+1)/2).

Cf. A006666 (halving steps), A006667 (tripling steps), A006577 (total).

(c(n,c=[0,0])=while(n>1,bittest(n,0)&&c[1]++&&(n=n*3+1)&&next;n\=2;c[2]++);c); m=1;for(n=1,oo,m<<(t=c(n))[2]>n*3^t[1]||next;m=n*3^t[1]/2^t[2];print1(n", ")) \\ M. F. Hasler, May 07 2018

New name from M. F. Hasler, May 07 2018

A206374 a(n) = (7*4^n - 1)/3.

Original entry on oeis.org

2, 9, 37, 149, 597, 2389, 9557, 38229, 152917, 611669, 2446677, 9786709, 39146837, 156587349, 626349397, 2505397589, 10021590357, 40086361429, 160345445717, 641381782869, 2565527131477, 10262108525909, 41048434103637, 164193736414549, 656774945658197

Offset: 0

Brad Clardy, Feb 07 2012

First bisection of A062092 and A081253, second bisection of A097163. - Bruno Berselli, Feb 12 2012

Except a(0)=2, this is the 3rd row of table A178415. - Michel Marcus, Apr 13 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Mike Warburton, Ulam-Warburton Automaton - Counting Cells with Quadratics, arXiv:1901.10565 [math.CO], 2019. See Table 1.
Index entries for linear recurrences with constant coefficients, signature (5,-4).

Cf. A002450, A006666, A072197; A002042 (first differences), A178415, A347834.

Magma
```
[(7*4^n-1)/3 : n in [0..30]];
```

Table[(7(4^n) - 1)/3, {n, 0, 24}] (* Alonso del Arte, Feb 11 2012 *)
CoefficientList[Series[(2-x)/(1-5*x+4*x^2),{x,0,30}],x] (* or *) LinearRecurrence[{5,-4},{2,9},30] (* Vincenzo Librandi, Mar 20 2012 *)

vector(20,n,(7*4^(n-1)-1)/3) \\ Derek Orr, Apr 12 2015

G.f.: (2-x)/(1-5*x+4*x^2). - Bruno Berselli, Feb 12 2012
a(n) = A083597(n)+1. - Bruno Berselli, Feb 12 2012
a(n) = 4*a(n-1)+1 for n>0, a(0)=2. - Bruno Berselli, Oct 22 2015
a(n) = 7*A002450(n) + 2. - Yosu Yurramendi, Jan 24 2017
A006666(a(n)) = 2*n+11 for n > 0. - Juan Miguel Barga Pérez, Jun 18 2020
a(n) = 5*a(n-1) - 4*a(n-2) for n >= 2. - Wesley Ivan Hurt, Jun 30 2020
a(n) = A178415(3, n) = A347834(4, n-1), arrays, for n >= 1.- Wolfdieter Lang, Nov 29 2021

A332452 Starting from sigma(n), number of halving steps to reach 1 in '3x+1' problem, or -1 if this never happens.

Original entry on oeis.org

0, 5, 2, 11, 6, 7, 3, 12, 7, 14, 7, 13, 12, 8, 8, 67, 14, 23, 6, 7, 5, 15, 8, 14, 67, 7, 7, 14, 13, 16, 5, 68, 9, 71, 9, 59, 15, 14, 14, 13, 7, 10, 12, 8, 24, 16, 9, 69, 22, 13, 16, 18, 71, 15, 16, 15, 8, 13, 14, 9, 68, 10, 10, 31, 8, 17, 11, 69, 10, 17, 16, 76, 16, 23, 69, 12, 10, 9, 8, 14, 61, 69, 8, 16, 72, 20, 15, 14, 13, 16, 15, 9, 7, 17

Offset: 1

Antti Karttunen, Feb 16 2020

nonn

Antti Karttunen, Table of n, a(n) for n = 1..8192
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for sequences related to sigma(n)

Cf. A000203, A006666, A332209, A332453, A332454.

A006666(n) = { my(t); while(n>1, if(n%2, n=3*n+1, n>>=1; t++)); (t); }; \\ From A006666
A332452(n) = A006666(sigma(n));

a(n) = A006666(A000203(n)).

a(n) = A332209(n) - A332453(n).

A375267 Number of iterations of the A375265 map to reach 1 starting from n, or -1 if 1 is never reached.

Original entry on oeis.org

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

Offset: 1

Paolo Xausa, Aug 09 2024

			a(10) = 6 because the trajectory 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 consists of 6 steps.

(Row lengths of A375266) - 1.

Cf. A006577, A006666, A375280.

A375265[n_] := Which[Divisible[n, 3], n/3, Divisible[n, 2], n/2, True, 3*n + 1];
Array[Length[NestWhileList[A375265, #, # > 1 &]] - 1 &, 100]

A082984 Numbers k for which the 3x+1 problem takes at least k halving and tripling steps to reach 1.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 11, 14, 15, 18, 19, 27, 31, 41, 47, 54, 55, 62, 63, 71, 73, 82, 83, 91, 94, 95, 97, 108, 109, 110

Offset: 1

Hauke Worpel (hw1(AT)email.com), May 29 2003

This sequence is almost certainly full; there are no more terms below 100000.
An exhaustive search revealed no further results up to 10^9. - Gonzalo Ciruelos, Aug 02 2013
If we do not count the initial number, then 1 and 2 do not appear in this sequence. It appears that no number k has a Collatz iteration requiring more than 5*k iterations. - T. D. Noe, Feb 21 2014

			3 is in the list because it takes A006577(3) = 7 steps to reach 1 (3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1).

Gonzalo Ciruelos, Python script to generate the sequence.
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A006577, A006666, A008908.

Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Select[Range[1000], Length[Collatz[#]] >= # &] (* T. D. Noe, Feb 21 2014 *)

A332454 Starting from sigma(n)+1, number of halving steps to reach 1 in '3x+1' problem, or -1 if this never happens.

Original entry on oeis.org

1, 2, 4, 3, 11, 7, 13, 4, 12, 14, 7, 13, 12, 16, 16, 5, 14, 7, 6, 20, 18, 15, 16, 14, 5, 20, 69, 22, 67, 73, 18, 6, 17, 71, 17, 13, 23, 14, 22, 59, 20, 75, 12, 8, 24, 73, 17, 69, 14, 67, 73, 18, 71, 61, 73, 61, 16, 59, 14, 33, 68, 75, 26, 7, 8, 74, 11, 31, 75, 74, 73, 19, 11, 23, 69, 12, 75, 33, 16, 30, 15, 31, 8, 35, 72, 20

Offset: 1

Antti Karttunen, Feb 16 2020

nonn

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for sequences related to sigma(n)

Cf. A000203, A006666, A088580, A332452, A332455.

Cf. also A324294.

A006666(n) = { my(t=0); while(n>1, if(n%2, n=3*n+1, n>>=1; t++)); (t); }; \\ From A006666
A332454(n) = A006666(1+sigma(n));

a(n) = A006666(A088580(n)) = A006666(1+sigma(n)).

A375909 Number of iterations of the Farkas map (A349407) to reach 1 starting from 2*n - 1.

Original entry on oeis.org

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

Offset: 1

Paolo Xausa, Sep 02 2024

			a(10) = 5 because the trajectory 19 -> 29 -> 15 -> 5 -> 3 -> 1 takes 5 steps.

Paolo Xausa, Table of n, a(n) for n = 1..10000

(Row lengths of A350279) - 1.

Cf. A006577, A006666, A349407, A375267, A375911.

FarkasStep[x_] := Which[Divisible[x, 3], x/3, Mod[x, 4] == 3, (3*x + 1)/2, True, (x + 1)/2];
Array[Length[FixedPointList[FarkasStep, 2*# - 1]] - 2 &, 100]

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

A064433 Number of iterations of A064455 to reach 2 (or 1 in the case of 1).

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A092892 Smallest starting value in a Collatz '3x+1' sequence such that the sequence contains exactly n halving steps.

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A127789 Record indices of 2^h(n)/(3^t(n)*n), where h and t are the number of halving resp. tripling steps in the '3x + 1' problem.

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A206374 a(n) = (7*4^n - 1)/3.

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Magma

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A332452 Starting from sigma(n), number of halving steps to reach 1 in '3x+1' problem, or -1 if this never happens.

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A375267 Number of iterations of the A375265 map to reach 1 starting from n, or -1 if 1 is never reached.

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Mathematica

A082984 Numbers k for which the 3x+1 problem takes at least k halving and tripling steps to reach 1.

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A332454 Starting from sigma(n)+1, number of halving steps to reach 1 in '3x+1' problem, or -1 if this never happens.

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