A126241 -id:A126241 - cp's OEIS frontend

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

0, 2, 1, 5, 2, 8, 3, 11, 4, 14, 5, 17, 6, 20, 7, 23, 8, 26, 9, 29, 10, 32, 11, 35, 12, 38, 13, 41, 14, 44, 15, 47, 16, 50, 17, 53, 18, 56, 19, 59, 20, 62, 21, 65, 22, 68, 23, 71, 24, 74, 25, 77, 26, 80, 27, 83, 28, 86, 29, 89, 30, 92, 31, 95, 32, 98, 33, 101, 34, 104

Offset: 0

Mohammad K. Azarian

This is the function usually denoted by T(n) in the literature on the 3x+1 problem. See A006370 for further references and links.
Intertwining of sequence A016789 '2,5,8,11,... ("add 3")' and the nonnegative integers.
a(n) = log_2(A076936(n)). - Amarnath Murthy, Oct 19 2002
The average value of a(0), ..., a(n-1) is A004526(n). - Amarnath Murthy, Oct 19 2002
Partial sums are A093353. - Paul Barry, Mar 31 2008
Absolute first differences are essentially in A014681 and A103889. - R. J. Mathar, Apr 05 2008
Only terms of A016789 occur twice, at positions given by sequences A005408 (odd numbers) and A016957 (6n+4): (1,4), (3,10), (5,16), (7,22), ... - Antti Karttunen, Jul 28 2017
a(n) represents the unique congruence class modulo 2n+1 that is represented an odd number of times in any 2n+1 consecutive oblong numbers (A002378). This property relates to Jim Singh's 2018 formula, as n^2 + n is a relevant oblong number. - Peter Munn, Jan 29 2022

			a(3) = -3*(-1) - 2*1 - 1*(-1) - 0*1 + 1*(-1) + 2*1 + 3*(-1) + 4*1 + 5*(-1) + 6*1 = 5. - _Bruno Berselli_, Dec 14 2015

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.

N. J. A. Sloane (terms 0..1000) & Antti Karttunen, Table of n, a(n) for n = 0..100000
C. J. Everett, Iteration of the number-theoretic function f(2n) = n, f(2n + 1) = 3n + 2, Advances in Mathematics, Volume 25, Issue 1, July 1977, Pages 42-45.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
Keenan Monks, Kenneth G. Monks, Kenneth M. Monks, and Maria Monks, Strongly sufficient sets and the distribution of arithmetic sequences in the 3x+1 graph, arXiv:1204.3904v1 [math.DS], Apr 17 2012.
R. Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252.
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A002378, A004526, A076936, A139391, A016116, A126241, A060412, A060413, A006370, A070168 (iterations), A005408, A016957, A064455, A153285, A008619, A193356.
Bisections: A001477, A016789.
Cf. A093353.

a014682 n = if r > 0 then div (3 * n + 1) 2 else n'
            where (n', r) = divMod n 2
-- Reinhard Zumkeller, Oct 03 2014

[IsOdd(n) select (3*n+1)/2 else n/2: n in [0..52]]; // Vincenzo Librandi, Sep 28 2018

T:=proc(n) if n mod 2 = 0 then n/2 else (3*n+1)/2; fi; end; # N. J. A. Sloane, Jan 31 2011
A076936 := proc(n) option remember ; local apr,ifr,me,i,a ; if n <=2 then n^2 ; else apr := mul(A076936(i),i=1..n-1) ; ifr := ifactors(apr)[2] ; me := -1 ; for i from 1 to nops(ifr) do me := max(me, op(2,op(i,ifr))) ; od ; me := me+ n-(me mod n) ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; if a = A076936(n-1) then me := me+n ; a := 1 ; for i from 1 to nops(ifr) do a := a*op(1,op(i,ifr))^(me-op(2,op(i,ifr))) ; od ; fi ; RETURN(a) ; fi ; end: A014682 := proc(n) log[2](A076936(n)) ; end: for n from 1 to 85 do printf("%d, ",A014682(n)) ; od ; # R. J. Mathar, Mar 20 2007

Collatz[n_?OddQ] := (3n + 1)/2; Collatz[n_?EvenQ] := n/2; Table[Collatz[n], {n, 0, 79}] (* Alonso del Arte, Apr 21 2011 *)
LinearRecurrence[{0, 2, 0, -1}, {0, 2, 1, 5}, 70] (* Jean-François Alcover, Sep 23 2017 *)
Table[If[OddQ[n], (3 n + 1) / 2, n / 2], {n, 0, 60}] (* Vincenzo Librandi, Sep 28 2018 *)

a(n)=if(n%2,3*n+1,n)/2 \\ Charles R Greathouse IV, Sep 02 2015

a(n)=if(n<2,2*n,(n^2-n-1)%(2*n+1)) \\ Jim Singh, Sep 28 2018

def a(n): return n//2 if n%2==0 else (3*n + 1)//2
print([a(n) for n in range(101)]) # Indranil Ghosh, Jul 29 2017

From Paul Barry, Mar 31 2008: (Start)
G.f.: x*(2 + x + x^2)/(1-x^2)^2.
a(n) = (4*n+1)/4 - (2*n+1)*(-1)^n/4. (End)
a(n) = -a(n-1) + a(n-2) + a(n-3) + 4. - John W. Layman
For n > 1 this is the image of n under the modified "3x+1" map (cf. A006370): n -> n/2 if n is even, n -> (3*n+1)/2 if n is odd. - Benoit Cloitre, May 12 2002
O.g.f.: x*(2+x+x^2)/((-1+x)^2*(1+x)^2). - R. J. Mathar, Apr 05 2008
a(n) = 5/4 + (1/2)*((-1)^n)*n + (3/4)*(-1)^n + n. - Alexander R. Povolotsky, Apr 05 2008
a(n) = Sum_{i=-n..2*n} i*(-1)^i. - Bruno Berselli, Dec 14 2015
a(n) = Sum_{k=0..n-1} Sum_{i=0..k} C(i,k) + (-1)^k. - Wesley Ivan Hurt, Sep 20 2017
a(n) = (n^2-n-1) mod (2*n+1) for n > 1. - Jim Singh, Sep 26 2018
The above formula can be rewritten to show a pattern: a(n) = (n*(n+1)) mod (n+(n+1)). - Peter Munn, Jan 29 2022
Binary: a(n) = (n shift left (n AND 1)) - (n shift right 1) = A109043(n) - A004526(n). - Rudi B. Stranden, Jun 15 2021
From Rudi B. Stranden, Mar 21 2022: (Start)
a(n) = A064455(n+1) - 1, relating the number ON cells in row n of cellular automaton rule 54.
a(n) = 2*n - A071045(n).
(End)
E.g.f.: (1 + x)*sinh(x)/2 + 3*x*cosh(x)/2 = ((4*x+1)*e^x + (2*x-1)*e^(-x))/4. - Rénald Simonetto, Oct 20 2022
a(n) = n*(n mod 2) + ceiling(n/2) = A193356(n) + A008619(n+1). - Jonathan Shadrach Gilbert, Mar 12 2023
a(n) = 2*a(n-2) - a(n-4) for n > 3. - Chai Wah Wu, Apr 17 2024

Edited by N. J. A. Sloane, Apr 26 2008, at the suggestion of Artur Jasinski

Edited by N. J. A. Sloane, Jan 31 2011

A020914 Number of digits in the base-2 representation of 3^n.

Original entry on oeis.org

1, 2, 4, 5, 7, 8, 10, 12, 13, 15, 16, 18, 20, 21, 23, 24, 26, 27, 29, 31, 32, 34, 35, 37, 39, 40, 42, 43, 45, 46, 48, 50, 51, 53, 54, 56, 58, 59, 61, 62, 64, 65, 67, 69, 70, 72, 73, 75, 77, 78, 80, 81, 83, 85, 86, 88, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 105, 107

Offset: 0

Clark Kimberling

Also, numbers k such that the first digit in the ternary expansion of 2^k is 1. - Mohammed Bouayoun (Mohammed.bouayoun(AT)sanef.com), Apr 24 2006
a(n) is the smallest integer such that n/a(n) < log_2(3). - Trevor G. Hyde (thyde12(AT)amherst.edu), Jul 31 2008
This sequence represents allowable values of the "dropping time" in the Collatz (3x+1) problem when iterated according to the function f(n) := n/2 if n is even, (3n+1)/2 otherwise, as tabulated in A126241. There is one exception, A126241(1), which has been set to zero by convention. - K. Spage, Oct 22 2009
An integer k is a term of A020914 if and only if floor(k*(1 + log(2)/log(3))) - abs(k-1)*(1 + log(2)/log(3)) - 1 >= 0. - K. Spage, Oct 22 2009
Also smallest k such that ceiling(2^k / 3^n) = 2. - Michel Lagneau, Jan 31 2012
For n > 0, first differences of A022330. - Michel Marcus, Oct 03 2013
Also the number of powers of two less than or equal to 3^n. - Robert G. Wilson v, May 25 2014
Except for 1, A020914 is the complement of A054414 and therefore these two form a pair of Beatty sequences. - Robert G. Wilson v, May 25 2014

T. D. Noe, R. J. Mathar, Table of n, a(n) for n = 0..20000
Mike Winkler, On the structure and the behaviour of Collatz 3n + 1 sequences, 2014.
Mike Winkler, New results on the stopping time behaviour of the Collatz 3x + 1 function, arXiv:1504.00212 [math.GM], 2015.
Mike Winkler, The algorithmic structure of the finite stopping time behavior of the 3x + 1 function, arXiv:1709.03385 [math.GM], 2017.

Cf. A056576, A054414, A070939, A000244, A227048, A022330, A022921 (first differences), A126241.
Cf. A020857 (decimal expansion of log_2(3)).
Cf. A020915.
Cf. A204399 (essentially the same).

a020914 = a070939 . a000244  -- Reinhard Zumkeller, Jun 30 2013

A020914 :=n->nops(convert(3^n,base,2)):
seq(A020914(n),n=0..70); # Emeric Deutsch, Apr 30 2006
seq(ilog2(3^n)+1, n=0 .. 100); # Robert Israel, Dec 12 2014

Table[Length[IntegerDigits[3^n, 2]], {n, 0, 100}] (* Stefan Steinerberger, Apr 19 2006 *)
a[n_] := Floor[ Log2[3^n] + 1]; Array[a, 105, 0] (* Robert G. Wilson v, May 25 2014 *)

for(n=0,100,print1(floor(1+n*log(3)/log(2)),",")) \\ K. Spage, Oct 22 2009

a(n)=exponent(3^n)+1 \\ Charles R Greathouse IV, Nov 03 2022

def A020914(n): return (3**n).bit_length() # Chai Wah Wu, Oct 09 2024

a(n) = floor(1 + n*log(3)/log(2)). - K. Spage, Oct 22 2009
a(0) = 1, a(n+1) = a(n) + A022921(n). - K. Spage, Oct 23 2009
a(n) = A122437(n-1) - n. - K. Spage, Oct 23 2009
A098294(n) = a(n) + n for n > 0. - Mike Winkler, Dec 31 2010
a(n) = A070939(A000244(n)) = length of n-th row in triangle A227048. - Reinhard Zumkeller, Jun 30 2013
a(n) = 1 + floor(n*log_2(3)) = 1 + A056576(n) = 1 + floor(n*A020857). - L. Edson Jeffery, Dec 12 2014
A020915(a(n)) = n + 1. - Reinhard Zumkeller, Mar 28 2015

More terms from Stefan Steinerberger, Apr 19 2006

A060412 In the '3x+1' problem, these values for the starting value set new records for the "dropping time", number of steps to reach a lower value than the start.

Original entry on oeis.org

2, 3, 7, 27, 703, 10087, 35655, 270271, 362343, 381727, 626331, 1027431, 1126015, 8088063, 13421671, 20638335, 26716671, 56924955, 63728127, 217740015, 1200991791, 1827397567, 2788008987, 12235060455

Offset: 1

N. J. A. Sloane, Apr 06 2001; b-file added Nov 27 2007

nonn

The (3x+1)/2 steps and the halving steps are counted. - Don Reble, May 13 2006

Where records occur in A102419 (could be prefixed by an initial 1). - N. J. A. Sloane, Oct 20 2012

			See A102419.

N. J. A. Sloane, Table of n, a(n) for n = 1..35 (from the web page of Tomás Oliveira e Silva)
Tomás Oliveira e Silva, Tables
Eric Roosendaal, On the 3x + 1 problem
N. J. A. Sloane, First 36 terms of A217934 and A060412 [From Roosendaal web site]
Index entries for sequences related to 3x+1 (or Collatz) problem

A060413 gives associated "dropping times", A060414 the maximal values and A060415 the steps at which the maxima occur. See also A217934.

Cf. A060445, A008884, A161021, A161022, A161023, A014682, A126241.

dcoll[n_]:=Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>=n&]]; t={max=2}; Do[If[(y=dcoll[n])>max,max=y; AppendTo[t,n]],{n,3,1130000,4}]; t (* Jayanta Basu, May 28 2013 *)

A102419 "Dropping time" in 3x+1 problem starting at n (number of steps to reach a lower number than starting value); a(1) = 0 by convention. Also called glide(n).

Original entry on oeis.org

0, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 96, 1, 3, 1, 91, 1, 3, 1, 6, 1, 3, 1, 13, 1, 3, 1, 8, 1, 3, 1, 88, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 88, 1, 3, 1, 6, 1, 3, 1, 83, 1, 3, 1, 8, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 73, 1, 3, 1, 13, 1, 3, 1, 6

Offset: 1

N. J. A. Sloane, Sep 15 2006

nonn

			1: 0 steps
2 1: 1 step
3 10 5 16 8 4 2 1: 6 steps (before it drops below n)
4 2 1: 1 step
5 16 8 4 2 1: 3 steps
6 3 ...: 1 step
7 22 11 34 17 52 26 13 40 20 10 5 ...: 11 steps
...
Records: 0.1.6.11.96.132...171... (A217934)
at.......1.2.3..7.27.703.10087... (A060412)

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
Eric Roosendaal, On the 3x + 1 problem
N. J. A. Sloane, First 36 terms of A217934 and A060412 [From Roosendaal web site]

Cf. A060445, A074473, A126241.

For records see A060412, A217934. - N. J. A. Sloane, Oct 20 2012

Prepend[Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>=n&]],{n,2,99}],1]-1 (* Jayanta Basu, May 28 2013 *)

def a(n):
    if n<3: return n - 1
    N=n
    x=0
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        x+=1
        if nIndranil Ghosh, Apr 22 2017

a(2n) = 1; a(2n+1) = A060445(n).
a(n) = A074473(n)-1 for n>1.
a(n) = floor(A126241(n)*(1+log(2)/log(3))). - K. Spage, Oct 22 2009

A060413 In the '3x+1' problem, take the sequence of starting values which set new records for the "dropping time" (A060412); sequence gives associated dropping times.

Original entry on oeis.org

1, 4, 7, 59, 81, 105, 135, 164, 165, 173, 176, 183, 224, 246, 287, 292, 298, 308, 376, 395, 398, 433, 447, 547, 550, 606, 688, 712, 722, 728, 886, 902, 966, 990, 1005

Offset: 1

N. J. A. Sloane, Apr 06 2001; b-file added Nov 27 2007

nonn

N. J. A. Sloane, Table of n, a(n) for n = 1..35 (from the web page of Tomás Oliveira e Silva)
Tomás Oliveira e Silva, Tables
Index entries for sequences related to 3x+1 (or Collatz) problem

Cf. A126241, A060412.

A186107 Numerator of the frequency of the n-th dropping time in the Collatz iteration.

Original entry on oeis.org

1, 1, 1, 1, 3, 7, 3, 15, 85, 173, 119, 961, 663, 8045, 17637, 51033, 54475, 312455, 663535, 950235, 5936673, 1684037, 39993895, 87986917, 128989251, 205059181, 949737339, 2861515293, 400296173, 19018424205

Offset: 1

T. D. Noe, Feb 12 2011

The possible dropping times are in A020914. The denominators are in A186108. The cumulative frequency of the dropping time is A186109(n)/A186110(n).

			The frequencies are 1/2, 1/4, 1/16, 1/16, 3/128, 7/256, 3/256, 15/2048, 85/8192,....

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

Cf. A126241 (dropping times)

a(n) = numerator of A186009(n) / 2^A020914(n-1).

A186108 Denominator of the frequency of the n-th dropping time in the Collatz iteration.

Original entry on oeis.org

2, 4, 16, 16, 128, 256, 256, 2048, 8192, 32768, 16384, 262144, 262144, 2097152, 8388608, 16777216, 33554432, 134217728, 536870912, 1073741824, 4294967296, 2147483648, 34359738368, 137438953472, 274877906944, 274877906944, 2199023255552, 4398046511104, 1099511627776, 35184372088832

Offset: 1

T. D. Noe, Feb 12 2011

The numerators are in A186107.

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

Cf. A126241 (dropping times)

a(n) = denominator of A186009(n) / 2^A020914(n-1).

A186109 Numerator of the cumulative frequency of the dropping time in the Collatz iteration.

Original entry on oeis.org

1, 3, 13, 7, 115, 237, 15, 1935, 7825, 31473, 31711, 254649, 15957, 2050541, 8219801, 16490635, 33035745, 132455435, 530485275, 1061920785, 4253619813, 4256987887, 34095896991, 136471574881, 273072139013, 136638599097, 2187167322891, 4377196161075, 4378797345767, 35049397190341

Offset: 1

T. D. Noe, Feb 12 2011

The possible dropping times are in A020914. The denominators are in A186110. The frequency of the n-th dropping time is A186107(n)/A186108(n).

Riho Terras' classic paper about the Collatz problem shows the decimal values of 2(1-c(k)) in Table A, where c(k) is the cumulative frequency of dropping times <= k.

			The cumulative frequencies are 1/2, 3/4, 13/16, 7/8, 115/128, 237/256, 15/16, 1935/2048, 7825/8192, ... .

Index entries for sequences related to 3x+1 (or Collatz) problem
Riho Terras, A stopping time problem on the positive integers, ACTA Arith. 30 (1976), 241-252.

Cf. A126241 (dropping times).

a(n) = numerator of Sum_{k=1..n} A186009(k) / 2^A020914(k-1).

A186008 Irregular triangle T(n,k) read by rows, in which row n has the pattern of conjectured dropping times in the Collatz iteration.

Original entry on oeis.org

2, 4, 16, 12, 32, 8, 52, 128, 40, 56, 84, 136, 160, 180, 256, 60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024, 152, 232, 384, 648, 704, 788, 856, 1000, 1204, 1416, 1472, 1556, 1592, 1624, 1800, 1972, 2008, 2120, 2356, 2360, 2676, 2744, 2888, 2912, 3064, 3328, 3444, 3680, 3832, 4096

Offset: 1

T. D. Noe, Feb 09 2011

Consider A126241, the sequence of dropping times in the Collatz iteration. Only zero and the numbers in A020914 can be dropping times. The dropping times in A126241 have a definite pattern. For example, 1 appears at positions n = 2 + 2*i, for i=0,1,2,3,... Similarly, 2 appears at positions n = 5 + 4*i; 4 appears at n = 3 + 16*i; 5 appears at n = 11 + {12,32}*i; and 7 appears at 7 + {8, 52, 128}*i. In general, if we let s=A020914(r) be the r-th possible stopping time, then A126241(n) = s for n = A122442(r) + T(r)*i, where T(r) is the r-th row of this triangle.  The length of row n is A186009(n). The n-th row ends with 2^A020914(n).
The frequency of the r-th dropping time s=A020914(r) can be computed as A186009(r)/2^s. The first few frequencies are 1/2, 1/4, 1/16, 1/16, 3/128, 7/256, 3/256, 15/2048, and 85/8192.
The term "stopping time" is sometimes used instead of "dropping time", but the former usually refers to A006666.
This sequence is closely related to A177789.

			The triangle begins
2
4
16
12, 32
8, 52, 128
40, 56, 84, 136, 160, 180, 256
60, 80, 136, 220, 288, 296, 448, 528, 636, 688, 712, 1024

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010. See pp. 33, 35ff.

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

Cf. A014682, A126241, A006666, A186009.

cp's OEIS Frontend

A014682 The Collatz or 3x+1 function: a(n) = n/2 if n is even, otherwise (3n+1)/2.

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A020914 Number of digits in the base-2 representation of 3^n.

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A060412 In the '3x+1' problem, these values for the starting value set new records for the "dropping time", number of steps to reach a lower value than the start.

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A102419 "Dropping time" in 3x+1 problem starting at n (number of steps to reach a lower number than starting value); a(1) = 0 by convention. Also called glide(n).

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A060413 In the '3x+1' problem, take the sequence of starting values which set new records for the "dropping time" (A060412); sequence gives associated dropping times.

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A186107 Numerator of the frequency of the n-th dropping time in the Collatz iteration.

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A186108 Denominator of the frequency of the n-th dropping time in the Collatz iteration.

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