A102419 -id:A102419 - cp's OEIS frontend

A217934 Records in A102419.

0, 1, 6, 11, 96, 132, 171, 220, 267, 269, 282, 287, 298, 365, 401, 468, 476, 486, 502, 613, 644, 649, 706, 729, 892, 897, 988, 1122, 1161, 1177, 1187, 1445, 1471, 1575, 1614, 1639

Offset: 1

N. J. A. Sloane, Oct 20 2012

If the "1" is omitted, also records in A060445.

			See A102419.

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]

Cf. A102419, A060445, A060412.

A074473 Dropping time for the 3x+1 problem: for n >= 2, number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated starting at n; a(1)=1 by convention.

Original entry on oeis.org

1, 2, 7, 2, 4, 2, 12, 2, 4, 2, 9, 2, 4, 2, 12, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 97, 2, 4, 2, 92, 2, 4, 2, 7, 2, 4, 2, 14, 2, 4, 2, 9, 2, 4, 2, 89, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 12, 2, 4, 2, 89, 2, 4, 2, 7, 2, 4, 2, 84, 2, 4, 2, 9, 2, 4, 2, 14, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 74, 2, 4, 2, 14, 2, 4, 2, 7

Offset: 1

Labos Elemer, Sep 19 2002

nonn

Here we call the starting value iteration number 1, although usually the count is started at 0, which would subtract 1 from the values for n >= 2 - see A060445, A102419.

			n=2k: then a(2k)=2 because the second iterate is k<n=2k, the first iterate below 2k; n=4k+1, k>1: the list = {4k+1, 12k+4, 6k+2, 3k+1, ...} i.e. the 4th term is always the first below initial value, so a(4k+1)=4;
n=15: the list={15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1} and 12th term is first sinks below iv=15, so a(15)=12; relatively larger values occur at n=4k+3.
n=3: the list is {3, 10, 5, 16, 8, 4, 2, 1, ..}, the 7th term is 2, which is the first smaller than 3, so a(3)=7.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000

Cf. A006370, A075476, A075477, A075478, A075479, A075480, A075481, A075482, A075483, A060445, A060412, A217934.

Equals A102419(n)+1.

nextx[x_Integer] := If[OddQ@x, 3x + 1, x/2]; f[1] = 1; f[n_] := Length@ NestWhileList[nextx, n, # >= n &]; Array[f, 83] (* Bobby R. Treat (drbob(at)bigfoot.com), Sep 16 2006 *)

A074473(n) = if (n<3, n,  my(N=n, x=1); while (1, if (n%2==0, n/=2, n = 3*n + 1); x++; if (nMichel Marcus, Aug 15 2025

def a(n):
    if n<3: return n
    N=n
    x=1
    while True:
        if n%2==0: n/=2
        else: n = 3*n + 1
        x+=1
        if nIndranil Ghosh, Apr 15 2017

Edited by N. J. A. Sloane, Sep 15 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 *)

A126241 Dropping times in the 3n+1 problem (or the Collatz problem). Let T(n):=n/2 if n is even, (3n+1)/2 otherwise (A014682). Let a(n) be the smallest integer k such that T^(k)(n)

Original entry on oeis.org

0, 1, 4, 1, 2, 1, 7, 1, 2, 1, 5, 1, 2, 1, 7, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 59, 1, 2, 1, 56, 1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 5, 1, 2, 1, 54, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 7, 1, 2, 1, 54, 1, 2, 1, 4, 1, 2, 1, 51, 1, 2, 1, 5, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 45, 1, 2, 1, 8, 1, 2, 1, 4

Offset: 1

Christof Menzel (christof.menzel(AT)hs-niederrhein.de), Mar 08 2007

nonn

Also called "stopping times", although that term is usually reserved for A006666.
From K. Spage, Oct 22 2009, corrected Aug 21 2014: (Start)
Congruency relationship: For n>1 and m>1, all m congruent to n mod 2^(a(n)) have a dropping time equal to a(n).
By refining the definition of the dropping time to "starting with x=n, iterate x until (abs(x) <= abs(n))" the above congruency relationship holds for all nonnegative values of n and all positive or negative values of m including zero.
By this refined definition, a(1)=2 rather than the usual zero set by convention. All other values of positive a(n) remain unchanged. (End)
Terras defines a coefficient stopping time (definition 1.5) tau(n) = d which is the smallest d for which 3^u/2^d < 1 where u is the number of tripling steps among the first d steps starting from n. Clearly tau(n) <= a(n), and Terras conjectures (conjecture 2.9) that tau(n) = a(n) for n>=2. - Olivier Rozier, May 13 2024

			s(15) = 7, since the trajectory {T^(k)(15)} (k=1,2,3,...) equals 23,35,53,80,40,20,10.

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

T. D. Noe, Table of n, a(n) for n = 1..10000
J. C. Lagarias, The 3x+1 Problem: An Annotated Bibliography (1963-1999), arXiv:math/0309224 [math.NT], 2003-2011.
Olivier Rozier and Claude Terracol, Paradoxical behavior in Collatz sequences, arXiv:2502.00948 [math.GM], 2025. See p. 2.
Riho Terras, A stopping time problem on the positive integers, Acta Arith. 30 (1976) 241-252, with definition 0.1 chi(n) = a(n).
Index entries for sequences related to 3x+1 (or Collatz) problem

See A074473, which is the main entry for dropping times.
Cf. A014682, A006666, A006577.
Records: A060412, A060413.
Cf. A020914 (allowable dropping times). - K. Spage, Aug 22 2014

Collatz2[n_] := If[n<2, {}, Rest[NestWhileList[If[EvenQ[#], #/2, (3 # + 1)/2] &, n, # >= n &]]]; Table[Length[Collatz2[n]], {n, 1, 1000}]

a(n) = ceiling(A102419(n)/(1+log(2)/log(3))). - K. Spage, Aug 22 2014

Broken link fixed by K. Spage, Oct 22 2009

A060445 "Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).

Original entry on oeis.org

0, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 96, 3, 91, 3, 6, 3, 13, 3, 8, 3, 88, 3, 6, 3, 8, 3, 11, 3, 88, 3, 6, 3, 83, 3, 8, 3, 13, 3, 6, 3, 8, 3, 73, 3, 13, 3, 6, 3, 68, 3, 8, 3, 50, 3, 6, 3, 8, 3, 13, 3, 24, 3, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 65, 3, 34, 3, 6, 3, 47, 3, 8, 3, 13, 3, 6, 3, 8, 3

Offset: 0

N. J. A. Sloane, Apr 07 2001

If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).

The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - Juhani Heino, Jul 21 2017

			3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2, taking 6 steps, so a(1) = 6.

T. D. Noe, Table of n, a(n) for n = 0..10000
Jason Holt, Plot of first 1 billion terms, log scale on x axis
Jason Holt, Plot of first 10 billion terms, log scale on x axis
Eric Roosendaal, On the 3x + 1 problem
Index entries for sequences related to 3x+1 (or Collatz) problem

A060565 gives the first lower number that is reached. Cf. A060412-A060415, A217934.
See A074473, A102419 for other versions of this sequence.
Cf. A122437 (allowable dropping times), A122442 (least k having dropping time A122437(n)).
Cf. A070165.

a060445 0 = 0
a060445 n = length $ takeWhile (>= n') $ a070165_row n'
            where n' = 2 * n + 1
-- Reinhard Zumkeller, Mar 11 2013

nxt[n_]:=If[OddQ[n],3n+1,n/2]; Join[{0},Table[Length[NestWhileList[nxt, n,#>=n&]]-1, {n,3,191,2}]]  (* Harvey P. Dale, Apr 23 2011 *)

def a(n):
    if n<1: return 0
    n=2*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

More terms from Jason Earls, Apr 08 2001 and from Michel ten Voorde Apr 09 2001

Still more terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2001

A352891 Number of iterations of map x -> A341515(x) needed to reach x < n when starting from x=n, or 0 if such number is never reached. Here A341515 is the Collatz or 3x+1 map (A006370) conjugated by unary-binary-encoding (A156552).

Original entry on oeis.org

0, 0, 1, 6, 1, 3, 1, 11, 1, 3, 1, 9, 1, 7, 1, 11, 1, 3, 1, 6, 1, 6, 1, 9, 1, 11, 1, 7, 1, 1, 1, 91, 1, 106, 1, 5, 1, 16, 1, 14, 1, 4, 1, 7, 1, 20, 1, 89, 1, 3, 1, 7, 1, 3, 1, 87, 1, 21, 1, 1, 1, 50, 1, 92, 1, 5, 1, 18, 1, 3, 1, 8, 1, 98, 1, 14, 1, 5, 1, 14, 1, 34, 1, 6, 1, 35, 1, 12, 1, 2, 1, 21, 1, 71, 1, 90, 1, 3

Offset: 1

Antti Karttunen, Apr 08 2022

nonn

This is one possible analog for A102419 ("Dropping time" sequence) when computed for A341515. See also A352894.

Cf. A000040, A005940, A006577, A064989, A156552, A329603, A352890, A352894.

Cf. also A102419.

A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t); };
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A156552(n) = { my(f = factor(n), p, p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res };
A329603(n) = A005940(2+(3*A156552(n)));
A341515(n) = if(n%2, A064989(n), A329603(n));
A352891(n) = if(n<=2, 0, my(k=0,x=n); while(x>=n, x = A341515(x); k++); (k));

For n >= 1, a(2n+1) = 1.

For n >= 1, A352894(n) <= a(n) <= A352890(n).

A290101 a(n) = dropping time for the modified Collatz problem, where x -> 3x+1 if x is odd, and x -> either x/2 or 3x+1 if x is even (minimal number of any such steps to reach a lower number than the starting value n); a(1) = 0 by convention.

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, 16, 1, 3, 1, 19, 1, 3, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 16, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 19, 1, 3, 1, 6, 1, 3, 1, 13, 1, 3, 1, 8, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 16, 1, 3, 1, 8, 1, 3

Offset: 1

Antti Karttunen, Aug 20 2017

nonn

In contrast to the "3x+1" problem (see A006577, A102419), here you are free to choose either step if x is even. The sequence counts the minimum number of optimally chosen steps which leads to a value smaller than the value we started from.

			Starting from n = 27, the following is a shortest path leading to a value smaller than 27: 27 -> 82 -> 41 -> 124 -> 373 -> 1120 -> 560 -> 280 -> 140 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20. It has 16 steps, thus a(27) = 16. Note the 3x+1 step from 124 to 373 which is not allowed in the ordinary Collatz problem.

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

Cf. A127885, A290100, A290102, A000012 (even bisection).

Differs from A102419 for the first time at n=27, where a(27) = 16, while A102419(27) = 96.

A290101(n) = { if(1==n,return(0)); my(S, k); S=[n]; k=0; while( S[1]>=n, k++; S=vecsort( concat(apply(x->3*x+1, S), apply(x->x\2, select(x->x%2==0, S) )), , 8);  ); k } \\ After Max Alekseyev's code for A127885

from sympy import flatten
def ok(n, L):
    return any(i < n for i in L)
def a(n):
    if n==1: return 0
    L=[n]
    i = 0
    while not ok(n, L):
        L=set(flatten([[3*k + 1, k//2] if k%2==0 else 3*k + 1 for k in L]))
        i+=1
    return i
print([a(n) for n in range(1, 121)]) # Indranil Ghosh, Aug 31 2017

a(n) <= A102419(n).

a(n) <= A127885(n) [apart from any hypothetical -1's in A127885].

A318489 Number of steps to reach a lower number than starting value in 7x+-1 problem, or 0 if never reached.

Original entry on oeis.org

0, 1, 12, 1, 8, 1, 4, 1, 4, 1, 42, 1, 8, 1, 4, 1, 4, 1, 23, 1, 20, 1, 4, 1, 4, 1, 12, 1, 16, 1, 4, 1, 4, 1, 282, 1, 12, 1, 4, 1, 4, 1, 229, 1, 50, 1, 4, 1, 4, 1, 8, 1, 35, 1, 4, 1, 4, 1, 8, 1, 50, 1, 4, 1, 4, 1, 46, 1, 8, 1, 4, 1, 4, 1, 225, 1, 8, 1, 4, 1, 4, 1, 35, 1, 16, 1, 4, 1, 4, 1, 46, 1, 27, 1, 4, 1, 4, 1, 16

Offset: 1

David Barina, Aug 27 2018

The least positive k for which the iterate A317640^k(n) < n.
Also called the dropping time, glide, or stopping time.
a(2n) = 1.

			a(5) = 8 because the trajectory is (5, 36, 18, 9, 64, 32, 16, 8, 4, 2, 1, ...) and the first lower number is 4. Thus 8 steps to reach the value 4 starting from the value 5.

David Barina, Table of n, a(n) for n = 1..10000
D. Barina, 7x+-1: Close Relative of Collatz Problem, arXiv:1807.00908 [math.NT], 2018.
K. Matthews, David Barina's 7x+1 conjecture.

Cf. A317640 (7x+-1 function), A102419 (3x+1 equivalent).

int a(int n0) {
        if( n0 == 1 ) return 0;
        int s = 0;
        for(int n = n0; n >= n0; s++) {
                switch(n%4) {
                        case 1: n = 7*n+1; break;
                        case 3: n = 7*n-1; break;
                        default: n = n/2;
                }
        }
        return s;
}

a7(n) = {my(m=(n+2)%4-2); if(m%2, 7*n + m, n/2)};
a(n) = if (n==1, 0, my(nb=1, m=n, nm); while((nm=a7(m)) >= n, m = nm; nb++); nb); \\ Michel Marcus, Aug 28 2018

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A217934 Records in A102419.

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A074473 Dropping time for the 3x+1 problem: for n >= 2, number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated starting at n; a(1)=1 by convention.

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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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A126241 Dropping times in the 3n+1 problem (or the Collatz problem). Let T(n):=n/2 if n is even, (3n+1)/2 otherwise (A014682). Let a(n) be the smallest integer k such that T^(k)(n)

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A060445 "Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).

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A352891 Number of iterations of map x -> A341515(x) needed to reach x < n when starting from x=n, or 0 if such number is never reached. Here A341515 is the Collatz or 3x+1 map (A006370) conjugated by unary-binary-encoding (A156552).

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A290101 a(n) = dropping time for the modified Collatz problem, where x -> 3x+1 if x is odd, and x -> either x/2 or 3x+1 if x is even (minimal number of any such steps to reach a lower number than the starting value n); a(1) = 0 by convention.

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