A135282 -id:A135282 - cp's OEIS frontend

A006577 Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.

0, 1, 7, 2, 5, 8, 16, 3, 19, 6, 14, 9, 9, 17, 17, 4, 12, 20, 20, 7, 7, 15, 15, 10, 23, 10, 111, 18, 18, 18, 106, 5, 26, 13, 13, 21, 21, 21, 34, 8, 109, 8, 29, 16, 16, 16, 104, 11, 24, 24, 24, 11, 11, 112, 112, 19, 32, 19, 32, 19, 19, 107, 107, 6, 27, 27, 27, 14, 14, 14, 102, 22

Offset: 1

N. J. A. Sloane, Bill Gosper

The 3x+1 or Collatz problem is as follows: start with any number n. If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach 1? This is a famous unsolved problem. It is conjectured that the answer is yes.
It seems that about half of the terms satisfy a(i) = a(i+1). For example, up to 10000000, 4964705 terms satisfy this condition.
n is an element of row a(n) in triangle A127824. - Reinhard Zumkeller, Oct 03 2012
The number of terms that satisfy a(i) = a(i+1) for i being a power of ten from 10^1 through 10^10 are: 0, 31, 365, 4161, 45022, 477245, 4964705, 51242281, 526051204, 5378743993. - John Mason, Mar 02 2018
5 seems to be the only number whose value matches its total number of steps (checked to n <= 10^9). - Peter Woodward, Feb 15 2021

			a(5)=5 because the trajectory of 5 is (5,16,8,4,2,1).

R. K. Guy, Unsolved Problems in Number Theory, E16.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
David Eisenbud and Brady Haran, UNCRACKABLE? The Collatz Conjecture, Numberphile video, 2016.
Geometry.net, Links on Collatz Problem
Christian Hercher, There are no Collatz m-Cycles with m <= 91, J. Int. Seq. (2023) Vol. 26, Article 23.3.5.
Jason Holt, Log-log plot of first billion terms
Jason Holt, Plot of 1 billion values of the number of steps to drop below n (A060445), log scale on x axis
Jason Holt, Plot of 10 billion values of the number of steps to drop below n (A060445), log scale on x axis
A. Krowne, Collatz problem, PlanetMath.org.
J. C. Lagarias, The 3x+1 problem and its generalizations, Amer. Math. Monthly, 92 (1985), 3-23.
J. C. Lagarias, How random are 3x+1 function iterates?, in The Mathemagician and the Pied Puzzler - A Collection in Tribute to Martin Gardner, Ed. E. R. Berlekamp and T. Rogers, A. K. Peters, 1999, pp. 253-266.
J. C. Lagarias, The 3x+1 Problem: an annotated bibliography, II (2000-2009), arXiv:0608208 [math.NT], 2006-2012.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.
Jeffrey C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
M. Le Brun, Email to N. J. A. Sloane, Jul 1991
Mathematical BBS, Biblography on Collatz Sequence
P. Picart, Algorithme de Collatz et conjecture de Syracuse
E. Roosendaal, On the 3x+1 problem
J. L. Simons, On the nonexistence of 2-cycles for the 3x+1 problem, Math. Comp. 75 (2005), 1565-1572.
N. J. A. Sloane, "A Handbook of Integer Sequences" Fifty Years Later, arXiv:2301.03149 [math.NT], 2023, p. 8.
G. Villemin's Almanach of Numbers, Cycle of Syracuse
Eric Weisstein's World of Mathematics, Collatz Problem
Wikipedia, Collatz Conjecture
Index entries for sequences related to 3x+1 (or Collatz) problem

See A070165 for triangle giving trajectories of n = 1, 2, 3, ....
Cf. A006370, A125731, A127885, A127886, A008908, A112695, A135282, A208981, A025586.
See also A008884, A161021, A161022, A161023.

import Data.List (findIndex)
import Data.Maybe (fromJust)
a006577 n = fromJust $ findIndex (n `elem`) a127824_tabf
-- Reinhard Zumkeller, Oct 04 2012, Aug 30 2012

A006577 := proc(n)
        local a,traj ;
        a := 0 ;
        traj := n ;
        while traj > 1 do
                if type(traj,'even') then
                        traj := traj/2 ;
                else
                        traj := 3*traj+1 ;
                end if;
                a := a+1 ;
        end do:
        return a;
end proc: # R. J. Mathar, Jul 08 2012

f[n_] := Module[{a=n,k=0}, While[a!=1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Table[f[n],{n,4!}] (* Vladimir Joseph Stephan Orlovsky, Jan 08 2011 *)
Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#!=1&]]-1,{n,80}] (* Harvey P. Dale, May 21 2012 *)

a(n)=if(n<0,0,s=n; c=0; while(s>1,s=if(s%2,3*s+1,s/2); c++); c)

step(n)=if(n%2,3*n+1,n/2);
A006577(n)=if(n==1,0,A006577(step(n))+1); \\ Michael B. Porter, Jun 05 2010

def a(n):
    if n==1: return 0
    x=0
    while True:
        if n%2==0: n//=2
        else: n = 3*n + 1
        x+=1
        if n<2: break
    return x
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 05 2017

def A006577(n):
    ct = 0
    while n != 1: n = A006370(n); ct += 1
    return ct # Ya-Ping Lu, Feb 22 2024

collatz<-function(n) ifelse(n==1,0,1+ifelse(n%%2==0,collatz(n/2),collatz(3*n+1))); sapply(1:72, collatz) # Christian N. K. Anderson, Oct 09 2024

a(n) = A006666(n) + A006667(n).
a(n) = A112695(n) + 2 for n > 2. - Reinhard Zumkeller, Apr 18 2008
a(n) = A008908(n) - 1. - L. Edson Jeffery, Jul 21 2014
a(n) = A135282(n) + A208981(n) (after Alonso del Arte's comment in A208981), if 1 is reached, otherwise a(n) = -1. - Omar E. Pol, Apr 10 2022
a(n) = 2*A007814(n + 1) + a(A085062(n)) + 1 for n > 1. - Wing-Yin Tang, Jan 06 2025

More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001

"Escape clause" added to definition by N. J. A. Sloane, Jun 06 2017

A347270 Square array T(n,k) in which row n lists the 3x+1 sequence starting at n, read by antidiagonals upwards, with n >= 1 and k >= 0.

Original entry on oeis.org

1, 2, 4, 3, 1, 2, 4, 10, 4, 1, 5, 2, 5, 2, 4, 6, 16, 1, 16, 1, 2, 7, 3, 8, 4, 8, 4, 1, 8, 22, 10, 4, 2, 4, 2, 4, 9, 4, 11, 5, 2, 1, 2, 1, 2, 10, 28, 2, 34, 16, 1, 4, 1, 4, 1, 11, 5, 14, 1, 17, 8, 4, 2, 4, 2, 4, 12, 34, 16, 7, 4, 52, 4, 2, 1, 2, 1, 2, 13, 6, 17, 8, 22

Offset: 1

Omar E. Pol, Aug 25 2021

This array gives all 3x+1 sequences.
The 3x+1 or Collatz problem is described in A006370.
Column k gives the image of n at the k-th step.
This infinite square array contains the irregular triangles A070165, A235795 and A347271.
For a piping diagram of the 3x+1 problem see A235800.

			The corner of the square array begins:
   1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, ...
   2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, ...
   3,10, 5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, ...
   4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, ...
   5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, ...
   6, 3,10, 5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, ...
   7,22,11,34,17,52,26,13,40,20,10, 5,16, 8, 4, 2, 1, 4, 2, 1, ...
   8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, ...
   9,28,14, 7,22,11,34,17,52,26,13,40,20,10, 5,16, 8, 4, 2, 1, ...
  10, 5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, ...
  11,34,17,52,26,13,40,20,10, 5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, ...
  12, 6, 3,10, 5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, ...
  13,40,20,10, 5,16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, ...
  14, 7,22,11,34,17,52,26,13,40,20,10, 5,16, 8, 4, 2, 1, 4, 2, ...
...

Paolo Xausa, Table of n, a(n) for n = 1..11325 (antidiagonals 1..150 of the array, flattened)
J. C. Lagarias, The 3x+1 Problem: An Overview, arXiv:2111.02635 [math.NT], 2021.
Index entries for sequences related to 3x+1 (or Collatz) problem

Main diagonal gives A347272.
Parity of this sequence is A347283.
Largest value in row n gives A056959.
Number of nonpowers of 2 in row n gives A208981.
Some rows n are: A153727 (n=1), A033478 (n=3), A033479 (n=9), A033480 (n=15), A033481 (n=21), A008884 (n=27), A008880 (n=33), A008878 (n=39), A008883 (n=51), A008877 (n=57), A008874 (n=63), A258056 (n=75), A258098 (n=79), A008876 (n=81), A008879 (n=87), A008875 (n=95), A008873 (n=97), A008882 (n=99), A245671 (n=1729).
First four columns k are: A000027 (k=0), A006370 (k=1), A075884 (k=2), A076536 (k=3).
Cf. A006877, A014682, A057716, A070165, A078719, A135282, A235795, A235800, A235801, A347265, A347267, A347268, A347269, A347271, A347519.

T:= proc(n, k) option remember; `if`(k=0, n, (j->
      `if`(j::even, j/2, 3*j+1))(T(n, k-1)))
    end:
seq(seq(T(d-k, k), k=0..d-1), d=1..20);  # Alois P. Heinz, Aug 25 2021

T[n_, k_] := T[n, k] = If[k == 0, n, Function[j,
     If[EvenQ[j], j/2, 3*j + 1]][T[n, k - 1]]];
Table[Table[T[d - k, k], {k, 0, d - 1}], {d, 1, 20}] // Flatten (* Jean-François Alcover, Mar 02 2022, after Alois P. Heinz *)

A208981 Number of iterations required to reach a power of 2 in the 3x+1 sequence starting at n.

Original entry on oeis.org

0, 0, 3, 0, 1, 4, 12, 0, 15, 2, 10, 5, 5, 13, 13, 0, 8, 16, 16, 3, 1, 11, 11, 6, 19, 6, 107, 14, 14, 14, 102, 0, 22, 9, 9, 17, 17, 17, 30, 4, 105, 2, 25, 12, 12, 12, 100, 7, 20, 20, 20, 7, 7, 108, 108, 15, 28, 15, 28, 15, 15, 103, 103, 0, 23, 23, 23, 10, 10, 10

Offset: 1

L. Edson Jeffery, Mar 04 2012

The original name was: Number of iterations of the Collatz recursion required to reach a power of 2.
The statement that all paths must eventually reach a power of 2 is equivalent to the Collatz conjecture.
A006577(n) - a(n) gives the exponent for the first power of 2 reached in the Collatz trajectory of n. - Alonso del Arte, Mar 05 2012
Number of nonpowers of 2 in the 3x+1 sequence starting at n. - Omar E. Pol, Sep 05 2021

			a(7) = 12 because the Collatz trajectory for 7 is 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... which reached 16 = 2^4 in 12 steps.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
John Smith, Collatz sequence, PlanetMath.
Eric Weisstein's World of Mathematics, Collatz problem.
Index entries for sequences related to 3x+1 (or Collatz) problem

Row sums of A347519.
Cf. A006577 (and references therein).
Cf. A347270 (gives all 3x+1 sequences).
Cf. A070165, A010120, A057716, A135282, A209229.

a208981 = length . takeWhile ((== 0) . a209229) . a070165_row
-- Reinhard Zumkeller, Jan 02 2013

a:= proc(n) option remember; `if`(n=2^ilog2(n), 0,
      1+a(`if`(n::odd, 3*n+1, n/2)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Sep 05 2021

Collatz[n_?OddQ] := 3*n + 1; Collatz[n_?EvenQ] := n/2; Table[-1 + Length[NestWhileList[Collatz, n, Not[IntegerQ[Log[2, #]]] &]], {n, 50}] (* Alonso del Arte, Mar 04 2012 *)

ispow2(n)=n>>=valuation(n,2); n==1
a(n)=my(s); while(!ispow2(n), n=if(n%2, 3*n+1, n/2); s++); s \\ Charles R Greathouse IV, Jul 31 2016

For x>0 an integer, define f_0(x)=x, and for r=1,2,..., f_r(x)=f_{r-1}(x)/2 if f_{r-1}(x) is even, else f_r(x)=3*f_{r-1}(x)+1. Then a(n) = min(k such that f_k(n) is equal to a power of 2).

a(n) = A006577(n) - A135282(n) (after Alonso del Arte's comment), if A006577(n) is not -1. - Omar E. Pol, Apr 10 2022

Name clarified by Omar E. Pol, Apr 10 2022

A232503 Largest power of 2 in the Collatz (3x+1) trajectory of n.

Original entry on oeis.org

1, 2, 16, 4, 16, 16, 16, 8, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 64, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 32, 16, 16, 16, 16, 16, 16, 16, 16, 16, 64, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 64, 16, 16, 16, 16

Offset: 1

Lance Luttrell, Nov 24 2013

nonn

All terms are powers of 4 (A000302) unless n is an odd-indexed power of 2 (A004171), as explained in one of the comments for A135282. - Alonso del Arte, Nov 26 2013

			a(8) = 8 because then there are only halving steps and thus 8 is the largest power of 2 in its Collatz trajectory.
a(9) = 16 because the Collatz sequence for 9 goes 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1 (see A033479) and the largest power of 2 there is 16.

Antti Karttunen, Table of n, a(n) for n = 1..1024
Index entries related to the 3x+1 (Collatz) problem

Cf. A006370, A135282, A226123.

Collatz[n_?OddQ] := 3n + 1; Collatz[n_?EvenQ] := n/2; Table[NestWhile[Collatz, n, Not[IntegerQ[Log[2, #]]] &], {n, 64}] (* Alonso del Arte, Nov 26 2013 based on T. D. Noe's program for A135282 *)
Table[With[{p2=2^Range[0,10]},Select[NestWhileList[If[EvenQ[#], #/2, 3#+1]&, n,#>1&], MemberQ[ p2,#]&]]//Max,{n,70}] (* Harvey P. Dale, May 07 2016 *)

(define (A232503 n) (let loop ((n n) (m 1)) (if (= 1 n) m (loop (A006370 n) (if (= 1 (A209229 n)) (max n m) m)))))
(define (A006370 n) (if (even? n) (/ n 2) (+ 1 n n n)))
;; Antti Karttunen, Aug 18 2017

a(n) = 2^A135282(n). - Antti Karttunen, Aug 18 2017

A347409 Longest run of halving steps in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.

Original entry on oeis.org

0, 1, 4, 2, 4, 4, 4, 3, 4, 4, 4, 4, 4, 4, 5, 4, 4, 4, 4, 4, 6, 4, 5, 4, 4, 4, 5, 4, 4, 5, 5, 5, 4, 4, 5, 4, 4, 4, 4, 4, 5, 6, 4, 4, 4, 5, 5, 4, 4, 4, 4, 4, 5, 5, 5, 4, 4, 4, 4, 5, 5, 5, 5, 6, 4, 4, 4, 4, 4, 5, 5, 4, 5, 4, 8, 4, 4, 4, 4, 4, 5, 5, 5, 6, 8, 4, 4

Offset: 1

Paolo Xausa, Aug 30 2021

If the longest run of halving steps occurs as the final part of the trajectory, a(n) = A135282(n), otherwise a(n) > A135282(n).
Every nonnegative integer appears in the sequence at least once, since for any k >= 0 a(2^k) = k.
Conjecture: every integer >= 4 appears in the sequence infinitely many times.

			a(15) = 5 because the Collatz trajectory starting at 15 contains, as the longest halving run, a 5-step subtrajectory (namely, 160 -> 80 -> 40 -> 20 -> 10 -> 5).

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. A006370, A070165, A135282.

Cf. A347668 (indices of records), A347669 (indices of first occurrences), A348007.

nterms=100;Table[c=n;sm=0;While[c>1,If[OddQ[c],c=3c+1,If[(s=IntegerExponent[c,2])>sm,sm=s];c/=2^s]];sm,{n,nterms}]

a(n)=my(nb=0); while (n != 1, if (n % 2, n=3*n+1, my(x = valuation(n, 2)); n /= 2^x; nb = max(nb, x));); nb; \\ Michel Marcus, Sep 03 2021

def A347409(n):
    m, r = n, 0
    while m > 1:
        if m % 2:
            m = 3*m + 1
        else:
            s = bin(m)[2:]
            c = len(s)-len(s.rstrip('0'))
            m //= 2**c
            r = max(r,c)
    return r # Chai Wah Wu, Sep 29 2021

A347668 Indices of records in A347409.

Original entry on oeis.org

1, 2, 3, 15, 21, 75, 151, 1365, 5461, 7407, 14563, 87381, 184111, 932067, 5592405, 13256071, 26512143, 357913941, 1431655765, 3817748707, 22906492245, 91625968981, 244335917283, 1466015503701, 5212499568715, 10424999137431, 93824992236885

Offset: 1

Paolo Xausa, Sep 10 2021

Conjecture 1: A347409(a(n)) is even for n >= 11. Conjecture 2: all even numbers > 2 appear as A347409(a(n)) for some n. - Chai Wah Wu, Sep 29 2021

If conjectures 1 and 2 are true, then A347409(a(n)) = 2n - 6 for n >= 11, and hence a(n) <= (4^(n-3)-1)/3 for n >= 11 since A347409((4^(n-3)-1)/3) = 2n - 6. - Charles R Greathouse IV, Oct 25 2022

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

Cf. A006370, A070165, A135282, A347409, A347669, A002450.

A347409[n_]:=(c=n;sm=0;While[c>1,If[OddQ[c],c=3c+1,If[(s=IntegerExponent[c,2])>sm,sm=s];c/=2^s]];sm)
upto=100000;a={};rec=-1;Do[If[(r=A347409[i])>rec,rec=r;AppendTo[a,i]],{i,upto}];a

f(n)=my(nb=0); while (n != 1, if (n % 2, n=3*n+1, my(x = valuation(n, 2)); n /= 2^x; nb = max(nb, x)); ); nb; \\ A347409
lista(nn) = my(r=-1, m); for (n=1, nn, if ((m=f(n)) > r, print1(n, ", "); r = m);); \\ Michel Marcus, Sep 10 2021

a(15) from Michel Marcus, Sep 10 2021
a(16)-a(17) from Alois P. Heinz, Sep 10 2021
a(18)-a(20) from Michael S. Branicky, Sep 28 2021
a(21)-a(22) from Michael S. Branicky, Sep 30 2021
a(23) from Michael S. Branicky, Oct 04 2021
a(24)-a(27) from Kevin P. Thompson, Apr 14 2022

A226123 Number of terms of the form 2^k in Collatz(3x+1) trajectory of n.

Original entry on oeis.org

1, 2, 5, 3, 5, 5, 5, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 9, 5, 5, 5, 5, 5, 5, 5, 5, 7, 9, 5, 5

Offset: 1

Jayanta Basu, May 27 2013

nonn

a(n) = sum(A209229(A070165(n,k)): k=1..A006577(n)). - Reinhard Zumkeller, May 30 2013

			a(3)=5 since Collatz trajectory of 3 contains terms 1,2,4,8 and 16.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A070165.

Cf. A135282, A208981.

a226123 = sum . map a209229 . a070165_row
-- Reinhard Zumkeller, May 30 2013

coll[n_]:=NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]; Table[Length[Select[coll[n],IntegerQ[Log[2,#]]&]],{n,87}]

A347532 a(n) is the sum of the nonpowers of 2 in the 3x+1 sequence that starts at n.

Original entry on oeis.org

0, 0, 18, 0, 5, 24, 257, 0, 308, 15, 228, 36, 88, 271, 663, 0, 183, 326, 488, 35, 21, 250, 602, 60, 627, 114, 101409, 299, 411, 693, 101073, 0, 810, 217, 509, 362, 504, 526, 2313, 75, 101300, 63, 1307, 294, 466, 648, 100948, 108, 775, 677, 1099, 166, 368, 101463, 102285, 355

Offset: 1

Omar E. Pol, Sep 05 2021

nonn

a(n) is the sum of the nonpowers of 2 in the n-th row of A347270.

a(n) = 0 if and only if n is a power of 2.

			For n = 6 the 3x+1 sequence starting at 6 is 6, 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, ... Only the first four terms are nonpowers of 2. The sum of them is 6 + 3 + 10 + 5 = 24, so a(6) = 24.

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

Cf. A208981 (number of nonpowers of 2).

Cf. A000079, A006370, A033493, A033495, A057716, A135282, A232503, A347270, A347519.

a:= proc(n) option remember; `if`(n=2^ilog2(n), 0,
      n+a(`if`(n::odd, 3*n+1, n/2)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Sep 05 2021

a[n_] := Plus @@ Select[NestWhileList[If[OddQ[#], 3*# + 1, #/2] &, n, # > 1 &], # != 2^IntegerExponent[#, 2] &]; Array[a, 50] (* Amiram Eldar, Sep 06 2021 *)

From Alois P. Heinz, Sep 05 2021: (Start)
a(n) = A033493(n) - 2 * A232503(n) + 1.
a(n) = A033493(n) - 2^(A135282(n)+1) + 1. (End)

A347669 Indices of first occurrences of n in A347409.

Original entry on oeis.org

1, 2, 4, 8, 3, 15, 21, 128, 75, 512, 151, 2048, 1365, 8192, 5461, 7407, 14563, 131072, 87381, 524288, 184111, 2097152, 932067, 6213783, 5592405, 33554432, 13256071, 134217728, 26512143, 530242875, 357913941, 1899273247, 1431655765, 8589934592, 3817748707, 34359738368

Offset: 0

Paolo Xausa, Sep 10 2021

			a(5) = 15 because 5 occurs for the first time at position 15 in A347409.

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

Cf. A006370, A070165, A135282, A347409, A347668.

A347409[n_]:=A347409[n]=(c=n;sm=0;While[c>1,If[OddQ[c],c=3c+1,If[(s=IntegerExponent[c,2])>sm,sm=s];c/=2^s]];sm)
nterms=20;Table[i=0;While[A347409[++i]!=n];i,{n,0,nterms-1}]

f(n) = {my(nb=0); while (n != 1, if (n % 2, n=3*n+1, my(x = valuation(n, 2)); n /= 2^x; nb = max(nb, x)); ); nb; } \\ A347409
a(n) = my(k=1); while (f(k) != n, k++); k; \\ Michel Marcus, Sep 10 2021

a(21)-a(22) from Michel Marcus, Sep 10 2021
a(23)-a(28) from Alois P. Heinz, Sep 11 2021
a(29)-a(34) from Michael S. Branicky, Sep 28 2021
a(35) from Chai Wah Wu, Oct 02 2021

A348007 Starting value of the longest run of halving steps in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.

Original entry on oeis.org

2, 16, 4, 16, 16, 16, 8, 16, 16, 16, 16, 16, 16, 160, 16, 16, 16, 16, 16, 64, 16, 160, 16, 16, 16, 160, 16, 16, 160, 160, 32, 16, 16, 160, 16, 112, 16, 304, 16, 160, 64, 112, 16, 16, 160, 160, 48, 112, 16, 16, 16, 160, 160, 160, 16, 112, 16, 304, 160, 160, 160

Offset: 2

Paolo Xausa, Sep 24 2021

nonn

In case of ties (two or more longest runs of same length), the highest starting value is picked. The first n for which the longest run of halving steps occurs at two different subtrajectories is 37, where the Collatz map contains the 4-step subtrajectories 112 -> 56 -> 28 -> 14 > 7 and 16 -> 8 -> 4 -> 2 -> 1. a(37) is therefore 112 (highest starting value).

If the Collatz conjecture (i.e., all trajectories reach 1) is true then, except for n = 2, 4 and 8, a(n) mod 16 = 0, since all trajectories contain (at least) 4 consecutive halvings.

			a(2) = 2 because the Collatz trajectory from 2 to 1 is simply 2 -> 1 (one halving step, starting at 2).
a(3) = 16 because the trajectory from 3 to 1 is 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1. Here, the longest halving run is the 4-step subtrajectory 16 -> 8 -> 4 -> 2 -> 1, which starts at 16.
a(15) = 160 because the longest halving run in the trajectory from 15 to 1 (the 5-step subtrajectory 160 -> 80 -> 40 -> 20 -> 10 -> 5) starts at 160.

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

Cf. A006370, A070165, A135282, A347409.

nterms=100;Table[c=n;cm=sm=0;While[c>1,If[OddQ[c],c=3c+1,s=IntegerExponent[c,2];If[s>sm,sm=s;cm=c,If[s==sm,cm=Max[cm,c]]];c/=2^s]];cm,{n,2,nterms+1}]

A348007(n) = { my(m2v=valuation(n,2), mx=n, t); while(n>1, if((t=valuation(n,2))>m2v, m2v=t; mx=n, if(t==m2v && n>mx, mx=n)); if(!(n%2),n/=2,n+=(n+n+1))); (mx); }; \\ Antti Karttunen, Oct 13 2021

a(2^k) = 2^k, for integers k >= 1.

a(n) mod 2^A347409(n) = 0.

cp's OEIS Frontend

A006577 Number of halving and tripling steps to reach 1 in '3x+1' problem, or -1 if 1 is never reached.

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A347270 Square array T(n,k) in which row n lists the 3x+1 sequence starting at n, read by antidiagonals upwards, with n >= 1 and k >= 0.

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A208981 Number of iterations required to reach a power of 2 in the 3x+1 sequence starting at n.

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A232503 Largest power of 2 in the Collatz (3x+1) trajectory of n.

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A347409 Longest run of halving steps in the trajectory from n to 1 in the Collatz map (or 3x+1 problem), or -1 if no such trajectory exists.

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A347668 Indices of records in A347409.

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