A060322 Consider the version of the Collatz or 3x+1 problem where x -> x/2 if x is even, x -> (3x+1)/2 if x is odd. Define the stopping time of x to be the number of steps needed to reach 1. Sequence gives the number of integers x with stopping time n.

1, 1, 1, 1, 2, 3, 4, 5, 6, 8, 12, 18, 24, 31, 39, 50, 68, 91, 120, 159, 211, 282, 381, 505, 665, 885, 1187, 1590, 2122, 2829, 3765, 5014, 6682, 8902, 11878, 15844, 21122, 28150, 37536, 50067, 66763, 89009, 118631, 158171, 210939, 281334, 375129, 500106, 666725

Offset: 1

Author

Bo T. Ahlander (ahlboa(AT)isk.kth.se), Mar 29 2001

Keywords

• nonn

Comments

The Mathematica function StoppingTime[n] is the length of the Collatz sequence starting at n before reaching 1.

Suppose we have a list L of the numbers with StoppingTime n. Then the list LL of StoppingTime n+1 can be produced as: First. Add to LL all numbers in L multiplied by 2. Second. For the numbers x now in LL, if x == 1 (mod 3), AppendTo LL the number (x-1)/3 (if (x-1)/3 != 1). These two steps make LL complete.

I think the offset, examples, formula and code are all off by 1 -- they all treat the stopping time of 1 to be 1, rather than 0. - David Applegate, Oct 16 2008

a(n+1), n >= 0, is the row length of A248573(n,m) (Collatz-Terras tree). For the first differences see A131450(n+1), but with A131450(2) = 1 (the number of 2 (mod 3) numbers in row n, for n >= 0, of A248573). - Wolfdieter Lang, May 04 2015

Examples

			StoppingTime = 1: L = {1}, a(1)=1.
StoppingTime = 2: L = {2}, a(2)=1.
StoppingTime = 3: L = {4}, a(3)=1.
StoppingTime = 4: L = {8}, a(4)=1.
StoppingTime = 5: L = {5, 16}, a(5)=2. First, LL = {10, 32} (= 2*L). Second, 10 == 1 (mod 3), so we AppendTo LL also (10-1)/3 = 3. We get LL = {3, 10, 32}. So a(6) = 3.
		

Links

• Joe Slater, Table of n, a(n) for n = 1..70
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

See A005186 for another version.

Cf. A248573, A131450.

Programs

Mathematica

(*** Program #1 ***) For[v = 1, v <= 12, v++, lst = {}; For[n = 1, n < 2^v, n++, If[StoppingTime[n] == v, AppendTo[lst, n]]]; Print[lst]; Print[Length[lst]]; ]
(*** Program #2 ***) lst1 = {1}; For[v = 1, v <= 12, v++, L1 = Length[lst1]; Print["Number of numbers with StoppingTime ", v, ": ", L1]; Print["List of numbers: ", lst1]; (* Numbers with StoppingTime n *) Print["Control of StoppingTime: ", Map[StoppingTime, lst1]]; (* Controll *) Print[""]; lst2 = 2 lst1; For[i = 1, i <= L1, i++, x = (lst2[[i]] - 1)/3; If[IntegerQ[x] && x != 1, AppendTo[lst2, x]]; ]; lst1 = Sort[lst2]; ]
(*** Program #3 ***) lst0 = {}; lst1 = {1}; For[v = 1, v <= 35, v++, L1 = Length[lst1]; AppendTo[lst0, L1]; lst2 = 2 lst1; For[i = 1, i <= L1, i++, x = (lst2[[i]] - 1)/3; If[IntegerQ[x], AppendTo[lst2, x]]; ]; lst1 = Complement[lst2, {1}]; ]; lst0

PARI

first(N) = my(a=Vec([1, 1, 1, 1, 2], N), p=[], q=[5]); for(n=6, N, my(r=List()); foreach(p, x, listput(r, 4*x+1); if(1==x%6, listput(r, x+(x-1)/3))); foreach(q, x, if(5==x%6, listput(r, x-(x+1)/3))); a[n]=a[n-1]+#r; p=q; q=Vec(r)); a; \\ Ruud H.G. van Tol, Aug 14 2024

Perl

# code to calculate terms after a(4):
@x=(8,0);for($n=5;$n<=60;$n++){do{$q=2*shift(@x);push(@x,($q-1)/3)if($q%3==1);push @x,$q}while $q;print($#x,", ");} # Carl R. White, Oct 03 2006

Formula

Conjecture: lim_{n->oo} a(n) = a(n-1)*4/3. - Joe Slater, Jan 27 2024

Extensions

More terms from Carl R. White, Oct 03 2006

Edited by N. J. A. Sloane, Sep 15 2007

More terms from Joe Slater, Apr 10 2025