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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. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

COMMENTS

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

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

LINKS

Table of n, a(n) for n=1..47.

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

FORMULA

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 Mod[x, 3]==1, AppendTo LL the number (x-1)/3 (if (x-1)/3!=1). These two steps make LL complete.

EXAMPLE

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, Mod[10, 3]==1, so we AppendTo LL also (10-1)/3 == 3. We get LL = {3, 10, 32}. So a(6) == 3.

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

PROG

#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

CROSSREFS

See A005186 for another version. A248573, A131450.

Sequence in context: A134677 A104419 A092232 * A143285 A019532 A008537

Adjacent sequences:  A060319 A060320 A060321 * A060323 A060324 A060325

KEYWORD

nonn

AUTHOR

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

EXTENSIONS

More terms from Carl R. White, Oct 03 2006

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

STATUS

approved

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Last modified January 24 16:46 EST 2020. Contains 331209 sequences. (Running on oeis4.)