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A060322
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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.
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1
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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
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OFFSET
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1,5
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COMMENTS
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The Mathematica function StoppingTime[n] is the length of the Collatz sequence starting at n before hitting 1.
Comment from David Applegate, Oct 16 2008: 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.
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LINKS
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Table of n, a(n) for n=1..47.
Index entries for sequences related to 3x+1 (or Collatz) problem
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FORMULA
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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 makes LL complete.
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EXAMPLE
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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.
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MATHEMATICA
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(*** 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
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PROG
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#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
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CROSSREFS
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See A005186 for another version.
Sequence in context: A134677 A104419 A092232 * A143285 A019532 A008537
Adjacent sequences: A060319 A060320 A060321 * A060323 A060324 A060325
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KEYWORD
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nonn
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AUTHOR
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Bo T. Ahlander (ahlboa(AT)isk.kth.se), Mar 29 2001
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EXTENSIONS
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More terms from Carl R. White, Oct 03 2006
Edited by N. J. A. Sloane, Sep 15 2007
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STATUS
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approved
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