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A126241
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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)<n, where T^(k) is the k-th iterate, or infinity otherwise; a(n) is called the dropping time of n.
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14
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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
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OFFSET
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1,3
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COMMENTS
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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)
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REFERENCES
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J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010. See p. 33.
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LINKS
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FORMULA
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EXAMPLE
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s(15) = 7, since the trajectory {T^(k)(15)} (k=1,2,3,...) equals 23,35,53,80,40,20,10.
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MATHEMATICA
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Collatz2[n_] := If[n<2, {}, Rest[NestWhileList[If[EvenQ[#], #/2, (3 # + 1)/2] &, n, # >= n &]]]; Table[Length[Collatz2[n]], {n, 1, 1000}]
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CROSSREFS
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See A074473, which is the main entry for dropping times.
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KEYWORD
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nonn
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AUTHOR
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Christof Menzel (christof.menzel(AT)hs-niederrhein.de), Mar 08 2007
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EXTENSIONS
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Broken link fixed by K. Spage, Oct 22 2009
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STATUS
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approved
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