

A126241


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 kth iterate, or infinity otherwise; a(n) is called the dropping time of n.


14



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,3


COMMENTS

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)


REFERENCES

J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010. See p. 33.


LINKS



FORMULA



EXAMPLE

s(15) = 7, since the trajectory {T^(k)(15)} (k=1,2,3,...) equals 23,35,53,80,40,20,10.


MATHEMATICA

Collatz2[n_] := If[n<2, {}, Rest[NestWhileList[If[EvenQ[#], #/2, (3 # + 1)/2] &, n, # >= n &]]]; Table[Length[Collatz2[n]], {n, 1, 1000}]


CROSSREFS

See A074473, which is the main entry for dropping times.


KEYWORD

nonn


AUTHOR

Christof Menzel (christof.menzel(AT)hsniederrhein.de), Mar 08 2007


EXTENSIONS

Broken link fixed by K. Spage, Oct 22 2009


STATUS

approved



