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 A224183 Number of halving and tripling steps to reach the last number of the cycle in the Collatz (3x+1) problem for the negative integers (the initial term is not counted). 4
 1, 1, 4, 2, 4, 5, 4, 3, 11, 4, 6, 6, 9, 4, 9, 4, 17, 12, 7, 4, 23, 7, 18, 7, 17, 10, 7, 5, 10, 10, 21, 5, 28, 17, 13, 13, 17, 8, 13, 5, 17, 24, 8, 8, 22, 19, 16, 8, 26, 17, 11, 11, 16, 8, 17, 6, 29, 11, 11, 11, 17, 22, 19, 6, 37, 29, 20, 17, 19, 14, 19, 14, 24 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS The Collatz problem with negative numbers is as follows: start with any number n = -1, -2, -3, ... If n is even, divide it by 2, otherwise multiply it by 3 and add 1. Do we always reach a last number of each cycle? It is conjectured that the answer is yes. A majority of last numbers of each cycle is -1, but it is possible to obtain also -5, -10, -14, -20, -34,.... LINKS Michel Lagneau, Table of n, a(n) for n = 1..10000 EXAMPLE a(10) = 4 because the trajectory of -10 is -10 -> -5 -> -14 -> -7 -> -20 -> -10 and -10 is the last term of the cycle, hence 4 iterations. MAPLE printf(`%d, `, 1): for n from -2 by -1 to -100 do:x:=n:lst:={n}:for it from 1 to 100 do:if irem(x, 2)=0 then x := x/2: lst:=lst union{x} :else x := 3*x+1: lst:=lst union{x}fi:od:d:=nops(lst): printf(`%d, `, d-1): od: MATHEMATICA Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, UnsameQ, All]; Table[s = Collatz[-n]; len = Length[s] - 2; If[s[[-1]] == 2, len = len - 1]; len, {n, 1, 100}] CROSSREFS Cf. A006577. Sequence in context: A368437 A258852 A021706 * A131953 A167508 A167507 Adjacent sequences: A224180 A224181 A224182 * A224184 A224185 A224186 KEYWORD nonn AUTHOR Michel Lagneau, Apr 01 2013 EXTENSIONS a(1) changed to 1 by Pontus von Brömssen, Jan 24 2021 STATUS approved

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Last modified August 11 23:45 EDT 2024. Contains 375082 sequences. (Running on oeis4.)