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A074473
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Dropping time for the 3x+1 problem: for n >= 2, number of iteration that first becomes smaller than the initial value if Collatz-function (A006370) is iterated starting at n; a(1)=1 by convention.
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19
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1, 2, 7, 2, 4, 2, 12, 2, 4, 2, 9, 2, 4, 2, 12, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 97, 2, 4, 2, 92, 2, 4, 2, 7, 2, 4, 2, 14, 2, 4, 2, 9, 2, 4, 2, 89, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 12, 2, 4, 2, 89, 2, 4, 2, 7, 2, 4, 2, 84, 2, 4, 2, 9, 2, 4, 2, 14, 2, 4, 2, 7, 2, 4, 2, 9, 2, 4, 2, 74, 2, 4, 2, 14, 2, 4, 2, 7
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OFFSET
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1,2
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COMMENTS
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Here we call the starting value iteration number 1, although usually the count is started at 0, which would subtract 1 from the values for n >= 2 - see A060445, A102419.
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LINKS
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EXAMPLE
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n=2k: then a(2k)=2 because the second iterate is k<n=2k, the first iterate below 2k; n=4k+1, k>1: the list = {4k+1, 12k+4, 6k+2, 3k+1, ...} i.e. the 4th term is always the first below initial value, so a(4k+1)=4;
n=15: the list={15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1} and 12th term is first sinks below iv=15, so a(15)=12; relatively larger values occur at n=4k+3.
n=3: the list is {3, 10, 5, 16, 8, 4, 2, 1, ..}, the 7th term is 2, which is the first smaller than 3, so a(3)=7.
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MATHEMATICA
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nextx[x_Integer] := If[OddQ@x, 3x + 1, x/2]; f[1] = 1; f[n_] := Length@ NestWhileList[nextx, n, # >= n &]; Array[f, 83] (* Bobby R. Treat (drbob(at)bigfoot.com), Sep 16 2006 *)
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PROG
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(Python)
def a(n):
if n<3: return n
N=n
x=1
while True:
if n%2==0: n/=2
else: n = 3*n + 1
x+=1
if n<N: return x
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CROSSREFS
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Cf. A006370, A075476, A075477, A075478, A075479, A075480, A075481, A075482, A075483, A060445, A060412, A217934.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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