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A060445
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"Dropping time" in 3x+1 problem starting at 2n+1 (number of steps to reach a lower number than starting value). Also called glide(2n+1).
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12
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0, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 96, 3, 91, 3, 6, 3, 13, 3, 8, 3, 88, 3, 6, 3, 8, 3, 11, 3, 88, 3, 6, 3, 83, 3, 8, 3, 13, 3, 6, 3, 8, 3, 73, 3, 13, 3, 6, 3, 68, 3, 8, 3, 50, 3, 6, 3, 8, 3, 13, 3, 24, 3, 6, 3, 11, 3, 8, 3, 11, 3, 6, 3, 8, 3, 65, 3, 34, 3, 6, 3, 47, 3, 8, 3, 13, 3, 6, 3, 8, 3
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OFFSET
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0,2
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COMMENTS
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If the starting value is even then of course the next step in the trajectory is smaller (cf. A102419).
The dropping time can be made arbitrarily large: If the starting value is of form n(2^m)-1 and m > 1, the next value is 3n(2^m)-3+1. That divided by 2 is 3n(2^(m-1))-1. It is bigger than the starting value and of the same form - substitute 3n -> n and m-1 -> m, so recursively get an increasing subsequence of m odd values. The dropping time is obviously longer than that. This holds even if Collatz conjecture were refuted. For example, m=5, n=3 -> 95, 286, 143, 430, 215, 646, 323, 970, 485, 1456, 728, 364, 182, 91. So the subsequence in reduced Collatz variant is 95, 143, 215, 323, 485. - Juhani Heino, Jul 21 2017
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LINKS
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EXAMPLE
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3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2, taking 6 steps, so a(1) = 6.
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MATHEMATICA
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nxt[n_]:=If[OddQ[n], 3n+1, n/2]; Join[{0}, Table[Length[NestWhileList[nxt, n, #>=n&]]-1, {n, 3, 191, 2}]] (* Harvey P. Dale, Apr 23 2011 *)
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PROG
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(Haskell)
a060445 0 = 0
a060445 n = length $ takeWhile (>= n') $ a070165_row n'
where n' = 2 * n + 1
(Python)
def a(n):
if n<1: return 0
n=2*n + 1
N=n
x=0
while True:
if n%2==0: n//=2
else: n = 3*n + 1
x+=1
if n<N: break
return x
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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Still more terms from Larry Reeves (larryr(AT)acm.org), Apr 12 2001
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STATUS
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approved
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