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A006666
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Number of halving steps to reach 1 in '3x+1' problem, or -1 if this never happens.
(Formerly M3733)
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51
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0, 1, 5, 2, 4, 6, 11, 3, 13, 5, 10, 7, 7, 12, 12, 4, 9, 14, 14, 6, 6, 11, 11, 8, 16, 8, 70, 13, 13, 13, 67, 5, 18, 10, 10, 15, 15, 15, 23, 7, 69, 7, 20, 12, 12, 12, 66, 9, 17, 17, 17, 9, 9, 71, 71, 14, 22, 14, 22, 14, 14, 68, 68, 6, 19, 19, 19, 11, 11, 11, 65, 16, 73, 16, 11, 16
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OFFSET
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1,3
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COMMENTS
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Equals the total number of steps to reach 1 under the modified '3x+1' map: T(n) = n/2 if n is even, (3n+1)/2 if n is odd (see A014682).
Pairs of consecutive integers of the same height occur infinitely often and in infinitely many different patterns (Garner 1985). - Joe Slater, May 24 2018
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, E16.
J. C. Lagarias, ed., The Ultimate Challenge: The 3x+1 Problem, Amer. Math. Soc., 2010.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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FORMULA
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a(2^k-1) = a(2^(k+1)-1)-1, for odd k>1. - Joe Slater, May 17 2018
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EXAMPLE
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2 -> 1 so a(2) = 1; 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1, with 5 halving steps, so a(3) = 5; 4 -> 2 -> 1 has two halving steps, so a(4) = 2; etc.
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MAPLE
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T:=proc(n) if n mod 2 = 0 then n/2 else (3*n+1)/2; fi; end;
t1:=[0]:
for n from 2 to 100 do
L:=1; p := n;
while T(p) <> 1 do p:=T(p); L:=L+1; od:
t1:=[op(t1), L];
od: t1;
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MATHEMATICA
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Table[Count[NestWhileList[If[OddQ[#], 3#+1, #/2]&, n, #>1&], _?(EvenQ[#]&)], {n, 80}] (* Harvey P. Dale, Sep 30 2011 *)
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PROG
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(Haskell)
a006666 = length . filter even . takeWhile (> 1) . (iterate a006370)
(Python)
def a(n):
if n==1: return 0
x=0
while True:
if not n%2:
n//=2
x+=1
else: n = 3*n + 1
if n<2: break
return x
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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More terms from Larry Reeves (larryr(AT)acm.org), Apr 27 2001
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STATUS
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approved
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