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A062060
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Numbers with 10 odd integers in their Collatz (or 3x+1) trajectory.
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14
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43, 86, 87, 89, 172, 173, 174, 177, 178, 179, 344, 346, 348, 349, 354, 355, 356, 357, 358, 385, 423, 688, 692, 693, 696, 698, 705, 708, 709, 710, 712, 714, 716, 717, 729, 761, 769, 770, 771, 777, 846, 847, 1376, 1384, 1386, 1392, 1393, 1396, 1397, 1410, 1411, 1415
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OFFSET
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1,1
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COMMENTS
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The Collatz (or 3x+1) function is f(x) = x/2 if x is even, 3x+1 if x is odd.
The Collatz trajectory of n is obtained by applying f repeatedly to n until 1 is reached.
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REFERENCES
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J. Shallit and D. Wilson, The "3x+1" Problem and Finite Automata, Bulletin of the EATCS #46 (1992) pp. 182-185.
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LINKS
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EXAMPLE
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The Collatz trajectory of 43 is (43, 130, 65, 196, 98, 49, 148, 74, 37, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1), which contains 10 odd integers.
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MATHEMATICA
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Collatz[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; countOdd[lst_] := Length[Select[lst, OddQ]]; Select[Range[1000], countOdd[Collatz[#]] == 10 &] (* T. D. Noe, Dec 03 2012 *)
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PROG
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(Haskell)
import Data.List (elemIndices)
a062060 n = a062060_list !! (n-1)
a062060_list = map (+ 1) $ elemIndices 10 a078719_list
(Python)
def a(n):
l=[n]
while True:
if n%2==0: n//=2
else: n = 3*n + 1
if n not in l:
l.append(n)
if n<2: break
else: break
return len([1 for i in l if i%2])
print([n for n in range(40, 1501) if a(n)==10]) # Indranil Ghosh, Apr 14 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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