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We define a function \(f(x)\) such that \(f(14)=7\), and if there exists an integer \(a\) such that \(f(a)=b\), then \(f(b)\) is defined and \(f(b)=3b+1 \)if \(b\) is odd

\(f(b)=\frac{b}2\) if \(b\) is even.

What is the smallest possible number of integers in the domain of \(f\)?

DanielCai  Jun 28, 2018
edited by DanielCai  Jun 29, 2018
edited by DanielCai  Jun 29, 2018
 #1
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If there exists an integer  a  such that  f(a) = b ,   then  f(b)  is defined and

 

f(b) = 3b + 1  if  b  is odd      and      f(b) = b/2  if  b  is even .

 

We must have at least these defined values of the function:

 

f(14) = 7

f(7) = 22

f(22) = 11

f(11) = 34

f(34) = 17

f(17) = 52

f(52) = 26

f(26) = 13

f(13) = 40

f(40) = 20

f(20) = 10

f(10) = 5

f(5) = 16

f(16) = 8

f(8) = 4

f(4) = 2

f(2) = 1

f(1) = 2

 

There must be at least 18  integers in the domain of  f .

hectictar  Jun 29, 2018
 #2
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Hmmm, this question is very interesting, because it is related to the collatz conjecture, a famous unsolved problem in math.

Guest Jun 29, 2018
 #3
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Hectictar casually solves Collatz conjecture, nominated for Fields Medal. Quote: “What’s all the fuss? It was easy!”  News @ 7:00   

Guest Jun 29, 2018

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