+0

Sorry for posting this many questions...

0
83
1
+798

We define a function (fx) such that f(11)=34, 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)=b/2 if $b$ is even. What is the smallest possible number of integers in the domain of $f$?

Jun 19, 2020

#1
+8341
0

This is seemingly easy, but it is extremely hard.

This one is the Collatz conjecture, which remains unsolved for years.

EDIT: Sorry, I read the question wrong. But it is still good to read about the conjecture.

f(11) = 34, so f(34) is defined.

f(34) = 34/2 = 17, so f(17) is defined.

f(17) = 3(17) + 1 = 52, so f(52) is defined.

f(52) = 52/2 = 26, so f(26) is defined,

f(26) = 26/2 = 13, so f(13) is defined.

f(13) = 3(13) + 1 = 40, so f(40) is defined.

f(40) = 40/2 = 20, so f(20) is defined.

f(20) = 20/2 = 10, so f(10) is defined.

f(10) = 10/2 = 5, so f(5) is defined.

f(5) = 3(5) + 1 = 16, so f(16) is defined.

f(16) = 16/2 = 8, so f(8) is defined.

f(8) = 8/2 = 4, so f(4) is defined.

f(4) = 4/2 = 2, so f(2) is defined.

f(2) = 2/2 = 1, so f(1) is defined.

f(1) = 3(1) + 1 = 4

Now we ended up in an infinite loop of 4, 2, 1, 4, 2, 1, etc.

Let's count how many numbers did we encounter.

For $$x\in\{11,34,17,52,26,13,40,20,10,5,16,8,4,2,1\}$$, f(x) is defined.

Therefore there are at least 15 integers in the domain of f.

Jun 19, 2020
edited by MaxWong  Jun 19, 2020