FREE POINTS J SOLVE THE Collatz conjecture
If the number is even, divide it by two.
If the number is odd, triple it and add one.
In modular arithmetic notation, define the function f as follows:

{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}}\\[4px]3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}{\displaystyle f(n)={\begin{cases}{\frac {n}{2}}&{\text{if }}n\equiv 0{\pmod {2}}\\[4px]3n+1&{\text{if }}n\equiv 1{\pmod {2}}.\end{cases}}}
Now form a sequence by performing this operation repeatedly, beginning with any positive integer, and taking the result at each step as the input at the next.

In notation:

{\displaystyle a_{i}={\begin{cases}n&{\text{fo r }}i=0\\f(a_{i-1})&{\text{for }}i>0\end{cases}}}a_{i}={\begin{ cases}n&{\text{for }}i=0\\f(a_{i-1})&{\text{for }}i>0\end{cases}}
(that is: ai is the value of f applied to n recursively i times; ai = fi(n)).

The Collatz conjecture is: This process will eventually reach the number 1, regardless of which positive integer is chosen initially.

That smallest i such that ai = 1 is called the total stopping time of n.[3] The conjecture asserts that every n has a well-defined total stopping time. If, for some n, such an i doesn't exist, we say that n has infinite total stopping time and the conjecture is false.

If the conjecture is false, it can only be because there is some starting number which gives rise to a sequence that does not contain 1. Such a sequence would either enter a repeating cycle that excludes 1, or increase without bound. No such sequence has been found.

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