The number below might serve as a good example: $$13532385396179.\tag{$\star$}$$ It turns out that if we want to find the prime factorisation of this number, we get that this is equal to $$13\times 53^2\times 3853\times 96179.$$ This number is a counter-example to one of John Conway's five problems $-$ the very last one, as a matter of fact. Problem $5$ is known as Climb to a Prime and is stated as follows:

Problem 5. Climb to a Prime${}$:

Let $n\in\mathbb{Z}^+$. Write the prime factorisation of $n$ in the usual way, e.g. $60 = 2^2\times 3\times 5$, in which the primes are written in increasing order, and exponents of $1$ are omitted. Then, bring the exponents down to the line and emoit all multiplication signs, obtaining a number $f(n)$. Now, repeat.

So for example, $f(60) = f(2^2\times 3\times 5) = 2235$. Next, because $2235 = 3\times 5\times 149$, it maps under $f$ to $35149$, and since $35149$ is prime, it maps to itself. Thus, $60\to 2235\to 35149$ $\to 35149\to 35149\to\ldots$, so we have climbed to a prime and we stop there forever.

The conjecture, in which John Conway seemed to be the only believer at the time, was that every number eventually climbs to a prime. The number $20$, however, has not been verified to do so. Observe that $20\to 225\to 3252\to 223271\to\ldots$, eventually getting to more than one hundred digits without yet reaching a prime!

Of course, the number $(\star)$ is a counter-example. It is the first counter-example thus far, and is larger than $10^{14}$, which in my opinion, would make this number fairly big.

The number below might serve as a good example: $$13532385396179.\tag{$\star$}$$ It turns out that if we want to find the prime factorisation of this number, we get that this is equal to $$13\times 53^2\times 3853\times 96179.$$ This number is a counter-example to one of John Conway's five problems $-$ the very last one, as a matter of fact. Problem $5$ is known as Climb to a Prime and is stated as follows:

Problem 5. Climb to a Prime${}$:

Let $n\in\mathbb{Z}^+$. Write the prime factorisation of $n$ in the usual way, e.g. $60 = 2^2\times 3\times 5$, in which the primes are written in increasing order, and exponents of $1$ are omitted. Then, bring the exponents down to the line and omit all multiplication signs, obtaining a number $f(n)$. Now, repeat.

So for example, $f(60) = f(2^2\times 3\times 5) = 2235$. Next, because $2235 = 3\times 5\times 149$, it maps under $f$ to $35149$, and since $35149$ is prime, it maps to itself. Thus, $60\to 2235\to 35149$ $\to 35149\to 35149\to\ldots$, so we have climbed to a prime and we stop there forever.

The conjecture, in which John Conway seemed to be the only believer at the time, was that every number eventually climbs to a prime. The number $20$, however, has not been verified to do so. Observe that $20\to 225\to 3252\to 223271\to\ldots$, eventually getting to more than one hundred digits without yet reaching a prime!

Of course, the number $(\star)$ is a counter-example. It is the first counter-example thus far, and is larger than $10^{14}$, which in my opinion, would make this number fairly big.