TITLE: Counterexamples to "Naive Induction"
QUESTION [6 upvotes]: I was teaching a nine-year-old friend about prime numbers. When I asked him if he thought there were finitely or infinitely many primes, he answered confidently that there must be an infinite number.  "How do you know?" I asked.  "Because I can keep thinking up larger and larger primes. It's easy!" By way of proof, he came up with a new larger prime.
I call this "Naive Induction" (there might be a better term). I am looking for not-too-complicated counterexamples where

It appears to be the case that there are infinitely many members of
a set, or (equivalently) that some property is true for all
integers, but
It can be demonstrated that there is a largest member of the set, or a largest number with some property.

Any suggestions? Thanks.

REPLY [6 votes]: I just came across an example that really remind me of this question. This comes from a kiddie colloquium talk given by T. Church at Stanford. I quote the abstract of the seminar.
Define a sequence of integers by 
$a_0 = 3$,
$a_1 = 0$,
$a_2 = 2$, and then recursively by
$a_{n+3} = a_n + a_{n+1}$.
Calculate out a few terms, or a few thousand, and you'll notice a curious pattern: the $n$-th term $a_n$ is divisible by $n$ exactly when $n$ is prime!
The first counter-example is $n = 271441$, for which $a_n$ has over thirty thousand digits.