15
$\begingroup$

Call a function $f : \mathcal{N} \rightarrow \mathcal{N}$ repetitive if for every finite sequence of natural numbers $(a_1, a_2, \cdots,a_n)$ there exists a number $k \in \mathcal{N}$ satisfying

$f(k) = a_1$, $f(k+1) = a_2$, ... , $f(k + n -1) = a_n$.

Show that if $f$ is repetitive, then for any combination of $a_i$'s there are actually infinitely many numbers $k$ with this property.

I've been trying to figure out a proof for this for a while now. I'm feeling that the fact that the number $k$ exists for every sequence allows us to manipulate other sequences or something like that and get infinitely many numbers that way. But I haven't been able to actually find a way to construct infinitely many numbers $k$ for a random sequence.

$\endgroup$

2 Answers 2

27
$\begingroup$

There's a $k$ with $f(k)=a_1,\ldots,f(k+n-1)=a_n$, $f(k+n)=1$.

There's a $k'$ with $f(k')=a_1,\ldots,f(k'+n-1)=a_n$, $f(k'+n)=2$ etc.

$\endgroup$
11
$\begingroup$

For every finite sequence there are infinitely many finite distinct sequences that extend it.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .