Prove that the sequence $012345678910111213...$ (all non-negative integers written one by one in natural order) is not periodic.
I want to know the shortest and most elegant way to prove it. Can you help?
My proof:
Assume the sequence is periodic. Let $A$ be periodic subsequence with length $p$. Let $n$ be position of first occurrence of $A$. $A$ must contain at least 2 different digits, otherwise the tail of original sequence contains only single digit which is impossible, because any 2 consecutive integers have at least one different digit - last one. Let us take number $n$ such that it has only single digit in decimal notation, number of digits is more than $2p-1$ and it is more than a number that a digit at position $p + n - 1$ belongs to. We will get the contradiction proving that $n$ can not have all different digits because decimal notation of $n$ contains $A$. It is equivalent to the following lemma:
Lemma. Non-negative integers $n$ are colored: $1,2,..,k$ have color A; $k+1,k+2,..,2k$ have color B, etc. Prove that any $2k-1$ subsequent integers contains $k$ integers of the same color.
Proof. Let $c$ be number of different colors in subseqnece of $2k-1$ integers. If $c>2$ we immediately get desired subsequence - just take a subsequence with a color different from the colors of first and last elements. If $c=2$ we can note that there are only 2 ways to choose $2k-1$ subsequent integers from from $2k$ subsequent integers and in both cases it is clear that lemma is true, because first $k$ or last $k$ elements have the same color.
Shortest proof I could think of: let the period be $k$, then $10^k$ contains $k$ sequential $0$s and thus all digits in the sequence are $0$s - which is absurd.