How do I find a range $[n, 2n]$ such that it contains the smallest positive integer $x$ such that $x^x$ contains the digits $2016$ in a row (consecutively), up to sufficiently high probability? I asked my older brother and one of his math postdoc friends, and they had no clue, so I figured this would be fair game here.
Edit: No computer program here please, I would like to some "by hand" insights.
You can calculate the probability that $x^x$ contains the digit sequence 2016. Let f(n) be the number of digits in $x^x$, which is about $f(n) = x \cdot \log_{10}(x)$, then the probability is about $0.9999^{f(x)-3}$.
The expected number of values x ≤ N where $x^x$ contains the digits 2016 is very roughly $0.0001 N^2 \log_{10}(N/2)/ 2$. So we should find the smallest x by checking say 200 values x, no problem with the right software.
If you don't have any software that can find the smallest x, you can estimate the probability that [N, 2N] contains the smallest x: Let p(N) be the probability that $x^x$ contains 2016 for some x ≤ N; this can be calculated from the expected number of x's. And then we just need p(2N) - p(N-1) and find for which N it is largest.