# Puzzles/Easy Sequence 10/Solution

If one indexes the sequence from t, then the function for each term is trivially

$a_{n}=\left\{{\begin{matrix}1\ \mathrm {if} \ n>t\\0\ \mathrm {otherwise} \end{matrix}}\right.$ Because this is the end of the easy sequences, you get something a bit more tricky - another solution:

$a_{n}=\left\{{\begin{matrix}1\ \mathrm {if} \ n Part of what mathematicians do is look for patterns. But be warned, just because it *looks* like it's doing something doesn't mean it's going to keep doing it forever! Showing that something does actually follow a pattern is some of what mathematicians spend their time on. But don't worry about that now, go look at some more sequences!

A bit too trivial for my liking, what about

$a_{n}=(n+4)\ \mathbf {div} \ 6$ resulting in 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2...

What about any of these?

$a_{n}=(n+k)\ \mathbf {div} \ (k+2)$ with k >= 4

The sequence wasn´t supposed to necessarily change from 1 to 2 exactly after the ellipsis ("...").

another solution: $a_{n}=(k*f(n))\ \mathbf {div} \ f(n)$ with $k\in \ ]1,2[$ f(n) = f(n-1) ! for n > 1 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...