# Puzzles/Easy Sequence 10/Solution

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

${\displaystyle a_{n}=\left\{{\begin{matrix}1\ \mathrm {if} \ n>t\\0\ \mathrm {otherwise} \end{matrix}}\right.}$

Because this is the end of the easy sequences, you get something a bit more tricky - another solution:

${\displaystyle 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

${\displaystyle a_{n}=(n+4)\ \mathbf {div} \ 6}$

resulting in 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2...

${\displaystyle a_{n}=(n+k)\ \mathbf {div} \ (k+2)}$ with k >= 4
another solution: ${\displaystyle a_{n}=(k*f(n))\ \mathbf {div} \ f(n)}$ with ${\displaystyle k\in \ ]1,2[}$