Puzzles/Easy Sequence 10/Solution

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

$a_n=\{\begin{array}{c}1\mathrm{i}\mathrm{f}n>t\\ 0\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$

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

$a_n=\{\begin{array}{c}1\mathrm{i}\mathrm{f}n<t\\ 1\mathrm{i}\mathrm{f}t<n<t+7\\ 1\mathrm{i}\mathrm{f}t+7<n\\ 0\mathrm{o}\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{r}\mathrm{w}\mathrm{i}\mathrm{s}\mathrm{e}\end{array}$

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!

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A bit too trivial for my liking, what about

$a_n=(n+4)𝐝𝐢𝐯6$

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

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What about any of these?

$a_n=(n+k)𝐝𝐢𝐯(k+2)$ with k >= 4

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

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another solution: $a_n=(k*f(n))𝐝𝐢𝐯f(n)$ with $k\in ]1,2[$

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f(n) = f(n-1) ! for n > 1 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...