Show that any integer can be expressed as $S^{k}_{n}$ Consider the following sequence: $a_1=6$, $a_2=4$, $a_3=1$, $a_4=2$, and $a_n=a_{n-4}$ for $n \geqslant 5$.
Let $S^{k}_{n}$ be the sum of $k + 1$ elements from $a_n$ of the previous sequence. For example:
$
  S^{0}_{3} = 1
\\
  S^{0}_{4} = 2
\\
  S^{1}_{3} = 2 + 1 = 3
\\
  S^{0}_{2} = 4
\\
  S^{1}_{2} = 4 + 1 = 5
\\
…
\\
  S^{6}_{2} = 4 + 1 + 2 + 6 + 4 + 1 + 2 = 20
\\
…
\\
$
Show that any positive integer can be expressed as $S^{k}_{n}$.

This question was on an entrance exam for graduation. I wrote it from my memory, as the exam is not available online yet, maybe the wording is not 100% precise but I hope the examples are clear.

As you notice, the sequence $a_n$ is by definition periodic: $6,4,1,2,6,4,1,2,6,4,1,2,...$
The sum of a whole "period" is $6+4+1+2=13$.
You can easily check that every number between $1$ and $12$ can be obtained as a sum of CONSECUTIVE numbers in the sequence $a_n$ (for instance, $9=1+2+6$). Then you just write any desired number $k$ (choose for example $k=139$) as a multiple of $13$ plus a number between $0$ and $12$ (in the example, $139 = 13 \cdot 10 + 9$). Then it just suffices to choose the right starting point and the right number of elements (in the example: to get $9$ you need to start at some $1$ in the sequence, so that you get $1+2+6=9$; for example, $S_3^{42} = 1+2+6+4+1+2+6+4+...+1+2+6= 13 \cdot 10 + 9 = 139$).
