TITLE: Variety of proofs of a simple proposition in arithmetic
QUESTION [8 upvotes]: I have a couple of simple proofs of a simple proposition and I'm curious to see the variety of different approaches others would take to prove the same thing.
Definition: The dilation of a sequence $a_1, a_2, a_3,\ldots$ by a factor $n$ is the sequence $b_1,b_2,b_3,\ldots$ for which
$$
\begin{cases}
b_{kn} = a_k & \text{for } k=1,2,3,\ldots, \\
b_j = 0 & \text{if $j$ is not a multiple of $n$.}
\end{cases}
$$
Thus the dilation of $a_1,a_2,a_3,\ldots$ by a factor of $3$ is
$$
0,\ 0,\ a_1,\ 0,\ 0,\ a_2,\ 0,\ 0,\ a_3,\ 0,\ 0,\ a_4,\ \ldots
$$
(This is not standard terminology as far as I know, so tell me if there's some standard name for this.)
Now let $\{a_k\}_{k=1}^\infty$ consist of an infinite sequence of repetitions of $1,0,0,0,1,0$ (a $\text{“}1\text{''}$ in the $1$st and $5$th positions and $\text{“}0\text{''s}$ elsewhere).  In other words $a_k$ is just the indicator that $k$ is coprime to $6$.
Now start by letting $c$ be an infinite sequence of $0$s and proceed as follows:
1. Let $n$ be the smallest index for which $c_n=0$ (thus initially $n=1$);
2. Let the new value of $c$ be $(c + \text{the dilation of } a \text{ by } n)$ (where the occurrence of $c$ inside the round brackets is the value of $c$ we had at the end of step $1$.);
3. Go back to step $1$.
This goes on forever.
Proposition: This process converges to an infinite sequence of $1$s.  (Thus, we never add $1$ to any position that was not $0$.)
As I said, I have a couple of simple proofs and I am curious to see the variety of different approaches others would take to prove the same thing.

REPLY [2 votes]: Alright, here's one way that I know of.
Suppose $N$ is an infinitely large integer (oh $\ldots$ um $\ldots$ TRIGGER WARNING: If you suffer pain and distress as a result of reading something whose rigorous definition has not been stated and might be problematic, then don't read the foregoing phrase) whose prime factorization contains infinitely many $2$s and infinitely many $3$s and nothing else.  Look at the sequence
$$
\frac 1 N, \frac 2 N, \frac 3 N,\ \ldots,\ \frac{N-3} N, \frac{N-2} N, \frac{N-1} N. \tag a
$$
The first time one comes to step 2. in the algorithm, one adds $1$ in every position in $(\mathbf a)$ in which the fraction is in lowest terms.  
The second time one comes to step 2., one adds a $1$ in every position in which one reduces the fraction to lowest terms only by dividing the numerator and denominator by $2$ (not including those cases in which division by $2$ reduces the fraction incompletely).
The third time one comes to step 2., one adds a $1$ in every position in which one reduces the fraction to lowest terms only by dividing the numerator and denominator by $3$ (not including those cases in which division by $3$ reduces the fraction incompletely).
The fourth time one comes to step 2., one adds a $1$ in every position in which one reduces the fraction to lowest terms only by dividing the numerator and denominator by $4$ (not including those cases in which division by $4$ reduces the fraction incompletely).
The fifth time one comes to step 2., one adds a $1$ in every position in which one reduces the fraction to lowest terms only by dividing the numerator and denominator by $6$ (not including those cases in which division by $6$ reduces the fraction incompletely).
and so on.  One reduces by dividing by $2$, or $3$, or $4$, or $6$, or $8$, or $9$, or $12$, etc.  This sequence contains all integers that have no prime factors except $2$ and $3$.  This process will never reduce a fraction that's already reduced, so it never adds a $1$ to a position more than once.
\begin{align}
& \frac{2520}{2\times2\times2\times\cdots \times 3\times3\times3\times\cdots} \\[10pt]
= {} & \frac{\overbrace{2\times2\times2}\times\overbrace{3\times3}\times35}{2\times2\times2\times\cdots \times 3\times3\times3\times\cdots} \\[10pt]
= {} & \frac{35}{\not2\times\not2\times\not2\times2\times2\times\cdots \times \not3\times\not3\times3\times3\times\cdots}
\end{align}
(It is not hard to rephrase all this in a way that doesn't mention an infinitely large integer.)
In a comment above I asked "How would you write this same argument in a way intended to make it comprehensible to secondary-school pupils while still keeping it simple?".  This is how I would do that.  So a question remains: Are there other ways, in some essential way different?