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(My actual question is at the very bottom of this posting.)

Suppose you're teaching a course in mathematics-for-liberal-arts majors and it's the last math course they'll ever take. It has almost no prerequisites---say no algebra beyond solving quadratic equations and probably not even that.

Over a period of three weeks you've gotten them accustomed to the idea that consecutive integers can never have any prime factors in common. You don't prove this by a method that uses algebra (since then their attention would be on struggling to understand the algebra) but rather you've pointed out that if 56 is a multiple of 7, then the next multiple of 7 isn't due until 7 units later, and the last 7 units earlier, so 55 and 57 certainly cannot be multiples of 7. They've turned in homework on this idea; they've done in-class quiz problems on it.

Then you tell them: say we start with some finite list of prime numbers, e.g. $5$ and $7$. Mutliply them and get 35. Consider two consecutive integers, 35, and 36. They can't have prime factors in common: $$ \begin{align} 35 & = 5\times 7 \\ 36 & = 2\times2\times3\times 3 \end{align} $$ So you get some additional prime numbers that you now add to your list: $$ 5, 7, 2, 3 $$ Now iterate $2\times3\times5\times 7 = 210$. Suppose this time we consider the consecutive integers 210 and 209, where I've chosen 209 rather than 211 because its prime factors aren't very big: $$ \begin{align} 210 & = 2\times3\times5\times 7 \\ 209 & = 11\times 19 \end{align} $$ Add this to the list: $$ 2, 3, 5, 7, 11, 19 $$ Iterate: $2\times3\times5\times7\times11\times19=43890$.

Here's an unpleasant fact: $43889$ and $43891$ are prime. I can't pick one to keep the arithmetic moderately comfortable.

Of course all of this that one presents in class will be a story with a moral: this is how you prove that the prime numbers will always keep on going; your finite list can never be complete.

My question: Is there some way to churn out examples of reasonable starting sets and choices of $\pm1$ (such as I made in the case of $210$) that will let this go on for a fairly large number of steps without getting really big primes?

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    If I understand the question correctly, you don't really need an *algorithm*; it would be enough to have the (finite!) list of good starting sets/paths that don't exceed the "reasonable"/"moderately comfortable" criterion. Right?2011-10-12
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    I notice that if I'd picked $34$ instead of $36$, I get: $5,7,2,17$, and then again subtracting $1$ rather than adding $1$ I get: $5, 7, 2, 17, 29, 41$. If on the next step I add $1$ rather than subtracting $1$, I get $5, 7, 2, 17, 29, 41, 3, 19, 103, 241$. One more step and I have the option of adding $13, 105727, 1456561$ to the list or else getting a 13-digit prime number. Could just doodling like this be the most efficient way to construct pedagogically good examples?2011-10-12
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    @ShreevatsaR: For the particular purpose mentioned here that's probably right. Another thought comes to mind: suppose some unpleasantly large number comes along, and we admit it, and go on from there; could it be that at the very next step we can get a long list of fairly small primes?2011-10-12

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