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I'll probably end up asking more programming questions on StackExchange forums than math questions, but I'll lead off with a math question.

In my Number Theory class this past semester, I worked on a project on Bertrand's Postulate (aka Chebyshev Theorem) and came across this paper towards the end, but didn't have time to go through it.

I find it very fun (because it uses very neat objects called Newton polygons which I hadn't heard of before) so I've looked at it over break, and I'm guessing it would be pretty easy for most people here to understand (though I think it's advanced enough that I didn't put it under elementary-number-theory based on topics listed for that). So I was wondering if anyone would be interested in discussing it with me? There are some things in it which I don't quite understand, but I could also maybe explain some things to others here if any of you would like; and that would help me know whether I truly understand what I think I do.

Anyways, the areas where I'm confused are as follows:

On page 10, it says: "It is easy to see that there is no loss of generality in restricting to $a_j = 1$ for $0 \le j \le m$." Am I correct in my thinking that, because $|a_0| = |a_m| = 1$ is a condition of the lemma rather than a conclusion of it, this sort of generalization is acceptable? Which is also why it can later be stated that $b_j = m!/j!$, since it doesn't violate a restriction on $b_j$? I hope that my question is clear.

I'm a little fuzzy on the reasoning used in page 13 for why certain terms remain when deleting other terms. Like when $k = 3$, is it true that we can conclude that either $x$ or $x-1$ must be $3$ because otherwise those three consecutive terms contain a prime factor larger than $3$ (a special case of Theorem 2 which obviously would've had to come from different reasoning)? I think I'm missing something pretty obvious here, and I hope it doesn't have something to do with Bertrand's Postulate but I don't see how it would.

On page 14 (under "For this, we observe") it states that $_xC_k$ equals the product of all $p^a$ for which $p \le k$. I must be missing something here, because (for example) $_{22}C_{11}$ has 17 as a factor but $17 > 11$. I'm accounting for the fact that $x \ge 2k$ and $k \ge 11$. What else is there?

On page 16, how are $17 \le k \le 29$ excluded by (2)? Does that mean $_xC_k \ge _{4k}C_k$ fails to hold for all $x \ge 4k$ when $k$ is in that range? And how would such verification be performed without testing all of those $x$, which is impossible.

Finally, I wish either the proof of (b) or the proof of (c) would have been finished, as I'm thinking those are more similar than they are to the proof of (a). I don't see how to finish (c) in a manner similar to (a), especially since I don't see another formula (besides the one at the end) relating $_xC_k$ to an expression in only $x$ the way it was related to an expression in only k for proving (a).

If anyone could address any of these points, I would really appreciate it. Have a nice day! :)

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In recent times, it is increasingly standard to denote binomial coefficients by $\binom{n}{k}$ instead of $_nC_k$. Would you mind if we make this change in the question? –  Srivatsan Jan 7 '12 at 22:56
    
Not at all. And thank you to Michael for making that change, as well as the others in formatting. –  Gregory Fowler Jan 8 '12 at 5:57
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