I’m studying for a number theory exam. Our review sheets offers the question: Under what conditions will $n$ divide $n \choose k$ for all 1 $ \leq k \leq n-1$? I can see that this will be true for any prime $n$, and don’t think that it would be true for any composite $n$, but […]

Suppose $$\sum_{k\ {\rm odd}}^n {n \choose k} 2^{(k-1)/2} = \sum_{k\ {\rm odd}}^m {m \choose k} 2^{(k-1)/2} 3^{(m-k)/2}.$$ Does $m=n=1$? Clearly $m \leq n$, and for every $n$ there is at most one $m$.

This was inspired by this post. Define, $$a_n = {_2F_1}\left(\tfrac{1}{2},-n;\tfrac{3}{2};\tfrac{1}{2}\right)$$ $$b_n = \sum_{k=0}^n \binom{-\tfrac{1}{2}}{k}\big(-\tfrac{1}{2}\big)^k$$ where $_2F_1$ is the hypergeometric function and binomial $\binom n k$. The first few numerators $N$ are, $$N_1(n) = \color{brown}{1, \,5, \,43, \,177, \,2867, \,11531, 92479}, \,74069, 2371495,\dots$$ $$N_2(n) = \color{brown}{1, \,5, \,43, \,177, \,2867, \,11531, 92479}, \,370345, 11857475,\dots$$ respectively, and […]

A box has $m$ blue balls and $n$ red balls. You are randomly drawing a ball from the box one by one until drawing $k$ red balls ($k < n$)? What would be the average number of trials needed? To me, the solution seems to be $$\sum_i i * \frac{\mbox{the number of cases where k-th […]

I’m trying to use the binomial coefficient: $$\binom{x}k=\begin{cases} \frac{x^{\underline k}}{k!},&\text{if }k\ge 0\\\\ 0,&\text{if }k<0\;, \end{cases}$$ To check that ${-1\choose 0}=1$. But it doesn’t make sense. I’m using, specifically: $$\binom a k = \frac{\alpha(a-1)(a-2)\cdots(a-k+1)}{k(k-1)(k-2)\cdots 1}$$ Then I guess it would be: $${-1 \choose 0}=\frac{(-1)(-1-0)}{0(?)}$$ The denominator should be a product of decreasing numbers from $k$ to […]

Let $p\in \mathbb{R}$ and $n\in \mathbb{N}$ and $$\binom{p}{n}=\frac{p(p-1)(p-2)…(p-n+1)}{n!}$$ b) Prove $$n\binom{p}{n}=p\binom{p-1}{n-1}$$ Thanks for all the help with a! I definitely understand it now. Now part b is simply asking me to multiply those, correct? I did that and it just seems to easy. Also I never used the fact that $$\binom{p}{0}=1.$$ Was I supposed to […]

Find $v_p\left(\binom{ap}{bp}-\binom{a}{b}\right)$, where $p>a>b>1$ and $p$ odd prime. Here $v_p(k)$ denotes the largest $\alpha\in\mathbb Z_{\ge 0}$ s.t. $p^\alpha\mid k$. We have $p\nmid\binom{ap}{bp}$ and $p\nmid \binom{a}{b}$. $$\binom{ap}{bp}-\binom{a}{b}=\frac{(ap)!b!(a-b)!-a!(bp)!(ap-bp)!}{(bp)!(ap-bp)!b!(a-b)!}$$

For which odd integers $n>1$ is it true that $2n \choose r$ where $1 \le r \le n$ is odd only for $r=2$ ? I know that $2n \choose 2$ is odd if $n$ is odd but I want to find those odd $n$ for which the only value of $r$ between $1$ and $n$ […]

Numerically calculating the sum of the squares of the $m$th row of Pascal’s triangle, I found that for at least the first $10$ or so cases $$\sum_{i=0}^m \binom{m}{i}^2=\frac{(4m-2)!!!!}{m!}$$ Where $(4k-2)!!!!=(4k-2)(4k-6)(4k-10)\cdots 6\cdot2=2^m(2m-1)!!$. Of course the problem reduces to, using the well-known formula, $$\frac{(2m)!}{(m!)^2}=\frac{2^m(2m-1)!!}{m!}$$ $$(2m-1)!!=\frac{(2m)!}{m!2^m}$$ Is there a direct proof for the above equation, or (second-best) is […]

Inspired by this question I’m trying to prove that $$\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!} \approx e^{\frac{x}{2}}$$ So I needed to find the value of $$\frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{e^{\frac{x}{2}}} = \frac{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{m! (2m-k)!}{(m-k)!(2m)!}\frac{x^k}{k!}}{\lim_{m \to \infty} \sum_{k = 0}^{m} \frac{\frac{x}{2}^k}{k!}} \\ = \lim_{m \to \infty} […]

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