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7

You'll need a list of primes out to around $71{,}000$. This is because if $n\lt10^{10}$ is a product of three distinct primes, say $n=pqr$ with $p\lt q\lt r$, then $q\lt\sqrt{10^{10}/p}\le10^5/\sqrt2\approx70{,}710.678$. Note also that $p\lt\sqrt[3]{10^{10}}\approx2154.43$, so there is a fairly straightforward brute-force algorithm that seems practical, at ...


4

By "the product converges", it is meant that it converges in $\mathbb{R}\setminus \{0\}$, otherwise the assertion is false, consider $a_n = \frac{1}{n+2}$ and $t_1 = -1,\, t_2 = -2$. Or it may be allowed that $1 + t a_n = 0$ for finitely many $n$, and the product formed by the nonzero terms converges in $\mathbb{R}\setminus \{0\}$. With the latter viewpoint ...


4

$$\prod_{i=1}^{n+1}\frac{n+1+i}{2i-3}=\prod_{i=1}^n\frac{n+i}{2i-3}\cdot\frac{(2n+1)(2n+2)}{(n+1)(2n-1)}\stackrel{\text{Inductive Hypotesis}}=$$$${}$$ $$=2^n\color{red}{(1-2n)}\frac{(2n+1)\cdot2\color{green}{(n+1)}}{\color{green}{(n+1)}\color{red}{(2n-1)}}=2^{n+1}\left(-1-2n\right)=2^{n+1}(1-2(n+1))$$ Explanation under request of the first equality above: ...


4

First, taking logarithm of the product and using estimates from Mertens' 2nd theorem, we get \begin{align*} \log[\prod_{2<p\le y}\left(1-\frac{2}{p}\right)] &=\sum_{2<p\le y}\log(1-\frac{2}{p})\\ &=-\sum_{2<p\le y}\frac{2}{p}-\sum_{2<p\le y}\sum_{j\ge 2}j^{-1}(\frac{2}{p})^j\\ &=-2\left(\log\log ...


3

Let $P_n=\prod_{i=1}^{n} \frac {n+1}{2i-3}$. Consider the ratio: $$\frac {P_{n+1}}{P_n}=\frac {2n+2}{2n-1}\;\prod_{i=1}^n \left(\frac {n+i+1}{2i-3}\times \frac {2i-3}{n+i}\right)=\frac {2n+2}{2n-1}\;\prod_{i=1}^n\frac {n+i+1}{n+i}$$ $$=\frac {2n+1}{n+1}\times \frac {2n+2}{2n-1}=2\times\frac {2n+1}{2n-1}$$ And the induction hypothesis now completes the ...


2

Note that we have $$\begin{align} ||x\pm y||^2&=\langle x\pm y,x\pm y\rangle\\\\ &=||x||^2+||y||^2\pm \left(\langle x,y\rangle+\langle y,x\rangle\right)\\\\ &=||x||^2+||y||^2\pm 2\text{Re} \{\langle x,y\rangle\}\\\\ \end{align}$$ Therefore, it is easy to see that $$||x+ y||^2-||x-y||^2=4\text{Re}\{\langle x,y\rangle\} \tag 1$$ whereupon ...


2

GAP has an undocumented attribute DirectFactorsOfGroup which is used by StructureDescription. With its help, I've calculated that there are 3079 groups for which DirectFactorsOfGroup can not find factorisation into the direct product, hence the remaining 3459 groups of order 2016 are direct products. Just in case of DirectFactorsOfGroup works under some ...


2

My comments from the other question apply here as well. But we actually can define $Z(a,n)$ even with such a mild class of functions as one including the ceiling function, addition, subtraction, and division. $$Z(a,n) = \left\lceil \frac {n-1}a - \left\lceil \frac{n-1}{a} \right\rceil \right\rceil.$$ Actually, the example you gave in the comments extends ...


1

Well $$\prod_{k=2}^{n} \frac{p_{k}-2}{p_{k}}<\prod_{k=2}^{n} \frac{p_{k}-1}{p_{k}}=2\prod_{k=1}^{n} \frac{p_{k}-1}{p_{k}}\sim \frac{2e^{-\gamma }}{\ln{n}}$$ This means that there $\exists \varepsilon>0$, constant such that $$\prod_{k=2}^{n} \frac{p_{k}-2}{p_{k}} < (1+\varepsilon) \frac{2e^{-\gamma }}{\ln{n}}$$ always. For more details see Mertens' ...


1

Sorry, I would like to make a comment but unfortunately I don't have enough reputation. If $\mathfrak{A}_i$ is strongly separable and it is not lower bounded then $\prod\mathfrak{A}_i$ is strongly separable. To simplify let's consider the case with n=2. Suppose that there is $(a,a'),(b,b') \in \mathfrak{A}_1 \times \mathfrak{A}_2$ such that $\star ...


1

Here is a quick way to bound many kinds of functions defined by a recurrence if one is interested in the function's behaviour for large $n$. Let $\lg x=\log_2 x$. Suppose we are given a recurrence of the form $f(n)=cf(n/2)+p(n), f(1)=1$, where $p(n)$ is a function depending on $n$ and $c$ is a constant. If $p(n)\ge 0$ for all $n\ge 1$, then $f(n)\ge ...



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