Search Results

I assume you have checked this for small $n$, so I will only consider large $n$. I will show: if $n \geq 14$, then there exists a product $\sigma = (i_1 j_1) (i_2 j_2) \dots (i_5 j_5)$ of five transpo …

One can take $r = 1 + \frac{d(d-1)}{2}$. Indeed, one can consider the map $$ (\alpha_i)_{i=1}^r \in [|1,N|]^{r} \mapsto (\sum_{i=1}^r \alpha_i^k)_{k=1}^{d-1} \in \prod_{i=1}^{d-1} [|1,rN^k|]. $$ The …

Most of the proofs of Dirichlet's theorem on primes in arithmetic progressions actually give a Mertens-like theorem, and then the (weaker) statement Chebyshev-like bound : if $(a,q) = 1$ then …

One has $S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where $$ C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right). $$ This follows for exam …

Of course since $R = \{ (a,b) \in \mathbb{N} \times \mathbb{N} \ | \ (a=1) \implies (b=1) \} $. Indeed for $a > 1$ one can take $n = a^k-b$ for $k$ large enough.

Your guess is correct. Since $2^{\omega(n)} = \sum_{d | n} \mu^2(d)$ one has $$ S = \sum_{(a,b,c) \in R(X)} 2^{\omega(4ac-b^2)} = \sum_{d \leq X} \mu^2(d) |R_d(X)| $$ where $R_d(X)$ is the set of inte …

Let $x_j = \sum_{n \geq 1} \mathrm{1}_{a_n=j} 3^{-n}$. Then $$ x_0 + x_1 + x_2 = \frac{1}{2} \\ x_1 + 2x_2 = x \\ x_0 + 2x_1 = f(x). $$ If $x$ and $f(x)$ are algebraic then the equations above would …

First, the Fontaine-Winterberger isomorphism can also be recovered from a theorem of Deligne, namely Thm 2.8 here. Deligne showed that if two local fields $K_1$ and $K_2$ (possibly of different charac …

Let $R > 1$ and $\lambda \in \mathbb{R}$ be such that $$ \int_{1}^R \mathrm{tanh}(\frac{\lambda x}{2}) \frac{d x}{x} = \log R -2. $$ Then standard techniques in large deviation theory yield $$ \frac{ …

The simplest explanation is that it is a mistake. One can however complete the proof as follows: If $X$ is large enough then $u(m) \in \{ 0,1,2 \}$ for all $m$ since $e < 3$. Thus the number of disti …

First note that $\prod_{\xi} (1 - \xi^n q^n)$ is equal to $(1- q^n)^5$ if $5 | n$ and to $1 - q^{5n}$ otherwise. Moreover $$ \varphi(q) = \prod_{n \geq 1} (1 - q^n)^{e_n} $$ where $e_n = 1,-2,3,$ or $ …

Yes. If we set $\alpha = (a+b)/2$ and $\beta=(a-b)/2$, then the lattice generated by the four vectors $$ \binom{\alpha}{\beta},\binom{-\beta}{\alpha},\binom{\alpha+\beta}{\alpha -\beta},\binom{\beta …

Yes. The coefficient of $x^n$ in $F_r$ is the $2n$-th derivative at $t = 0$ of the function $$ t \mapsto (\cos t)^{r} = \frac{1}{2^r} \sum_{k=0}^r \binom{r}{k} e^{it(2k-r)}. $$ But the successive deri …

The following is only a partial answer. The number $H(n,k)$ is not a $2$-adic integer "for most $n$". I will stick to the case $k=2$ for convenience. I claim that there is a sequence $(a_j)_{j \geq 0} …

Two very nice answers have already been given, but I would like to add that Michel Balazard has a book in preparation on this topic (written for undergrads), in which he tries to give a deduction of t …