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14

Since $$2^n\equiv -n\pmod{2^n+n}$$ we deduce $$8^n = (2^n)^3 \equiv (-n)^3 \pmod {2^n+n}$$ So $2^n+n\mid 8^n+n$ if and only if $2^n+n\mid n-n^3$. For $n\geq 10$, $2^n>n^3$ so $2^n+n$ cannot divide $n^3-n=-(n-n^3)$. Clearly, if $n=0,1$, $n^3-n=0$ so $2^n+n\mid n^3-n$. So you really only need to check additionally $n=2,3,4,5,6,7,8,9$ by hand. You get ...


10

It converges to $0$, in fact $\frac{1}{k^7\sin{k}}$ already converges to $0$, see Theorem 2 here. This theorem gives a nice characterization of the irrationality measure of $\pi$ as the borderline number $\mu$ such that $\frac{1}{k^{u-1}\sin{k}}$ converges to $0$ for $u>\mu$, and diverges for $u<\mu$. So $\frac{1}{k^7\sin{k}}$ converges because $\mu$ ...


9

The predominant reason for interest in the Riemann Hypothesis is for its importance to other mathematical results. For example, proving the Riemann Hypothesis would immediately tell you a lot about the distribution of prime numbers. It is known to be equivalent to the following statement: $$\left|\pi(x) - \int_0^x \frac 1{\log ...


8

$4\uparrow\uparrow\uparrow\uparrow\uparrow4$ is far far larger than $20\uparrow\uparrow\uparrow\uparrow20$. For convenience, I will use $\uparrow^n$ for $\underset{n}{\underbrace{\uparrow\uparrow\cdots\uparrow}}$. We want to know the following: ...


7

Hint $\,\ n^2\!+2\mid 2\!+\!an\,\Rightarrow\,n^2\!+2\mid (2\!+\!an)(2\!-\!an)+a^2(n^2\!+2) \,=\, 4+2a^2\,$ Remark $\ $ Intuitively we took the "norm" of a quadratic number, i.e. $\qquad\qquad\begin{eqnarray} {\rm mod}\ n^2\!+2\!:\,\ n\equiv \sqrt{-2}\ \ \ {\rm so}\ \ \ 0 &\equiv& \alpha \ =\ \ \, 2 + a\sqrt{-2}\,\equiv\, 2+a\,n \\ \Rightarrow\ ...


7

$$\frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\frac{7}{2^4}+\cdots=\sum_{n=1}^{\infty} \frac{2n-1}{2^n}$$


7

First, if $G$ is connected, then $G/H$ is connected (the map $G\to G/H$ defining the ``quotient" is faithfully flat, in particular surjective). Take a connected linear algebraic group with disconnected center, e.g., $\mathrm{SL}_2$ in characteristic zero. The quotient by the center is connected, but the center isn't connected (it's ├ętale and non-trivial). In ...


5

There are an endless number of variants of statistical conjectures that can be made about the prime numbers. The Goldbach conjecture (at least in my opinion) doesn't get to the heart of what makes primes tick. Indeed, there are many different heuristic justification of it that are based solely on the growth of the primes. So in a manner of speaking, Goldbach ...


5

The associated Dirichlet series $\sum c(n^2)/n^{2s}$ is obtained by integrating $f$ against the product of the weight $1/2$ theta series and a half-integral-weight Eisenstein series, giving Shimura's (c. 1975) Rankin-Selberg-style integral representation of the symmetric square $L$-function attached to $f$. Gelbart-Jacquet showed a case of "functoriality", ...


4

Let $n=p^2$. Then $\varphi(n)=p^2-p=n-\sqrt{n}$. So we have equality for infinitely many $n$. If $n=p^e$ where $e\gt 2$, then $\varphi(n)=p^e-p^{e-1}=n-n^{1-1/e}$. Worst case is $e=3$, $p=2$. In this case we have $n^{2/3}\ge kn^{1/2}$ where $k=2^{1/6}$. Thus $\varphi(n)\le n-2^{1/6}n^{1/2}$. If $n=ab$ where $a$ and $b$ are greater than $1$ and ...


4

Your proof is basically valid, but there are special circumstances for this particular ring of (algebraic) integers. You first need to check that $y$ can't be divisible by $3.$ Note that ( in the usual ring of integers), if we have $3$ divides $1 + x + x^{2},$ then we must have $x \equiv 1$ (mod $3$),say $x = 3z+1$ for some integer $z.$ Then $x^{2}+x +1 = ...


4

For real numbers, yes. One infinite family of solutions is for instance $$ (a,b,c,d) = \left(r\frac{l+k}{2}, r\frac{l-k}{2}, \frac{1}{r}, \frac{1}{r}\right), $$ for $r \neq 0$. For integers, if $k = l = 0$, then $(a,b,c,d) = (n,n,0,0)$ gives infinite number of solutions. If $k \neq 0$ or $l \neq 0$, $$ 0 <k^2+l^2 = (ac-bd)^2+(ab+cd)^2= ...


3

This scaling is usually referred to as a power law. To see why, note that one assumes that for every nonnegative integer $i$, $k=a/r^i$ solves the equation $f(k)=2^ik$, for some fixed $a\gt0$ and $r\gt1$, that is, $f(a/r^i)=a2^i/r^i$ for every $i$. If $x=a/r^i$ then $i=\log(a/x)/\log r$ and $a2^i/r^i=x2^i=x\mathrm e^{i\log2}=x(a/x)^{\log2/\log r}$, hence ...


3

I'm going to caracterize the way to represent $n$ as sum of natural numbers raised to different powers. I'll be using the residue $\text{mod} 32$ so I won't be using exponents larger than $4$. We can represent $n - (n \mod 32)$ using different powers of $2$ (raising to powers larger than $4$). $$\begin{array}{|c|c|} \hline n \mod 32 & \text{Use:}\\ ...


3

Another approach is to use summation by parts. This method works as follows. If $$S_N = \sum_{n=1}^{N} a_n b_n$$ then we can define $$B_n = \sum_{k=1}^{n} b_k$$ Then $$S_N = a_N B_N - \sum_{n=1}^{N-1} B_n (a_{n+1} - a_n)$$ In this problem, we can set $a_n = 2n-1$ and $b_n = 1/2^n$. Then $$B_n = \sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}$$ and $a_{n+1} ...


3

$$\sum_{n=1}^{+\infty}\frac{2n-1}{2^n}=-1+2\sum_{n=1}^{+\infty}\frac{n}{2^n}=-1+2\sum_{n=1}^{+\infty}\sum_{m\geq n}\frac{1}{2^m}=-1+4\sum_{n=1}^{+\infty}\frac{1}{2^n}=-1+4=3.$$


3

In this answer, I give a pictorial proof that $$\frac14+\frac28+\frac3{16}+\frac{4}{32}+\frac{5}{64}+\cdots=1$$ With your sum after dividing by $2$ we have $$\frac14+\frac38+\frac5{16}+\frac{7}{32}+\frac{9}{64}+\cdots$$ and the same picture can be used if we expand the original $\frac12\times\frac12$ square to the left as well as to the right and upward. The ...


3

The ring $\mathbb{Z}[\sqrt{-5}]$ of complex numbers of the form $a+b\sqrt{-5}$ with $a,b\in \mathbb{Z}$ is not a UFD because $6=2\cdot 3$ and $6=(1+\sqrt{-5})(1-\sqrt{5})$; none of $2,3,1+\sqrt{-5},1-\sqrt{-5}$ are associates, so even when we make these irreducible factorizations (they actually are already), they won't be the same.


3

Yes, there is. Since $X=\frac{10^N-1}{9}$ and we can find the inverse of $9$ in $(\mathbb{Z}/M\mathbb{Z})^*$ in $O(\log M)$ time with the extended Euclidean algorthm, we just need to find $10^N\pmod{M}$. This can be done in $O(\log N)$ time with the repeated squaring algorithm. If you know the factorization of $M$ or a large portion of primes dividing ...


3

Consider $$u_n = \frac{10^n - 1}{3} \hspace{1cm}v_n = 10^{n}$$ Then: $$\frac{u_n}{v_n} = 0.\underbrace{3333\cdots 3}_{n \text{ times}}$$ What means that $$\lim_{n\to\infty} \frac{u_n}{v_n} = 0.333\cdots = \frac{1}{3}$$ So no, that condition is not sufficient.


3

Here's a similar condition that is sufficient: there exists a sequence of integers $u_n, v_n \to\infty$ such that $(u_n, v_n) = 1$ and $$\lim_{n\to\infty} v_n a - u_n = 0.$$ (Alternatively, we could just require that $a$ not be exactly equal to any $u_n/v_n$ rather than $(u_n,v_n)=1$.)


3

Here is my webpage with information about the 290-Theorem. It has links to the preprint, all escalator form datafiles and computer code. In Spring 2014 I had a student (Kate Thompson) graduate from UGA after learning about the analytic techniques involved in doing similar computations for totally positive definite $\mathcal{O}_F$-valued quadratic forms ...


3

Let's look at how the method works for the example sequence. The sequence $a_n$ is generated by the recurrence $a_{n+1}=100a_n+67$. Applying this three times, we get: $$a_{n+3}=100(100(100a_{n} + 67)+67)+67=10^6a_n+676767$$ (Which really we could also have realized intuitively) The prime factors of $676767$ are $3, 7, 13, 37$ and $67$, and it so happens ...


3

We can write $$\tau_{r+1}(n)= \sum_{d\mid n} \tau_r(d),$$ which leads to your $$\sum_{n\leqslant x} \tau_{r+1}(n) = \sum_{d\leqslant x} \left\lfloor \frac{x}{d}\right\rfloor \tau_r(d),$$ but we can also write $$\tau_{r+1}(n) = \sum_{d\mid n} \tau_r\left(\frac{n}{d}\right),$$ and that gives us $$\begin{align} \sum_{n\leqslant x} \tau_{r+1}(n) &= ...


2

Note that $N=\frac{10^{2008}-1}{9}$. One can use the Taylor series of $\sqrt{x}$ at $a=10^{2008}$ to solve this problem. More precisely, let $f(a)=\sqrt a$, $$f(x)=f(a)+(x-a)f'(a)+\frac{f''(a)}2(x-a)^2+\cdots$$ Or $$\sqrt{9N}=10^{1004}-\frac{1}{2\cdot 10^{1004}}-\frac{3}{4\cdot 10^{3\cdot 1004}}+\cdots$$ That means $$3\sqrt{N}=10^{1004}-5\cdot ...


2

Similar to casting out nines to find the remainder when divided by 9, you can cast out 999's to find the remainder when divided by 999. The reason this works is the same: 1000 = 1 modulo 999, so xyz000...000 = xyz modulo 999. But instead of just adding up the digits, you have to add three-digit groups. Take for instance the number 12345678. Break this into ...


2

We have the formula $$ \phi(n)=n\prod_{\substack{p\mid n\\p\text{ prime}}}\left(1-\frac1p\right)\tag{1} $$ For a composite $n$, the smallest prime $p_0\mid n$ is at most $\sqrt{n}$, so $(1)$ implies $$ \begin{align} \phi(n) &\le n\left(1-\frac1{p_0}\right)\\ &\le n\left(1-\frac1{\sqrt{n}}\right)\\ &=n-\sqrt{n}\tag{2} \end{align} $$ Furthermore, ...


2

If $N$ is a power of $10$ (such as $100$ or $10^{18}$) you could count the complement: no prime digits in $N$ slots (leading $0$s allowed since they don't affect the answer). This could be done in $6^N$ ways. So your answer would be $10^N-6^N$. For non-powers of $10$ you could build the answer using the powers-of-ten idea. For instance let's say ...


2

Introducing the variables $a$ and $b$ leads to the condition $$4a^3+9a^2-4=(2b+3a)^2\ ,$$ which implies that $$4a^3+9a^2-4$$ has to be a perfect square. A quick computer search ($|a|\leq 10^6$) produced the solutions $$a=-2,\quad-1,\quad1,\quad 2,\quad 25\ .$$ This should be some material to work with. (The logic in the second half of your argument is ...



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