Hot answers tagged

6

Let $a=z$, $b=y-z$, $c=x-y$. We are looking for the number of solutions of $c+2b+3a=n$ with $a,b,c\geq 0$, so the answer is given by: $$[x^n](1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^3+x^6+x^9+\ldots)$$ i.e. by: $$ [x^n]\frac{1}{(1-x)(1-x^2)(1-x^3)} \tag{1}$$ that can be recovered through partial fraction decomposition. The meromorphic function ...


4

Let the 21 numbers be $x_1<x_2<\ldots <x_{21}$. It is not possible to have $x_{k+2}\ge 2x_k+1$ (or equivalently $x_{k+2}+1\ge 2(x_k+1)$) for all $k$ as that would lead to $2047 \ge x_{21}+1\ge 2^{10}(x_1+1)\ge 2^{11}=2048$. Thus we find $k$ such that $x_{k+2}\le 2x_k$. With $b:=x_k$, $a:=x_{k+1}$, $c:=x_{k+2}\le 2b$, we have $$ bc\le ...


3

Use induction. Check the base cases. Let's look at the partitions of $2n+1$. Every partition of $2n+1$ includes odd number of $1$s. If a partition contains $2k+1$ $1$s, all the other terms in it are even. This kind of partitions can be identified with partitions of $n-k$ by dividing the even numbers by $2$. So, $q(2n+1)=q(n)+q(n-1)+\ldots+q(2)+q(1)+1$, ...


2

Your computations are correct. Since $-17\equiv3\bmod{4}$ our ring of integers is $\mathbb{Z}[\sqrt{-17}]$, so we may factor the ideal $(3)$ in $\mathbb{Z}[\sqrt{-17}]$ by factoring $$x^2 + 17 \equiv x^2 - 1 \equiv (x+1)(x+2) \bmod{3}.$$ This yields the ideal $(3,1+\sqrt{-17})$, and since 3 splits the norm of this ideal is 3. To see this more ...


1

The triples that satisfy your condition are of two types. Type 1 are the triples with smallest element $1$ and Type 2 are the ones with smallest element $\gt 1$. There are just as many triples of Type 1 with sum $n$ as there are pairs $(y,x)$ with $0\le y\le x$ such that $y+x=n-1$. There are just as many triples of Type 2 with sum $n$ as there are triples ...


1

Note that for any three positive integers $a,b,c$, if $2^k\leqslant b<a<c<2^{k+1}$ for some $k\in\mathbb{Z}_{>0}$, then $$bc<ac<a\cdot 2^{k+1}=2a\cdot 2^k\leqslant 2a\cdot a=2a^2,$$ and $$bc>ba\geqslant 2^ka=\frac{1}{2}\cdot 2^{k+1}a>\frac{1}{2}\cdot a\cdot a=\frac{1}{2}a^2.$$ Thus, $2^k\leqslant b<a<c<2^{k+1}$ for some ...


1

Let set $S$ be the set of integers from $1$ to $2046$, we split $S$ to $S_i$s like this: $$S_1=\{1,2,3\}\\S_i=\{s\in\mathbb{Z}\,|\,2^{i}\le s\lt2^{i+1}\}\quad (2\le i\lt10)\\ S_{10}=\{1024,1025,\dots,2046\}$$ So we have $10$ sets in total, by Pigeonhole principle there's a set which has at least $\lceil\frac{21}{10}\rceil=3$ elements from those $21$ chosen ...


1

You can deliberately define an irrational number to have this non-repeating quality, if required. For example, for an irrational number that avoids 3-blocks of repeating digits, take the binary definition of $\pi$ and generate a new number such that each digit $x$ in $\pi$ is replaced by $11x00$. (This also avoids 4-blocks). To avoid any repetitions of ...


1

This is hopeless. For example, we can build a set $E$ of density $1/2$, such that $\pi E$ will not have density $1/2$, as follows: I'm going to include one half of the numbers in $N^2\le n<(N+1)^2$ in $E$, for each $N\ge 1$. Then $E$ will have density $1/2$, and for this it does not matter which numbers exactly I choose. Summary: The exact procedure is a ...


1

Considering the case when $n = 6$ only shows that $2^n - 1$ is not prime for all composite $n$ (i.e. at least one of $2^n - 1$ is composite). However it does not show that $2^n - 1$ is not prime for every composite $n$ (i.e. all of $2^n - 1$ are composite), which is what is required in the contrapositive form of the question. By invoking the stated ...



Only top voted, non community-wiki answers of a minimum length are eligible