666 猜想
相信大家都知道欧拉函数（$\def \phi{\varphi}\phi(n)$）吧，这是一个非常有趣的函数。
欧拉函数
在数论中，对正整数$n$，欧拉函数是小于$n$的正整数中与$n$互质的数的个数。
观察下列等式。
$$
\begin{aligned}
\phi(6) &= 2 = 2^{1}\\
\phi(\phi(66)) &= 8 = 2^{3}\\
\phi(\phi(\phi(666))) &= 24\\
\phi(\phi(\phi(\phi(6666)))) &= 128 = 2^{7}\\
\phi(\phi(\phi(\phi(\phi(66666))))) &= 256 = 2^{8}\\
\phi(\phi(\phi(\phi(\phi(\phi(666666)))))) &= 512 = 2^{9}\\
\phi(\phi(\phi(\phi(\phi(\phi(\phi(6666666))))))) &= 8192 = 2^{13}\\
\phi(\phi(\phi(\phi(\phi(\phi(\phi(\phi(66666666)))))))) &= 32768 = 2^{15}\\
\phi(\phi(\phi(\phi(\phi(\phi(\phi(\phi(\phi(666666666))))))))) &= 24576\\
\phi(\phi(\phi(\phi(\phi(\phi(\phi(\phi(\phi(\phi(6666666666)))))))))) &= 131072 = 2^{18}\\
\cdots
\end{aligned}
$$
猜想：等于$2$的正整数幂的等式不会穷尽。
为了方便陈述，现在这里定义函数$\def \Liu{\rm{Liu}}\Liu(x)=\frac{2}{3}(10^x-1)$ 。（也就是说， $\Liu(x)=\overset{x个}{\overbrace{66\cdots6}}$ ）
形式化地说，猜想：集合$\{(x,k)|\phi^x(\Liu(x))=2^k,x,k\in\mathbb{Z}^+\}$是无限集。
换句话说：满足$\phi^x(\Liu(x))=2^k$的正整数对$(x,k)$有无穷多个。
二幂有奇猜想
观察下列式子。
$$
\begin{aligned}
2^0&=1\\
2^1&=2\\
2^2&=4\\
2^3&=8\\
2^4&=16\\
2^5&=32\\
2^6&=64\\
2^7&=128\\
2^8&=256\\
2^9&=512\\
2^{10}&=1024\\
2^{11}&=2048\\
2^{12}&=4096\\
\cdots
\end{aligned}
$$
我们定义：$2$的自然数幂为二幂数。形式化的说：$\forall k \in \mathbb{N}$，$2^k$被称为二幂数；
猜想：$\forall k > 11 \in \mathbb{N}$，$2^k$在十进制表示下至少有一位是奇数。