Yes. It is true that $ a_1 > 500$. Proof: Let $ X = \{a_i|0 < i\le 10000\}$ $ Y = \{x|10000\le x < 20000\text{ and }x\in \mathbb N\}$ Let For a number $ x < 20000$ there exists an unique non-negative integer $ t$ such that $ x\cdot 2^t\in Y$ Let $ f(x)$ denote the unique positive integer $ x\cdot 2^t$ For example $ f(12345) = 12345,f(4321) = 17284,f(700) = 11200,f(1) = 16384$ etc. $ f(x) = f(y)\implies \frac {x}{y} = 2^k$ for some integer $ k$ (not necessarily positive) which implies one of $ x,y$ divides the other. So $ f(x)\neq f(y)\ \forall\ x,y\in X\text{ and }x\neq y$ This is equivalent to saying that $ f$ is an injective map from $ X\to Y$ also $ |X| = |Y|\implies f$ is a bijective map from $ X\to Y$ (For a set $ S$, $ |S|$ denotes the number of elements of $ S$) For some odd $ x\in (0,20000)$ there exists unique $ t$ such that $ x\cdot 2^t\in Y$. Then There exists unique $ u$ such that $ x\cdot 2^u\in X$ For some odd $ x\in (0,20000)$ denote $ g(x)$ that unique $ u$ If $ h(x) = x\cdot 2^{g(x)}$ then similarly it can be showed that $ h$ is a bijective map between the set of odd integers $ < 20000$ and $ X$. We claim to show that $ x\cdot 2^{g(x)} > 500\ \forall \text{ odd }x\in (0,20000)$ For odd $ x,y\in (0,20000)$ such that $ y\neq 1$, if $ g(x)\le g(xy)$ then $ x\cdot 2^{g(x)}|xy\cdot 2^{g(xy)}$. But both $ x\cdot 2^{g(x)},xy\cdot 2^{g(xy)}\in X$ and are distinct. Contradiction. So we have that for all $ x,y\in (0,20000)$ such that $ y\neq 1$, $ g(x) > g(xy)$ or $ g(x)\ge g(xy) + 1$ $ g(1)\ge g(3) + 1\ge \cdots \ge g(3^2) + 2\ge \cdots g(3^9) + 9\ge 9$ (note that $ 3^9 < 20000$) $ 1\cdot 2^{g(1)}\ge 512 > 500$ $ g(3)\ge g(3^2) + 1\ge \cdots \ge g(3^9) + 8\ge 8$ $ 3\cdot 2^{g(3)}\ge 768 > 500$ $ g(5)\ge g(5\cdot 3) + 1\ge g(5\cdot 3^2) + 2\ge \cdots \ge g(5\cdot 3^7) + 7\ge 7$ (note that $ 5\cdot 3^7 < 20000$) $ 5\cdot 2^{g(5)}\ge 640 > 500$ $ g(7)\ge g(7\cdot 3) + 1\ge \cdots \ge g(7\cdot 3^7) + 7\ge 7$ (note that $ 7\cdot 3^7 < 20000$) $ 7\cdot 2^{g(7)}\ge 896 > 500$ Suppose $ t$ is an arbitrary odd integer $ \ge 9$ and $ < 20000$ We need to show that $ t\cdot 2^{g(t)} > 500$ Let $ m$ be the unique non-negative integer such that $ 3^{m + 2}\le t < 3^{m + 3}$ $ t\cdot 3^{6 - m} < 3^{m + 3}\cdot 3^{6 - m} = 3^9 < 20000$ $ g(t)\ge g(t\cdot 3) + 1\ge \cdots \ge g(t\cdot 3^{6 - m}) + 6 - m\ge 6 - m$ $ t\cdot 2^{g(t)}\ge 3^{m + 2}\cdot 2^{6 - m} = 9\cdot 3^m2^{6 - m}\ge 9\cdot 2^m2^{6 - m} = 9\cdot 2^6 = 576 > 500$ $ QED$