4 for fort

Let $A$ be a set of numbers which can be expressed as $2^m$ where $m$ is a non-negative integer and $B$ be a set of numbers which can be expressed as $4^n$ where $n$ is a non-negative integer. Then, $A=\{2^0,2^1,2^2,2^3,\ldots\}$ $B=\{4^0,4^1,4^2,4^3,\ldots\}$ We can establish a correspondence relationship between the two sets: $2^0 \leftrightarrow 4^0$ $2^1 \leftrightarrow 4^1$ $2^2 \leftrightarrow 4^2$ $\ldots$ So, $a_1=4^0 \leftrightarrow 2^0=1$ $a_2=4^1 \leftrightarrow 2^1=2$ $a_3=4^0+4^1 \leftrightarrow 2^0+2^1=3$ $a_4=4^2 \leftrightarrow 2^2=4$ $a_5=4^0+4^2 \leftrightarrow 2^0+2^2=5$ $\ldots$ Notice that the corresponding values are always the same as the term number, hence $64=2^6 \leftrightarrow 4^6=4096=a_{64}$ So, the value of $a_{64}$ is 4096.

Note that $a_i$ expressed in base-4 is the same as $i$ expressed in binary (base-2). Problem solved.

We notice that $64 = 2^6 \implies a_{64}= 4^6 = 4096$

Let $a_n = b_0 \dot{}4^0 + b_1 \dot{}4^1+ b_2 \dot{}4^2 +...+ b_p \dot{}4^p$ , where $b_i = 0$ or $1$ are digits of $n$ in base $2$ ((n) {(10)} = \overline {b p ...b 2 b 1 b_0 _ {(2)}}.

For example:

$n = 1_{(10)} = 1_{(2)} \Rightarrow a_1 = b_0 \dot {} 4^0 = 1 \dot {} 1 = 1$

$n = 2_{(10)} = 10_{(2)} \Rightarrow a_2 = b_0 \dot {} 4^0 + b_1 \dot {} 4^1 = 0 \dot {} 1 + 1 \dot {} 4 = 4$

$n = 3_{(10)} = 11_{(2)} \Rightarrow a_3 = 1 \dot {} 4^0 + 1 \dot {} 4^1 = 5$

$n = 4_{(10)} = 100_{(2)} \Rightarrow a_4 = 0 + 0 + 1 \dot {} 4^2 = 16$

$n = 5_{(10)} = 101_{(2)} \Rightarrow a_5 = 1 + 0 + 16 = 17$

Therefore, $n = 64_{(10)} = 1000000_{(2)} \Rightarrow a_{64} = 4^6 = \boxed{4096}$