What is the sum of all positive integers that have twice as many digits when written in base 2 as they have when written in base 4? Express your answer in base 10.
+9519 Suppose the positive integer \(x\) satisfies the condition.
Then, for some positive integer \(n\), we have \(4^{n - 1} \leq x < 4^n\) and \(2^{2n - 1} \leq x < 2^{2n}\). i.e., \(\dfrac12 \cdot 4^n \leq x < 4^n\).
In fact, if there exists a positive integer \(n\) such that \(\dfrac12 \cdot 4^n \leq x < 4^n\), then the positive integer \(x\) satisfies the condition.
It turns out that there are infinitely many numbers satisfying the condition.
