New Year's Eve Pattern

$D(n)$ is equivalent to the number of non-negative integer $p$ such that $10^n>2^p\geq 10^{n-1}$ . $10^n>2^p\geq 10^{n-1} \\ n>p \times\log{2}\geq n-1 \\ \frac{n}{\log2}>p\geq \frac{n-1}{\log2} \\ 3.3219n>p\geq 3.3219(n-1)$ For $D(31)$ , it means the number of integer $p$ such that $102.98>p\geq 99.66$ . There exist only $\large 3$ integers satisfy it, which are $100,101,102$ , so $D(n)=3$ .