Difference between revisions of "2023 SSMO Team Round Problems/Problem 12"

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Latest revision as of 22:25, 15 December 2023

Problem

Let $T$ be the set of rationals of the form $\frac{a}{2^b}$ for nonnegative $a$ and $b$. Define the function $f \colon T \to \mathbb{Z}$ such that, for $t = \frac{a}{2^b}$ such $b$ is minimal, we have that \[f\left(\frac{a}{2^b}\right) = \begin{cases}     1 & b = 0 \\     f\left(\frac{a-1}{2^b}\right) + f\left(\frac{a+1}{2^b}\right) & b \ne 0 \\ \end{cases}\]

Suppose \[\sum_{i=0}^{2^{10} - 1} \frac{f\left(\frac{i+1}{2^{10}}\right)}{f\left(\frac{i}{2^{10}}\right)}\] equals $\frac{m}{n}$. Find $m+n$.

Solution